17 worked examples30 exercises6 figuresbölüm 9 / 14
09Fundamental Theorem of Calculus and indefinite integrals
Somebody hands you a machine. Feed it a number $x$ and it reports the exact area trapped under the curve $y=\sin(t^{2})$ between $t=0$ and $t=x$. Nobody, including the person who built it, can write down a formula for what the machine outputs. Question: how fast is its reading rising when you turn the dial past $x=2$?
By the end of this section you can answer that in one line without ever writing down the machine's formula, and you can turn any whose integrand you can antidifferentiate into two evaluations and a subtraction.
In 60 seconds
Accumulating and differentiating undo each other: differentiate an area that grows from a fixed left end and you get the height of the curve back, and running that backwards turns every definite integral into $F(b)-F(a)$.
Fundamental Theorem, Part 1
$\frac{d}{dx}\int_{a}^{x} f(t)\,dt = f(x)$
anything of the form: differentiate a function defined by an integral
a region is described by two curves and you know where they cross
Three most common mistakes
Differentiating the integrand instead of evaluating it. $\frac{d}{dx}\int_{0}^{x}\sin(t^{2})\,dt$ is $\sin(x^{2})$, not $2x\cos(x^{2})$: Part 1 hands you the height, not the slope.
Changing the variable but not the limits. If $u=x^{2}+1$ and $x$ runs from $0$ to $3$, then $u$ runs from $1$ to $10$, and an integral sign carrying $0$ and $3$ next to a $du$ is a mixture of two different problems.
Applying Part 2 across a point where the integrand is not defined. $\int_{-1}^{1} x^{-2}\,dx$ is not $-2$; the integrand is positive everywhere it exists, and no positive integrand has a negative integral.
Midterm 1, Midterm 2 and the Final are worth 28 percent each, quizzes 10 and homework 6. Almost everything on this page is a line or two of work, so the marks here are lost to slips rather than to not knowing: write the substitution and its new limits on their own line before you touch the arithmetic.
How much time do you have?
10 minutes
You leave able to differentiate an and to evaluate a definite integral by , which together cover the two shortest question types in this material.
In 60 seconds card, Differentiating an area nobody can write down, From the Riemann limit to a subtraction, Formula card
45 minutes
Add moving limits, substitution and the area between two curves. That is the full spread of the standard long question, and you can now check your own answers.
everything in the 10 minute path, When both ends of the integral move, Substitution: the chain rule read backwards, The area trapped between two curves, Scaffolding comes off, Full exam-style question, Practice C
full read
, the counterexamples and the interleaved set, where the questions no longer say which tool they want and the mixed set reaches across all of accumulation, substitution, area and separable equations. Deciding which one a fresh question needs is the actual exam skill.
Prerequisites and pretest, all six concept blocks, Separating the variables, Practice A, Practice B, Practice C, Practice D, Mistake ledger, Self audit
By the end of this section
Differentiate a function defined as an integral with a fixed lower limit, and state which hypothesis on the integrand licenses it.
Apply the moving-limit rule when one or both limits of the integral are functions of $x$, keeping the chain factor and the sign.
Evaluate a definite integral as the change in an antiderivative across the two limits, and say when that is not allowed.
Compute a definite integral by substitution, converting the limits with the variable and leaving no $x$ behind.
Set up and evaluate the area between two curves, splitting the interval wherever the curves swap places.
Solve a separable first-order equation with an , and report the interval on which the solution actually lives.
Syllabus coverage
4.4
The Fundamental Theorem of Calculus and
Both halves of the theorem plus the moving-limit extension occupy the first three blocks; the indefinite integral is introduced there as the bookkeeping that makes the second half usable.
covered
4.5
The
Both playbooks are given, the one that converts the limits and the one that back-substitutes, together with the even and odd shortcut that substitution produces.
covered
9.3
Separable differential equations
Treated as the first application of antidifferentiation to an equation rather than to a function, with the initial condition and the .
covered
5.1
Areas between curves
, the crossing case that forces a split, and the horizontal-strip alternative.
covered
improper integrals
Integrals with an unbounded integrand or an infinite limit
Named here only because the second half of the theorem is often misapplied to one. The definition and the convergence tests come later in the term, and no question in this section needs them.
deferred
Recall first
The definite integral
$\int_{a}^{b} f(x)\,dx$ is the number the Riemann sums $\sum f(x_{i}^{*})\Delta x$ approach as the rectangles get thin. It is a signed total: pieces of the graph below the axis count negative.
Everything on this page is a shortcut for computing that number. If you forget what the number is, the shortcuts become spells.
Antiderivative and the constant
$F$ is an antiderivative of $f$ on an interval when $F'=f$ there. Two antiderivatives of the same $f$ on the same interval differ by a constant, so the whole family is $F(x)+C$.
The second half of the theorem says any antiderivative works. That is only safe because they all differ by a constant, and the constant cancels in a subtraction.
Zero derivative on an interval
If $G'(x)=0$ for every $x$ in an interval, then $G$ is constant on that interval. This is a consequence of the Mean Value Theorem, not of common sense: it is false on a domain made of two separated pieces.
It is the single step that turns the first half of the theorem into the second.
Splitting and bounding an integral
$\int_{a}^{c} f + \int_{c}^{b} f = \int_{a}^{b} f$, and if $m \le f \le M$ on $[a,b]$ then $m(b-a) \le \int_{a}^{b} f \le M(b-a)$.
The splitting rule isolates the thin sliver in the proof of Part 1; the bounds are what squeeze it.
Half of this section is the chain rule seen from one side or the other: forwards it gives the moving-limit rule, backwards it gives substitution.
Try it yourself first (3 questions)
1§09.0 — signed total against geometric area●●○○○
Before the new material, one check on what the previous section's number actually measures. Getting this wrong is not a problem; it tells you which line of the conventions box to read twice.
Given
$f(x)=x-1$ on the interval $[0,2]$
Find
True or false: $\int_{0}^{2}(x-1)\,dx$ equals the area of the region trapped between the graph of $f$ and the $x$-axis over $[0,2]$.
Hint 1/4
Sketch the line. Notice that part of it sits below the axis and ask what the Riemann sum does with a negative height.
Hint 2/4
A Riemann sum adds $f(x_{i}^{*})\Delta x$, and $\Delta x>0$, so a negative height contributes a negative term.
Hint 3/4
With $f(x)=x-1$ on $[0,2]$: the graph is below the axis on $[0,1]$ and above it on $[1,2]$, and the two triangles have the same size, $\tfrac12$ each.
Hint 4/4
The two contributions cancel to $0$, while the geometric area is $1$.
the sign of the integrand changes at $x=1$, and that is the only place it can
$=-\tfrac12+\tfrac12=0$
each piece is a triangle with legs $1$ and $1$; the left one lies below the axis
Geometric area ignores the sign
$A=\tfrac12+\tfrac12=1$
area counts both triangles as positive, which is what $\int_{0}^{2}\lvert x-1\rvert\,dx$ computes
Answer $$\int_{0}^{2}(x-1)\,dx=0,\qquad A=1$$
Check
Independent check: $f$ is odd about the point $(1,0)$, so the interval $[0,2]$ is symmetric about that point and the integral has to vanish without any computation.
Whenever a question says area, ask first whether the integrand keeps one sign.
2§09.0 — the antiderivative family●○○○○
A one-line warm-up on the object the second half of the theorem consumes. You need to be fluent here before the rest of the page is worth reading.
Given
$f(x)=6x^{2}-4x+5$
Find
(a) Write the general antiderivative of $f$.
(b) Pick the one whose value at $x=0$ is $7$.
Hint 1/4
You are looking for a function whose derivative is the given polynomial, so run the power rule in reverse, term by term.
Hint 2/4
Reverse power rule: an antiderivative of $x^{n}$ is $\dfrac{x^{n+1}}{n+1}$ for $n\neq-1$, and antiderivatives add term by term.
Hint 3/4
With $f(x)=6x^{2}-4x+5$: the terms give $6\cdot\frac{x^{3}}{3}$, $-4\cdot\frac{x^{2}}{2}$ and $5x$.
Hint 4/4
The family is $2x^{3}-2x^{2}+5x+C$, and $F(0)=C$, so $C=7$.
the outer function is evaluated at the inner one and left alone otherwise
$=3x^{2}\cos(x^{3})$
the inner derivative is the factor that gets forgotten
Answer $$3x^{2}\cos(x^{3})$$
Check
Independent check at $x=0$: $\sin(x^{3})$ is flat there because $x^{3}$ is, and the formula gives $3\cdot 0\cdot\cos 0=0$.
Write the inner derivative on its own line before multiplying. That factor is the one that vanishes when the chain rule is done in your head, and Part 1 will hand you the same pattern back the moment a limit stops being a bare $x$.
Notation
symbol
reads as
means
watch out
$\int_{a}^{x} f(t)\,dt$
the integral of f of t, dt, from a to x
the signed area accumulated from the fixed left end $a$ up to the movable right end $x$
A function of $x$, not of $t$. Feeding it a number means putting that number in the upper limit.
$\int f(x)\,dx$
the indefinite integral of f of x, dx
the family of all antiderivatives of $f$ on the interval in question
Not a number and not one function. It is a family, and it always carries $+C$.
$\bigl[F(x)\bigr]_{a}^{b}$
F of x, evaluated from a to b
the number $F(b)-F(a)$
The order is fixed: top minus bottom. Reversing it flips the sign of your whole answer.
$u=g(x),\quad du=g'(x)\,dx$
u equals g of x, du equals g prime of x, dx
the renaming used by substitution, together with the record of how lengths are stretched
The second equation is not optional decoration; it is what lets you replace $g'(x)\,dx$ by $du$.
$\int_{g(a)}^{g(b)} f(u)\,du$
the integral of f of u, du, from g of a to g of b
the same number as the original integral, now written in the new variable
The limits are $u$-values. Seeing an old $x$-limit next to a $du$ means one half of the substitution was skipped.
$\dfrac{dy}{dx}=g(x)h(y)$
dy by dx equals g of x times h of y
a first-order equation whose right-hand side factors into a part in $x$ alone and a part in $y$ alone
$x+y$ does not factor this way. Separability is a property of the formula, not a wish.
$y(x_{0})=y_{0}$
y of x nought equals y nought
the initial condition: one point the solution curve is required to pass through
It is applied after integrating, never before, and it fixes $C$ rather than removing it.
$\lvert f-g\rvert$
the absolute value of f minus g
the vertical distance between two curves, whichever one is on top
You cannot integrate this symbol directly. You split at the crossings and drop the bars piece by piece.
Conventions used here
Undefined is not the same as discontinuous
Continuity is only ever claimed or denied at numbers that are in the domain. For $f(x)=x^{-2}$ the correct sentence about $x=0$ is "$f$ is not defined there", and the useful consequence is that $f$ is not continuous on $[-1,1]$, so a theorem that asks for continuity on $[-1,1]$ does not apply.
The one hypothesis in this section that gets skipped is the one about continuity.
A definite integral is a number, an indefinite integral is a family
$\int_{a}^{b} f(x)\,dx$ is a number and carries no $+C$. $\int f(x)\,dx$ is the whole family of antiderivatives and always carries one. The same symbol with and without limits means two different kinds of object.
A stray $+C$ in a definite answer, or a missing one in an indefinite answer, is a marked error.
Radians throughout
Every trigonometric function on this page takes radians.
The antiderivatives of $\sin$ and $\cos$ are false in degrees, and this section uses them constantly.
Signed area and geometric area
$\int_{a}^{b} f$ is signed. Geometric area is $\int_{a}^{b}\lvert f\rvert$, which you compute by splitting at the zeros of $f$ and flipping the sign on the pieces where $f<0$. We say "the integral" for the first and "the area" for the second.
Most wrong area answers in this material are right integrals of the wrong integrand.
The
In $\int_{a}^{x} f(t)\,dt$ the letter $t$ lives and dies inside the integral, while $x$ survives as the variable of the answer. Writing $\int_{a}^{x} f(x)\,dx$ uses one letter for two jobs and is not accepted.
It is the single most common notation error at this point in the course.
Only one constant survives
When both sides of an equation are antidifferentiated, one constant is written, on the right. Two constants would immediately collapse into their difference anyway.
Carrying two constants makes the initial condition look underdetermined when it is not.
Maximal interval of existence
The solution of an initial value problem is reported on the largest open interval that contains the initial point and on which the formula is defined and satisfies the equation. A formula that is defined further away but on the other side of a blow-up is a different solution, not the same one.
A formula written without its interval is not yet an answer to an initial value problem.
Differentiating an area nobody can write down
The previous section defined the number $\int_{a}^{b} f$ and computed it in a few cases. Now we let the right-hand end move.
Solvable with what we have
$\int_{0}^{3} 5\,dt = 15$, the area of a rectangle.
$\int_{0}^{4} t\,dt = 8$, the area of a triangle.
$\int_{-2}^{2}\sqrt{4-t^{2}}\,dt = 2\pi$, half a disc of radius $2$.
$\int_{0}^{2} t^{2}\,dt = \tfrac{8}{3}$, from the Riemann limit and $\sum_{k=1}^{n}k^{2}=\tfrac{n(n+1)(2n+1)}{6}$.
Not solvable yet
$\int_{0}^{\pi/2}\sin t\,dt$: the Riemann limit needs a trigonometric identity nobody memorises.
$\int_{0}^{1}\sqrt{1+t^{3}}\,dt$: no elementary formula for this area exists.
$F(2)$ for $F(x)=\int_{0}^{x}\sin(t^{2})\,dt$: the same problem, with a movable end.
The natural move is to find a formula for the area and then differentiate it. For $F(x)=\int_{0}^{x}t^{2}\,dt$ that works: the Riemann limit gives $F(x)=\tfrac{x^{3}}{3}$, so $F'(x)=x^{2}$. For $F(x)=\int_{0}^{x}\sin(t^{2})\,dt$ it dies at the first step, and $F'$ looks out of reach with it.
Why it fails
That conclusion is wrong: you do not need a formula for a quantity in order to know its rate of change. Look at what the successful case produced — $F(x)=\tfrac{x^{3}}{3}$ and $F'(x)=x^{2}$, the integrand back again. That is no coincidence about $t^{2}$.
TheoremThe
Conditions
$f$ is continuous on the closed interval $[a,b]$
$F$ is defined by $F(x)=\int_{a}^{x} f(t)\,dt$ for $x$ in $[a,b]$
$$\boxed{\,F(x)=\int_{a}^{x} f(t)\,dt \,\Longrightarrow\, F'(x)=f(x) \quad\text{for } a
Accumulate area from a fixed left end. The rate at which the total is growing, when the right end sits at $x$, is exactly the height of the graph at $x$.
Why it is true, in four lines
Splitting at $x$ gives $F(x+h)-F(x)=\int_{x}^{x+h} f(t)\,dt$: the difference is the sliver alone.
For $h>0$ the continuous $f$ has a least value $m_{h}$ and a greatest value $M_{h}$ on $[x,x+h]$, so $m_{h}h \le \int_{x}^{x+h} f \le M_{h}h$.
Divide by $h$: $m_{h} \le \dfrac{F(x+h)-F(x)}{h} \le M_{h}$.
Continuity drives $m_{h}$ and $M_{h}$ to $f(x)$ as $h\to 0^{+}$, and the squeeze theorem does the rest. For $h<0$ the same argument runs on $[x+h,x]$.
Moving the right-hand end from $x$ to $x+h$ leaves the orange area untouched and only adds the purple sliver, of area about $f(x)\cdot h$. Divide by $h$ and shrink it.
Looks like this, but is not
Take $f(t)=0$ for $t<1$ and $f(t)=1$ for $t\ge 1$, and set $F(x)=\int_{0}^{x} f(t)\,dt$, so that $F(x)=0$ on $[0,1]$ and $F(x)=x-1$ afterwards. This $F$ is continuous everywhere, and $F'=f$ at every $x$ but one.
At $x=1$ the left-hand derivative of $F$ is $0$ and the right-hand one is $1$, so $F'(1)$ does not exist and there is nothing for $f(1)$ to equal. What failed is continuity of $f$: integration smooths a jump into a corner, and a corner has no derivative.
The machine from the opening, differentiated
$F(x)=\int_{0}^{x}\sin(t^{2})\,dt$ has no elementary formula. Find $F'(x)$, then $F'(2)$ and $F'(\sqrt{\pi})$.
Given
$F(x)=\int_{0}^{x}\sin(t^{2})\,dt$
the lower limit $0$ is fixed
the upper limit is $x$ itself
Find
$F'(x)$, and its values at $x=2$ and $x=\sqrt{\pi}$
SolutionCheck the hypothesis before using the theorem
$t\mapsto\sin(t^{2})\ \text{is continuous on }\mathbb{R}$
a composition of continuous functions; this is the only condition Part 1 asks for, so it is worth ten seconds
Independent check on the last value: near $x=\sqrt{\pi}$ the integrand $\sin(t^{2})$ is crossing zero, so the accumulated area momentarily stops changing. A graph of $F$ has a horizontal tangent there, which is what $F'(\sqrt{\pi})=0$ says.
One application of Part 1 and no algebra at all — that is the whole point of the theorem.
The formula for $F$ was never needed and never existed. Read the opening question again: the answer is $\sin 4$, and it took one line.
Where the accumulation of t² − 4 rises and falls
Let $F(x)=\int_{0}^{x}(t^{2}-4)\,dt$. Find the intervals on which $F$ increases and decreases, and locate its local minimum.
Given
$F(x)=\int_{0}^{x}(t^{2}-4)\,dt$
$F$ is defined for every real $x$
Find
the intervals of increase and decrease and the local minimum of $F$
SolutionGet F′ without computing F
$F'(x)=x^{2}-4=(x-2)(x+2)$
Part 1 again; the point is that a sign line for $F'$ is available even though $F$ itself has not been written down
Sign line
$F'(-3)=5>0,\quad F'(0)=-4<0,\quad F'(3)=5>0$
one test value strictly inside each piece, which is enough because $F'$ is continuous and has no other zero
positive derivative means rising; the accumulated area falls on $(-2,2)$ because the integrand is negative there
$x=2:\ -\to+\ \Rightarrow\ \text{local minimum}$
the First Derivative Test, applied to $F$ rather than to a formula
Answer $$F\ \text{increases on }(-\infty,-2)\text{ and }(2,\infty),\ \text{decreases on }(-2,2);\ \text{local minimum at }x=2$$
Check
Independent check: here $F$ can be written down, as $F(x)=\tfrac{x^{3}}{3}-4x$, and differentiating that gives $x^{2}-4$ — the same $F'$, reached without the theorem.
Every graph-shape tool from earlier in the term applies to a function defined by an integral, because Part 1 hands you its derivative for free.
Checkpoint
§09.1 — reading Part 1 off a fixed lower limit●●○○○
Thirty seconds. The lower limit is a number and the upper limit is $x$, which is the exact shape Part 1 covers.
Given
$F(x)=\int_{3}^{x}\cos t\,dt$
Find
What is $F'(x)$?
Hint 1/4
Ask what Part 1 returns: the integrand, or an antiderivative of it.
Hint 2/4
Part 1: $\dfrac{d}{dx}\int_{a}^{x} f(t)\,dt=f(x)$, with $a$ any fixed number.
Hint 3/4
Here $f(t)=\cos t$ and $a=3$, so $f(x)=\cos x$.
Hint 4/4
The lower limit contributes nothing at all: $F'(x)=\cos x$.
Show solutionApply Part 1
$F'(x)=\cos x$
the derivative is the integrand at the moving end; the fixed end never appears
splitting shows the lower limit only subtracts a constant, and constants have zero derivative
Answer $$F'(x)=\cos x$$
Check
Independent check: $\sin x-\sin 3$ is a formula for this $F$, and its derivative is $\cos x$.
A fixed lower limit only shifts the accumulation up or down by a constant, and shifting a graph never changes its slope. Look at what moves before you compute anything.
⚠ Differentiating the integrand instead of evaluating it
The word derivative is in the question, the integrand is the only visible function, and the reflex fires before the theorem is read.
Put the top limit into the integrand and multiply by the speed of the top limit; do the same with the bottom limit and subtract. The top end pays area in, the bottom end takes area out, and nothing else about the interval matters.
Why it is true, in two lines
Fix a convenient number $c$ in the interval and let $\Phi(u)=\int_{c}^{u} f(t)\,dt$. Part 1 says $\Phi'=f$.
Splitting at $c$ gives $\int_{a(x)}^{b(x)} f=\Phi(b(x))-\Phi(a(x))$, and one chain rule on each term produces $f(b(x))b'(x)-f(a(x))a'(x)$. So this rule is not a new theorem; it is Part 1 with the chain rule bolted on.
The shaded number changes only at its two ends. The orange sliver on the right is being added at rate $\textcolor{#d1690a}{f(b(x))b'(x)}$ and the purple sliver on the left is being given up at rate $\textcolor{#6f42c1}{f(a(x))a'(x)}$; the difference of the two is the whole derivative.
Looks like this, but is not
$\dfrac{d}{dx}\displaystyle\int_{1}^{x^{2}} x\sin t\,dt$ looks like a job for the boxed rule, and the rule would give $x\sin(x^{2})\cdot 2x$.
The rule assumes the integrand contains no $x$ — only the limits may. Here $x$ is inside as well, and it is a constant as far as the $t$-integration is concerned, so pull it out first: $x\int_{1}^{x^{2}}\sin t\,dt$. Now the product rule applies, and the answer carries an extra term, $\int_{1}^{x^{2}}\sin t\,dt + 2x^{2}\sin(x^{2})$, which the boxed rule alone would have lost.
the bottom term is absent because $a(x)=1$ has $a'(x)=0$
$f(u(x))=\sin\bigl((x^{2})^{2}\bigr)=\sin(x^{4})$
substitute the whole moving end into the integrand, which is where the fourth power comes from
$G'(x)=2x\sin(x^{4})$
multiply by the speed of the top end
Answer $$G'(x)=2x\sin(x^{4})$$
Check
Independent structural check: replace $\sin(t^{2})$ by the constant $1$. Then $G(x)=x^{2}-1$ exactly, so $G'(x)=2x$ — and the formula gives $1\cdot 2x=2x$ as well. The chain factor survives a case where the answer is known outright.
Two things to keep: the substitution into the integrand and the factor $2x$. Most lost marks here are the second one.
If the upper limit is not a bare $x$, its derivative is a factor in your answer. No exceptions.
Both limits moving, checked against a closed form
$J(x)=\displaystyle\int_{\sin x}^{x^{2}} t\,dt$. Find $J'(x)$ using the boxed rule, then confirm it by computing $J$ outright.
The two routes are genuinely independent — one never writes $J$ down, the other never uses the boxed rule — and they agree. As a second check, at $x=0$ both limits collapse to $0$ and the formula gives $0-\sin 0\cos 0=0$: the interval has zero length there, so nothing is being accumulated and the rate has to be zero. Just to the right of $0$ the rule gives a negative number, and that is right too, because $x^{2}<\sin x$ for small $x>0$, so $J(x)=\int_{\sin x}^{x^{2}}t\,dt<0$ while $J(0)=0$.
When an integrand is easy enough to antidifferentiate, computing $J$ and differentiating is a free check on the moving-limit rule. Use it while the rule is still new.
Checkpoint
§09.2 — the moving limit is the bottom one●●○○○
Thirty seconds. The upper limit is the fixed number this time, which flips which term survives and which sign it carries.
Given
$K(x)=\int_{x}^{5}\cos t\,dt$
Find
What is $K'(x)$?
Hint 1/4
Decide first which of the two ends is moving, and remember which of the two terms in the rule belongs to that end.
Hint 2/4
The rule is $f(b(x))b'(x)-f(a(x))a'(x)$; here $b$ is constant and $a(x)=x$.
Hint 3/4
So the first term is $0$ and the second is $\cos x\cdot 1$, entering with a minus sign.
Hint 4/4
Therefore $K'(x)=-\cos x$.
Show solutionIdentify the two ends
$b(x)=5,\ b'(x)=0;\qquad a(x)=x,\ a'(x)=1$
writing both ends out is what stops the minus sign from being dropped
Assemble
$K'(x)=\cos 5\cdot 0-\cos x\cdot 1=-\cos x$
the fixed end contributes nothing because it is not moving
Answer $$K'(x)=-\cos x$$
Check
Independent check: $K(x)=\sin 5-\sin x$ outright, and differentiating that gives $-\cos x$.
Before assembling anything, write down which end moves and how fast. Then the sign is produced by the rule instead of being recalled under pressure.
⚠ Dropping the derivative of the moving limit
Part 1 is remembered as "the answer is the integrand", and with a bare $x$ on top that is true. The extra factor only appears when the top is something else, so it is exactly the case that is never rehearsed.
Part 1 turned an integral into a derivative. Run that sentence backwards and it turns a derivative into an integral, which is what actually gets numbers out.
TheoremThe
Conditions
$f$ is continuous on the closed interval $[a,b]$
$F$ is any antiderivative of $f$ on $[a,b]$, that is $F'=f$ there
The total accumulated between the two ends equals the of any antiderivative across those ends. The infinitely many rectangles collapse into two evaluations and one subtraction.
Why any antiderivative is allowed
Let $\Phi(x)=\int_{a}^{x} f(t)\,dt$. Part 1 gives $\Phi'=f$ on $(a,b)$.
Let $F$ be any other antiderivative, so $F'=f$ too. Then $(\Phi-F)'=f-f=0$ on the interval.
A function with zero derivative on an interval is constant, so $\Phi(x)=F(x)+C$ for one fixed $C$.
At $x=a$ the accumulated area is $0$, so $0=F(a)+C$ and $C=-F(a)$. Putting $x=b$ then gives $\int_{a}^{b} f=\Phi(b)=F(b)-F(a)$. Nothing in the argument cared which antiderivative $F$ was, which is why the $+C$ never has to be written in a definite integral.
The two halves of the theorem run in opposite directions around the same square. Part 1 goes from the accumulation function back to $f$; Part 2 uses that to replace the Riemann limit by $\textcolor{#d1690a}{F(b)-F(a)}$.
Looks like this, but is not
$\displaystyle\int_{-1}^{1}\frac{1}{x^{2}}\,dx$: the integrand has the antiderivative $-\dfrac{1}{x}$, so Part 2 seems to give $\Bigl[-\tfrac{1}{x}\Bigr]_{-1}^{1}=-1-1=-2$.
The answer is absurd before it is wrong: $x^{-2}>0$ wherever it exists, and a positive integrand cannot produce a negative integral. What failed is the first condition. The integrand is not defined at $x=0$, so it is not continuous on $[-1,1]$, and $-1/x$ is an antiderivative on $(0,1]$ and on $[-1,0)$ separately but on no interval containing both. This is the kind of integral that needs its own definition and its own limit, which this course sets up later.
$f(x)$
$\int f(x)\,dx$
Condition
$x^{n}$
$\dfrac{x^{n+1}}{n+1}+C$
$n\neq-1$
$k$ (a constant)
$kx+C$
always
$\sin x$
$-\cos x+C$
always
$\cos x$
$\sin x+C$
always
$\sec^{2}x$
$\tan x+C$
$\cos x\neq 0$
$\sec x\tan x$
$\sec x+C$
$\cos x\neq 0$
$\csc^{2}x$
$-\cot x+C$
$\sin x\neq 0$
Every row is a derivative fact from earlier in the term, read right to left; there is nothing new to memorise. Two things are worth noticing. The first row is silent at $n=-1$, and that gap is filled by a function this course meets later, so no question in this section needs it. And the conditions column is not decoration: an antiderivative of $\sec^{2}x$ is only valid on an interval where $\tan x$ exists, so it cannot be used across $x=\pi/2$.
The sine integral that the Riemann limit could not reach
Evaluate $\displaystyle\int_{0}^{\pi/2}\sin x\,dx$ — the integral that was on the cannot-do list at the top of this section.
Given
$f(x)=\sin x$
the interval $[0,\pi/2]$
Find
the exact value of the definite integral
SolutionCheck the hypothesis, find one antiderivative
$\sin\ \text{is continuous on }[0,\pi/2]$
the condition of Part 2, and here it costs nothing to confirm
$F(x)=-\cos x$
because $F'(x)=\sin x$; the $+C$ is deliberately dropped, since it would cancel anyway
the double negative on the lower limit is where most sign errors in this section are born
Answer $$\int_{0}^{\pi/2}\sin x\,dx = 1$$
Check
Independent check by pinning the area between two shapes it must lie between. The region sits inside the rectangle of width $\pi/2$ and height $1$, area $\approx 1.571$, and it contains the triangle with vertices $(0,0)$, $(\pi/2,0)$, $(\pi/2,1)$, area $\pi/4\approx 0.785$. The value $1$ lies between them.
No sums, no limits, three lines. Compare that with the Riemann-sum route, which needs a trigonometric identity that is not in this course.
A clean number like $1$ is the normal outcome once an antiderivative is available. That is the whole payoff of Part 2.
Displacement and distance for v(t) = t² − 4
A particle moves along a line with velocity $v(t)=t^{2}-4$ metres per second, for $t$ between $0$ and $3$ seconds. Find its and the total distance it travelled.
Given
$v(t)=t^{2}-4$ in metres per second
the time interval $0\le t\le 3$ seconds
Find
the displacement, and the total distance travelled
Independent check on the relationship rather than the arithmetic: distance can never be smaller than the size of the displacement, and it is equal only when the velocity keeps one sign. Here $\tfrac{23}{3}\approx 7.67$ is comfortably larger than $3$, and the velocity does change sign at $t=2$, so both facts are consistent.
Displacement is $\int v$; distance is $\int\lvert v\rvert$. The second one always costs you a sign hunt and a split, and a question that wants it will say total distance.
Checkpoint
§09.3 — the order of the subtraction●●○○○
Thirty seconds, and the arithmetic is trivial on purpose: what is being tested is the order of the two evaluations.
Given
$\displaystyle\int_{1}^{2} 3x^{2}\,dx$
an antiderivative of $3x^{2}$ is $x^{3}$
Find
What is the value of the integral?
Hint 1/4
The antiderivative is already handed to you, so the only decision left is which of the two numbers gets subtracted from which.
Hint 2/4
Part 2: $\int_{a}^{b} f = F(b)-F(a)$, the upper limit first.
Hint 3/4
With $F(x)=x^{3}$, $a=1$ and $b=2$: $F(2)=8$ and $F(1)=1$.
⚠ Applying Part 2 across a point where the integrand is not defined
The antiderivative formula keeps working symbolically on both sides of the bad point, so nothing on the page looks broken until the answer has the wrong sign.
right$x^{-2}\ \text{is undefined at }0,\ \text{so Part 2 does not apply on }[-1,1]$
Substitution: the chain rule read backwards
Part 2 is only as strong as your stock of antiderivatives, and the table above is short. Substitution is the one move that stretches it, and it stretches it a long way.
RuleThe Substitution Rule
Conditions
$u=g(x)$ is differentiable and $g'$ is continuous on $[a,b]$
$f$ is continuous on an interval containing all the values of $g$ on $[a,b]$
If the factor left over outside is exactly the derivative of the thing inside, then renaming the inside $u$ turns the whole integral into an integral in $u$ — and in the definite case the two limits get renamed with everything else, from $x$-values into $u$-values.
Why it is true, in three lines
Let $F$ be an antiderivative of $f$. The chain rule says $\dfrac{d}{dx}F(g(x))=f(g(x))g'(x)$.
So $F(g(x))$ is an antiderivative of the integrand on the left, and Part 2 turns the definite integral into $F(g(b))-F(g(a))$.
But $F(g(b))-F(g(a))$ is exactly what Part 2 gives for $\int_{g(a)}^{g(b)} f(u)\,du$. The two integrals are the same number because they have the same evaluation.
What $u=\sin x$ does to the interval $[0,\pi/2]$. The two endpoints travel to $\textcolor{#d1690a}{0}$ and $\textcolor{#d1690a}{1}$, and evenly spaced $x$-values arrive unevenly spaced — the stretching is what $du=\cos x\,dx$ records.
Looks like this, but is not
$\displaystyle\int \sin(x^{3})\,dx$ has an obvious inside function, $u=x^{3}$, so substitution looks available: $du=3x^{2}\,dx$.
There is no $x^{2}$ in the integrand to absorb into $du$. Forcing it gives $dx=\dfrac{du}{3x^{2}}=\dfrac{du}{3u^{2/3}}$, so the integral becomes $\int\dfrac{\sin u}{3u^{2/3}}\,du$ — a different problem, no easier than the first. Substitution is the chain rule run backwards, and the chain rule always leaves the inner derivative behind as a factor; if that factor is missing, there is nothing to run backwards. Compare $\int x^{2}\sin(x^{3})\,dx$, which is a two-line exercise.
∫ from 0 to π/2 of sin³x cos x dx, both playbooks
Evaluate $\displaystyle\int_{0}^{\pi/2}\sin^{3}x\cos x\,dx$ twice: once by converting the limits, once by back-substituting.
Given
$\int_{0}^{\pi/2}\sin^{3}x\cos x\,dx$
the interval $[0,\pi/2]$ in the variable $x$
Find
the exact value, computed two ways
SolutionSpot the inside function
$u=\sin x,\qquad du=\cos x\,dx$
the giveaway is that $\cos x\,dx$ is sitting there already, complete, with no constant to fix up
Independent check by differentiating the antiderivative from playbook B: $\dfrac{d}{dx}\dfrac{\sin^{4}x}{4}=\dfrac{4\sin^{3}x\cos x}{4}=\sin^{3}x\cos x$, which is the original integrand.
Playbook A is two lines shorter. Playbook B buys you the differentiate-back check.
Pick a playbook before you touch the numbers and stay in it. The sign errors in this topic come almost entirely from starting one and finishing the other.
∫ from 0 to 2 of x√(x²+5) dx
Evaluate $\displaystyle\int_{0}^{2} x\sqrt{x^{2}+5}\,dx$. Here the leftover factor is off by a constant, which is allowed.
Independent check by differentiating: $\dfrac{d}{dx}\Bigl[\tfrac13(x^{2}+5)^{3/2}\Bigr] =\tfrac13\cdot\tfrac32(x^{2}+5)^{1/2}\cdot 2x = x\sqrt{x^{2}+5}$. A size check agrees too: on $[0,2]$ the integrand runs from $0$ up to $2\cdot 3=6$, so a value of about $5.3$ over an interval of length $2$ is the right order.
A missing constant is repairable, a missing variable factor is not. That single sentence decides whether substitution is available before you write anything down.
∫ from −2 to 2 of x³/(1 + x⁴) dx, and where the symmetry shortcut comes from
Substitution does one more thing for us: on an interval symmetric about $0$ it can finish an integral before it starts. We derive that with $u=-x$ and then use it on $\displaystyle\int_{-2}^{2}\frac{x^{3}}{1+x^{4}}\,dx$.
Given
$\int_{-a}^{a} f(x)\,dx$ with $a>0$ and $f$ continuous on $[-a,a]$
$\int_{-2}^{2}\dfrac{x^{3}}{1+x^{4}}\,dx$
Find
the shortcut, then the value of the integral
SolutionFold the left half onto the right half with u = −x
Independent check by grinding it out: with $u=1+x^{4}$ we get $\tfrac14\,du=x^{3}\,dx$, and both endpoints land on the same value, $u=1+16=17$. So the integral becomes $\tfrac14\int_{17}^{17}\frac{du}{u}$, an integral over an interval of zero length, which is $0$ whatever the integrand is.
The shortcut costs one sign test. The substitution route above costs three lines and ends at the same zero.
On a symmetric interval, test $f(-x)$ before integrating anything. One line of algebra can replace the entire computation — and when the test fails, you have lost one line, not the question.
Checkpoint
§09.4 — what happens to the limits●●○○○
Thirty seconds, and nothing is asked about the integrand. The whole question is what the two numbers on the integral sign become.
A definite integral already measures the signed area between one curve and the axis. Two curves need exactly one new idea, and it is a picture rather than a formula.
RuleArea between two curves
Conditions
$f$ and $g$ are continuous on $[a,b]$
$f(x)\ge g(x)$ for every $x$ in $[a,b]$, checked with a test point rather than assumed
$$\boxed{\,A=\int_{a}^{b}\bigl[f(x)-g(x)\bigr]\,dx \quad\text{and, when the curves swap, split at every crossing}\,}$$
At each $x$ the region has a vertical strip in it whose height is the top curve minus the bottom curve; adding those heights across the interval gives the area. The axis plays no role at all — the region does not care where it is.
Where the formula comes from
Cut the region into thin vertical strips of width $\Delta x$. The strip at $x$ is very nearly a rectangle of height $f(x)-g(x)$, so it contributes about $[f(x)-g(x)]\Delta x$. Adding the strips is a Riemann sum for $\int_{a}^{b}[f-g]\,dx$, and the limit of that sum is the integral. If $g$ dips below the axis, $f(x)-g(x)$ is still the height of the strip, which is why no separate case is needed for regions that straddle the axis.
Between the crossings at $0$ and $1$ the curve $\textcolor{#1f6feb}{y=2x-x^{2}}$ is above $\textcolor{#d1690a}{y=x^{2}}$, so every strip has height $(2x-x^{2})-x^{2}$ and the area is the integral of that height.
Looks like this, but is not
$\displaystyle\int_{-1}^{1}\bigl(x^{3}-x\bigr)\,dx=0$, and $x^{3}-x$ is the difference of the two curves $y=x^{3}$ and $y=x$. It looks as though the two curves enclose no area on $[-1,1]$.
They enclose two lobes, one on each side of the origin, each of area $\tfrac14$. The curves swap places at $x=0$: on $[-1,0]$ the cubic is on top and on $[0,1]$ the line is. The integral of $x^{3}-x$ counts the second lobe negatively and the two cancel. The formula was applied without its condition, which is that one curve stays on top for the whole interval.
Area enclosed by y = x² and y = 2x − x²
Find the area of the region enclosed by the parabolas $y=x^{2}$ and $y=2x-x^{2}$.
Given
$y=x^{2}$
$y=2x-x^{2}$
no interval is given, so the curves must supply it themselves
Find
the area of the enclosed region
SolutionFind the crossings — they are the limits
$x^{2}=2x-x^{2} \iff 2x^{2}-2x=0 \iff 2x(x-1)=0$
when the region is described only by two curves, the crossings are the only candidates for $a$ and $b$
Part 2, with the lower limit contributing nothing here
Answer $$A=\dfrac{1}{3}$$
Check
Independent check with a geometry fact rather than more calculus: the height function $2x-2x^{2}$ is a downward parabola with roots $0$ and $1$ and maximum $\tfrac12$ at $x=\tfrac12$. The area under a parabolic arch is two thirds of its bounding rectangle, that is $\tfrac23\cdot 1\cdot\tfrac12=\tfrac13$.
Three moves: crossings, test point, one integral. Skipping the test point is what turns this into a sign error.
When the region is described by curves alone, the limits are the crossings and finding them is step one, not a detail.
Total area between y = x³ and y = x on [−1, 1]
Find the total geometric area of the region between $y=x^{3}$ and $y=x$ for $-1\le x\le 1$ — the case the counterexample above got wrong.
Given
$y=x^{3}$
$y=x$
the interval $[-1,1]$
Find
the total geometric area
SolutionFind every crossing inside the interval
$x^{3}=x \iff x(x-1)(x+1)=0 \iff x=-1,\,0,\,1$
the interior crossing at $0$ is the one that forces the split
Test each piece separately
$x=-\tfrac12:\ x^{3}=-\tfrac18 > x=-\tfrac12$
on $[-1,0]$ the cubic is on top; a negative number closer to zero is the larger one, which is where intuition slips
$x=\tfrac12:\ x=\tfrac12 > x^{3}=\tfrac18$
on $[0,1]$ the line is on top
One integral per piece, each with a positive integrand
Independent check by symmetry: $x^{3}-x$ is an odd function, so the two lobes are congruent and computing one of them and doubling must give the same answer. It does: $2\cdot\tfrac14=\tfrac12$. Notice that the same oddness is what made the signed integral $0$.
Two curves that cross inside the interval always cost you a split. Find the crossings before you decide what to integrate.
Checkpoint
§09.5 — which curve is on top●●●○○
Thirty seconds, on the half of the previous example that most people get backwards. Only the piece to the left of the origin is in question.
Given
the curves $y=x^{3}$ and $y=x$
the piece $-1\le x\le 0$
Find
Which expression gives the area of that piece?
Hint 1/4
Pick one number strictly inside the piece and compare the two heights there. Do not decide from the shape of the formulas.
Hint 2/4
The strip height is top minus bottom, and the integrand of an area must be positive on the piece.
Hint 3/4
At $x=-\tfrac12$: $x^{3}=-\tfrac18$ and $x=-\tfrac12$, and $-\tfrac18$ is the larger of the two.
Hint 4/4
So the cubic is on top there and the area is $\int_{-1}^{0}(x^{3}-x)\,dx$.
one value strictly inside is enough because the curves only meet at the ends of this piece
$-\tfrac18 > -\tfrac12 \Rightarrow x^{3}\ \text{is on top}$
on the negative side, cubing a number between $-1$ and $0$ moves it towards zero, so it rises above the line
Write the integral
$A=\int_{-1}^{0}(x^{3}-x)\,dx=\tfrac14$
top minus bottom, and the value is positive as an area must be
Answer $$A=\int_{-1}^{0}(x^{3}-x)\,dx=\tfrac14$$
Check
Independent check on the sign alone: the integrand $x^{3}-x$ at $x=-\tfrac12$ is $-\tfrac18+\tfrac12=\tfrac38>0$, and an area integrand must be positive throughout the piece.
One test point strictly inside the piece settles which curve is on top. Reading it off the picture is what produces negative areas.
⚠ Deciding which curve is on top by looking at the formulas
A cubic "grows faster" than a line, which is true far out and false on $[-1,1]$. The general fact gets applied to a particular interval where it does not hold.
Every antidifferentiation so far has been of the form: here is $y'$, find $y$. A separable equation is the same request with $y$ appearing on the right as well.
MethodSolving a separable first-order equation
Conditions
the equation can be written $\dfrac{dy}{dx}=g(x)h(y)$, with the two variables in separate factors
$h(y)\neq 0$ on the piece of the solution being found
Get everything with a $y$ onto one side next to $dy$ and everything with an $x$ onto the other next to $dx$, then antidifferentiate each side in its own variable. One constant is written, and the initial condition fixes it.
Why writing dy and dx on opposite sides is legal
The manipulation looks like algebra on a symbol that is not a fraction, so here is the honest version. Divide the equation by $h(y)$ to get $\dfrac{1}{h(y)}\dfrac{dy}{dx}=g(x)$, then antidifferentiate both sides with respect to $x$. The left side is $\int\dfrac{1}{h(y(x))}y'(x)\,dx$, and that is exactly the shape substitution eats, with $u=y(x)$ and $du=y'(x)\,dx$; it equals $\int\dfrac{du}{h(u)}$. So the separated form is a substitution in disguise, not a new liberty with symbols.
Two solutions of the same equation $y'=xy^{2}$. Through $\textcolor{#d1690a}{(0,1)}$ the solution runs into vertical asymptotes at $\pm\sqrt2$ and lives only on $(-\sqrt2,\sqrt2)$; through $\textcolor{#1f6feb}{(0,-1)}$ it is defined for every real $x$. Same family, different life expectancy.
Looks like this, but is not
$\dfrac{dy}{dx}=x+y$ has an $x$ and a $y$ on the right, so it looks like a candidate for separation: move the $y$ across and integrate.
Separable means the right-hand side is a product of a function of $x$ and a function of $y$, and $x+y$ is a sum that does not factor. Dividing by $x+y$ gives $\dfrac{dy}{x+y}=dx$, whose left side still contains $x$, so nothing has been separated. This equation is solvable, but by a method this course does not cover; separability is a property of the formula, and checking it is the first step, not a formality.
Solving y′ = xy² with y(0) = 1, and finding where it lives
Solve the initial value problem $\dfrac{dy}{dx}=xy^{2}$, $y(0)=1$, and give the largest open interval on which the solution exists.
Given
$\dfrac{dy}{dx}=xy^{2}$
$y(0)=1$
Find
$y$ as an explicit function of $x$, and its interval of existence
SolutionSeparate
$g(x)=x,\quad h(y)=y^{2}$
confirm the product form before dividing; this is the step that decides whether the method applies at all
$\frac{dy}{y^{2}}=x\,dx$
dividing by $y^{2}$ is legal wherever $y\neq0$, and $y(0)=1$ puts us on such a piece
Antidifferentiate each side in its own variable
$\int y^{-2}\,dy=\int x\,dx$
each side is now a one-variable antiderivative problem
$-\frac{1}{y}=\frac{x^{2}}{2}+C$
one constant, written on the right; a second one on the left would merge into this one immediately
Use the initial condition, then solve for y
$-\frac{1}{1}=0+C \Rightarrow C=-1$
apply the data point to the integrated equation, never to the separated one
Independent check by substitution into the original equation: $y'=\dfrac{4x}{(2-x^{2})^{2}}$ and $xy^{2}=x\cdot\dfrac{4}{(2-x^{2})^{2}}=\dfrac{4x}{(2-x^{2})^{2}}$, which agree, and $y(0)=\tfrac22=1$.
Four moves, and the last one is the one people skip. An answer without its interval is not finished.
The initial condition does two jobs: it fixes $C$ and it chooses which side of the blow-up you are on.
Solving y′ = −x/y with y(0) = 3
Solve $\dfrac{dy}{dx}=-\dfrac{x}{y}$ with $y(0)=3$, and say what the solution curves look like.
Given
$\dfrac{dy}{dx}=-\dfrac{x}{y}$
$y(0)=3$
Find
the explicit solution and its interval
SolutionSeparate and integrate
$y\,dy=-x\,dx$
the right side is $(-x)\cdot\tfrac1y$, a product, so the equation is separable
Independent check, geometric rather than algebraic: the slope $-x/y$ at a point is the negative reciprocal of the slope $y/x$ of the line from the origin, so every solution curve meets every radius at a right angle — and the curves with that property are exactly the circles centred at the origin. Differentiating also works: $\dfrac{d}{dx}\sqrt{9-x^{2}}=\dfrac{-x}{\sqrt{9-x^{2}}}=-\dfrac{x}{y}$.
An implicit relation is a legitimate stopping point; picking the branch is what the initial condition is for.
Checkpoint
§09.6 — which equation separates●●○○○
Thirty seconds. Nothing has to be solved; the only question is which right-hand side is a product of a function of $x$ and a function of $y$.
Given
four first-order equations
Find
Which one is separable?
Hint 1/4
You are looking for a product, not for the presence of both letters.
Hint 2/4
Separable means $\dfrac{dy}{dx}=g(x)\,h(y)$: one factor in $x$ alone, one in $y$ alone.
Hint 3/4
Test each right-hand side by asking whether it can be written as a product. A sum of $x$ and $y$ cannot; a product of a power of $x$ and a power of $y$ obviously can.
Hint 4/4
$x^{2}y$ is $g(x)=x^{2}$ times $h(y)=y$.
Show solutionTest each right-hand side
$x^{2}y=(x^{2})(y) \ \checkmark$
a genuine product, so the variables can be pulled apart
$x^{2}+y,\ \ x+\tfrac1y \ \text{are sums}$
no factoring turns a sum of that shape into a product
$\sin(xy) \ \text{mixes the variables inside one function}$
there is no identity that splits it into a product of a function of $x$ and one of $y$
Answer $$\frac{dy}{dx}=x^{2}y$$
Check
Independent check: separate it and see that the method actually runs. $\dfrac{dy}{y}=x^{2}\,dx$ has each variable alone on its own side, which is the definition being tested.
Separable means factored. The test is whether every $y$ can be moved to one side without inventing algebra, and it is faster to try than to argue about.
⚠ Carrying two constants of integration
Both sides were antidifferentiated, and each antiderivative genuinely has a family constant. Keeping both makes the initial condition look like one equation in two unknowns.
The algebra ends with a formula, the formula looks like an answer, and nothing on the page reminds you that it stops being a solution at the first blow-up.
Any $\int_{a}^{b} f$ with a continuous integrand and no unknown functions in it.
Check the interval
Is $f$ defined and continuous at every point of $[a,b]$, endpoints included? If it is not, stop — the theorem does not apply and the answer you would get is meaningless.
Simplify before antidifferentiating
Expand products, split fractions with a single term downstairs, and rewrite roots as powers. Every one of these is cheaper than the antiderivative it saves.
Find one antiderivative
One is enough, and the $+C$ is dropped on purpose, because it cancels.
Write the bracket
$\bigl[F(x)\bigr]_{a}^{b}$ on its own line, before any numbers go in. This is what stops the subtraction from happening backwards.
Subtract, then sanity-check
$F(b)-F(a)$. Then ask whether the sign matches the picture: an integrand that is positive on the whole interval cannot give a negative answer.
Where it goes wrong
Integrating across a point where the integrand is undefined, which produces a confidently wrong number.
Subtracting in the wrong order, which flips the sign of everything.
Evaluating $f$ at the limits instead of $F$, usually when the antiderivative was found in the margin.
Running a substitution
The integrand contains a composition and a factor that is, up to a constant, the derivative of the inside function.
Name the inside
Set $u$ equal to the expression that is inside something else: under a root, inside a power, inside a trigonometric function.
Differentiate it
Write $du=g'(x)\,dx$ and then look for $g'(x)\,dx$ in the integrand. A missing constant is repairable; a missing variable factor is not.
Convert the limits, on their own line
$u(a)$ and $u(b)$, written down before anything else moves. Skipping this line is the single most common error in this topic.
Rewrite with no x left
Every $x$ must be gone. If one survives, either express it through $u$ or abandon the substitution.
Integrate and evaluate
Finish in $u$ with the new limits, or back-substitute and finish in $x$ with the old ones. Choose one before you start.
Where it goes wrong
Converting the integrand but leaving the original limits next to a $du$.
Trying to force a substitution when the inner derivative is not present, which just relabels the difficulty.
Mixing the two playbooks: new limits with a back-substituted antiderivative.
Setting up an
A region is described by two or more curves, with or without an interval.
Sketch first
A rough sketch, thirty seconds. Everything after this depends on knowing the shape of the region, and no algebra recovers it.
Find the crossings
Solve $f(x)=g(x)$. If no interval was given, the crossings are the interval.
Test each piece
One value strictly inside each piece between consecutive crossings, to see which curve is on top there.
Choose strip direction
Vertical strips and $dx$ when the region has one top curve and one bottom curve over the whole interval; and $dy$ when it has one right curve and one left curve instead. The test is which choice avoids a split.
One integral per piece
Top minus bottom on each piece, each integrand positive, and add the results.
Where it goes wrong
Assuming the curve that is on top at the left end stays on top.
Using the crossings as limits when the question named its own interval instead.
Sticking with vertical strips through a region whose top curve changes halfway, when horizontal strips would need one integral.
Solving a separable initial value problem
A first-order equation whose right-hand side factors into a part in $x$ and a part in $y$, together with one data point.
Confirm the product form
Write the right-hand side as $g(x)h(y)$ explicitly. If you cannot, the method does not apply and no amount of algebra will make it.
Separate
$\dfrac{dy}{h(y)}=g(x)\,dx$, and note which $y$-values you just divided away.
Antidifferentiate both sides, one constant
Each side in its own variable, with a single $+C$ on the right.
Apply the initial condition
Put the data point into the integrated relation to get $C$. Doing it earlier gives nothing.
Solve for y, then state the interval
Make $y$ explicit if the algebra allows, pick the branch the initial point sits on, and give the largest open interval around $x_{0}$ on which the formula is defined.
Check what you divided away
Each root of $h$ gives a constant solution. Say whether it is the one you want.
Where it goes wrong
Keeping two constants and then declaring the problem underdetermined.
Reporting the formula with no interval, so that it silently claims to be a solution on the far side of a blow-up.
Losing the equilibrium solution, which matters whenever the initial value is exactly at a root of $h$.
Independent check that the bracket is not empty: the integrand is at least $\cos 1>0$ everywhere on $[0,1]$, so the integral is positive, and it is below $1$ because the integrand never reaches $1$ except at the single point $x=0$.
When no substitution exists, a bound is still an answer. Naming the largest and smallest value the integrand takes says something true, and it is worth marks where a wrong closed form is worth none.
The two integrands are built from the same inside function $x^{3}$ or $x^{2}$ and the same outer trigonometric function; the only difference is whether a factor of the inner derivative is sitting outside. One is a two-line exercise and the other has no elementary answer at all.
How to tell them apart
Before writing $u=\dots$, look for the inner derivative as a factor in the integrand. If it is there up to a constant, substitution will finish the job; if it is not there, no choice of $u$ will manufacture it, and a bound may be the only honest answer.
Scaffolding comes off
The common skeleton
Check that the integrand is defined and continuous at every point of the closed interval.
Look for an inside function whose derivative is present outside as a factor; name it $u$ and write $du$.
Convert both limits into $u$-values, on their own line, before anything else moves.
Rewrite the integral so that no $x$ survives anywhere, integrand or limits.
Antidifferentiate in $u$, evaluate top minus bottom, and check the sign against the picture.
1 · fully worked
∫ from 0 to 2 of x(x² + 1)³ dx, fully worked
Every step written out with its reason. Read it once, then cover the right-hand column and read it again.
Given
$\int_{0}^{2} x(x^{2}+1)^{3}\,dx$
Find
the exact value
SolutionContinuity
$x(x^{2}+1)^{3}\ \text{is a polynomial}$
polynomials are continuous everywhere, so the theorem applies on any interval
Independent check by brute force: expanding gives $x^{7}+3x^{5}+3x^{3}+x$, whose integral over $[0,2]$ is $32+32+12+2=78$. The substitution took four lines, the expansion took eight.
One substitution, one constant repair. The expansion route also works here and will not work at all on the next rung.
Count the lines before choosing a route: four here against eight for the expansion. When both work, the shorter one also has fewer places to lose a sign.
2 · you write the reasoning
Deliberately easier than the one above: the constant is already right, so there is no $\tfrac12$ to carry. The steps are written for you and the reasons are not. Say out loud why each line is allowed before you open the box under it, and evaluate $\displaystyle\int_{0}^{1}2x(x^{2}+1)^{2}\,dx$.
$u=x^{2}+1,\qquad du=2x\,dx$
reasoning
The inside function is whatever sits under the square. Its derivative is $2x$, and $2x$ is standing right there in the integrand — complete, with no constant to repair. That is why this problem is easier than the one above.
$x=0\Rightarrow u=1,\qquad x=1\Rightarrow u=2$
reasoning
The limits are $x$-values, and everything in the integral is about to become a $u$-expression, so they have to travel too. $u=x^{2}+1$ sends $0$ to $1$ and $1$ to $2$.
Now the rewrite is exact rather than approximate: $2x\,dx$ is $du$, and $(x^{2}+1)^{2}$ is $u^{2}$. The test that it was done correctly is that no $x$ appears anywhere on the right, limits included.
The reverse power rule applies to $u^{2}$ exactly as it would to $x^{2}$; the letter is irrelevant. Then top minus bottom, in that order.
3 · find the buried error
Harder than the rung above, and this time the work is done for you — badly. Exactly two of the four steps below contain an error. Find both. The problem: evaluate $\displaystyle\int_{0}^{\pi/2}\cos x\sqrt{1+\sin x}\,dx$.
Step 1. $u=1+\sin x$, so $du=\cos x\,dx$ — and $\cos x\,dx$ is already sitting in the integrand, complete.
Step 2. So the integral becomes $\displaystyle\int_{0}^{\pi/2}\sqrt{u}\,du$.
Step 3. An antiderivative of $\sqrt{u}=u^{1/2}$ is $\tfrac12 u^{3/2}$.
Step 4. Evaluating between the limits written in step 2: $\tfrac12\left(\tfrac{\pi}{2}\right)^{3/2}-\tfrac12(0)^{3/2}\approx 0.98$.
the two buried errors (2)
⚠ step 2
The integrand was converted to $u$ but the limits were left as $x$-values. They must go through the substitution as well: $x=0$ gives $u=1+\sin 0=1$, and $x=\pi/2$ gives $u=1+\sin(\pi/2)=2$. The correct line is $\int_{1}^{2}\sqrt{u}\,du$.
The limits sit outside the expression being rewritten, so the eye never goes back to them, and the resulting line looks perfectly well formed.
The reverse power rule was applied by dividing by the old exponent instead of by the new one. With $n=\tfrac12$ the new exponent is $\tfrac32$, so the antiderivative is $\dfrac{u^{3/2}}{3/2}=\tfrac23 u^{3/2}$, not $\tfrac12 u^{3/2}$.
The rule is remembered as "raise and divide", and the number you were just looking at is the old exponent. With whole-number powers the slip is visible; with a half it is not.
right
$\displaystyle\int u^{1/2}\,du=\frac{u^{3/2}}{3/2}+C=\frac{2}{3}u^{3/2}+C$. With both repairs the answer is $\tfrac23\bigl[u^{3/2}\bigr]_{1}^{2}=\tfrac23\bigl(2\sqrt2-1\bigr)\approx 1.219$, so you can check your own version against it.
4 · the bare problem
§09.4 — substitution with no scaffolding●●●○○
No steps, no hints on the page until you open them. Same skeleton as the three rungs above, with a negative power this time.
the reverse power rule works for $n=-2$ because the exception is only at $n=-1$
$=-\tfrac17+\tfrac14=\tfrac{3}{28}$
top minus bottom; the two minus signs are where this goes wrong if it goes wrong
Answer $$\frac{3}{28}$$
Check
Independent check by size: on $[1,2]$ the integrand runs from $\tfrac{2}{16}=0.125$ down to $\tfrac{4}{49}\approx 0.082$, so over an interval of length $1$ the value must sit between those, and $\tfrac{3}{28}\approx 0.107$ does.
A negative exponent changes nothing about the method; only $n=-1$ is special.
Full exam-style question
A full three-part question of the kind this week producesexam format
Three parts, about twelve minutes, and each part uses a different one of the tools on this page. This is the shape the long question takes.
Three independent checks, one per part. (a) Replacing $\sqrt{1+t^{3}}$ by the constant $1$ makes $G(x)=x^{2}$ and $G'(x)=2x$, which the formula reproduces. (b) The integrand is between $0$ and $1$ on an interval of length $\pi/2\approx1.571$, so a value of $\tfrac13$ is the right order of size. (c) The height function $2-2x^{2}$ is a parabolic arch of base $2$ and height $2$, and two thirds of the bounding rectangle is $\tfrac23\cdot 2\cdot 2=\tfrac83$.
Three different tools in twelve minutes. Deciding which one a part wants is most of the exam skill; each individual computation is short.
Notice that no part of this question was long. What makes the marks hard to collect is that the three parts want three different tools, and nothing on the paper announces which.
Practice
A · concept 4 questions
1§09.1 — what continuity of the integrand buys you●●○○○
A statement about every function of this shape at once, not about one example. Decide whether it can ever fail.
Given
$f$ is continuous on $[a,b]$
$F(x)=\int_{a}^{x} f(t)\,dt$
Find
True or false: $F$ is differentiable at every point of $(a,b)$, whatever the continuous $f$ is.
Hint 1/4
Ask what the theorem's conclusion is and what its hypothesis is, and check whether the hypothesis is satisfied here.
Hint 2/4
Part 1: if $f$ is continuous on $[a,b]$ and $F(x)=\int_{a}^{x}f$, then $F'(x)=f(x)$ on $(a,b)$.
Hint 3/4
The hypothesis given is exactly continuity of $f$ on $[a,b]$, which is what the theorem asks for.
Hint 4/4
So $F'$ exists everywhere in $(a,b)$, and equals $f$ there.
Show solutionMatch hypothesis to conclusion
$f\ \text{continuous on }[a,b] \Rightarrow F'(x)=f(x)\ \text{on}\ (a,b)$
nothing else about $f$ is required — it need not be differentiable, monotone or given by a formula
Note what would break it
$f\ \text{with a jump at }x_{0} \Rightarrow F\ \text{has a corner at }x_{0}$
the counterexample earlier in this section; the statement is true precisely because continuity was assumed
Answer $$\text{True}$$
Check
Independent check with a hard case: $f(t)=\lvert t\rvert$ is continuous but not differentiable at $0$, and $F(x)=\int_{0}^{x}\lvert t\rvert\,dt$ is still differentiable there, with $F'(0)=0$. Integration only ever makes things smoother.
The hypothesis is about $f$ and the conclusion is about $F$. Keep track of which function each condition is talking about; most wrong answers here are answers to the other question.
2§09.3 — a hypothesis that is easy to skip●●●○○
A calculation that looks complete and gives a specific number. Decide whether the number means anything.
True or false: the integral equals $\Bigl[-\tfrac1x\Bigr]_{-1}^{1}=-1-1=-2$.
Hint 1/4
Before checking the algebra, check the answer against the graph: what sign must this integral have?
Hint 2/4
Part 2 requires the integrand to be continuous on the whole closed interval $[a,b]$.
Hint 3/4
Here the integrand $x^{-2}$ is not defined at $x=0$, which is inside $[-1,1]$, so it is not continuous there.
Hint 4/4
The hypothesis fails, so the computation is not licensed — and the answer $-2$ is impossible for a positive integrand anyway.
Show solutionThe hypothesis
$f\ \text{is undefined at }0\in[-1,1]$
so $f$ is not continuous on $[-1,1]$, and $-1/x$ is an antiderivative on each side separately but on no interval spanning $0$
The sign, independently
$x^{-2}>0\ \text{wherever it is defined}$
every Riemann sum of a positive integrand is positive, so no limit of them can be $-2$
Answer $$\text{False}$$
Check
Independent check by shrinking the interval: $\int_{\varepsilon}^{1}x^{-2}\,dx=\tfrac1\varepsilon-1$, which grows without bound as $\varepsilon\to 0^{+}$. The area on the right half alone is already unbounded, so no finite number can be the answer.
A number produced by a correct-looking computation is still wrong if the hypothesis failed. Check that the integrand is defined on the whole interval before evaluating anything.
3§09.4 — which integrand the substitution fits●●●○○
Four integrals, one substitution. Only one of them is set up for it, and the difference is a single factor.
Given
the substitution $u=x^{2}+1$, so $du=2x\,dx$
Find
For which integral does this substitution finish the job?
Hint 1/4
Substitution needs the derivative of the inside function to be present as a factor outside. Ask which integrand has an $x$ available.
Hint 2/4
With $u=g(x)$ the integrand must have the form $f(g(x))\cdot g'(x)$, up to a constant factor.
Hint 3/4
Here $g(x)=x^{2}+1$ and $g'(x)=2x$, so the integrand needs a factor of $x$ next to a function of $x^{2}+1$.
Hint 4/4
Only $x(x^{2}+1)^{5}$ has that factor, and it becomes $\tfrac12\int u^{5}\,du$.
Independent check by differentiating the claimed antiderivative: $\dfrac{d}{dx}\tfrac{1}{12}(x^{2}+1)^{6}=\tfrac{6}{12}(x^{2}+1)^{5}\cdot 2x=x(x^{2}+1)^{5}$.
Is the inner derivative present as a factor? That is a yes-or-no test costing five seconds, and it is worth running on every candidate before starting any of them.
4§09.5 — what a zero difference integral forces●●●●○
A statement about two curves whose difference integrates to zero. It is tempting to say the curves must be equal; decide what is actually forced.
same argument with the sign reversed; both contradict the given value $0$
Answer $$\text{True}$$
Check
Independent check with the example from earlier in this section: $y=x^{3}$ and $y=x$ on $[-1,1]$ have difference integral $0$, and sure enough they meet — three times, at $-1$, $0$ and $1$.
A zero integral says the signed areas balance, which is much weaker than saying the curves coincide.
B · computation 7 questions
1§09.1 — accumulation with the moving end on the left of a●●○○○
The standard warm-up, with one twist: the point you are asked about lies below the fixed limit, which unsettles people who picture the integral as area only.
(a) Say why $F$ is differentiable at every real $x$.
(b) Find $F'(x)$.
(c) Evaluate $F'(0)$, and say whether $F$ is rising or falling there.
Hint 1/4
Two of the three parts are one line each. Start by naming which theorem applies and checking its single hypothesis.
Hint 2/4
Part 1: if $f$ is continuous then $\dfrac{d}{dx}\int_{a}^{x}f(t)\,dt=f(x)$, for $x$ on either side of $a$.
Hint 3/4
Here $f(t)=\sqrt{1+t^{4}}$ and $a=2$, and $1+t^{4}\ge1>0$ for every real $t$.
Hint 4/4
So $F'(x)=\sqrt{1+x^{4}}$, and $F'(0)=1>0$: $F$ is rising at $0$.
Show solution
Part 1 answers all three parts without an antiderivative, and that is not a preference: $\sqrt{1+t^{4}}$ has no elementary antiderivative, so the route through a formula for $F$ is not merely longer here, it does not exist.
Check continuity
$1+t^{4}\ge 1 \Rightarrow \sqrt{1+t^{4}}\ \text{is defined and continuous everywhere}$
the root never meets a negative number, so there is no domain restriction to worry about
Differentiate and evaluate
$F'(x)=\sqrt{1+x^{4}}$
Part 1; the fixed lower limit $2$ never appears in the derivative
$F'(0)=\sqrt{1}=1>0$
so $F$ is increasing at $0$, even though $F(0)<0$
Answer $$F'(x)=\sqrt{1+x^{4}},\qquad F'(0)=1$$
Check
Independent check on the apparent oddity: for $x<2$ the integral runs backwards, so $F(x)<0$ there. But $F$ is still rising, because moving $x$ to the right shortens the backwards interval. Negative value and positive derivative are perfectly compatible.
Part 1 does not care whether $x$ is to the left or the right of the fixed limit.
2§09.2 — a cube in the upper limit●●●○○
The most-set version of the moving-limit question. The only thing being tested is whether the chain factor survives to the final line.
Given
$H(x)=\displaystyle\int_{0}^{x^{3}}\cos t\,dt$
Find
(a) Find $H'(x)$.
(b) Evaluate $H'(0)$.
(c) Write $H$ in closed form and differentiate it, as a check on (a).
Hint 1/4
Name the moving end and its derivative before touching the integrand.
Hint 2/4
With only the upper limit moving, $\dfrac{d}{dx}\int_{a}^{u(x)}f(t)\,dt=f(u(x))\,u'(x)$.
Hint 3/4
Here $u(x)=x^{3}$, $u'(x)=3x^{2}$ and $f(t)=\cos t$, so $f(u(x))=\cos(x^{3})$.
Hint 4/4
Therefore $H'(x)=3x^{2}\cos(x^{3})$, and $H'(0)=0$.
Show solution
The moving-limit rule is the cheap route; computing $H$ in closed form first would work too, and it is used below only as the check, because that route is available just when the integrand happens to be easy to antidifferentiate.
Apply the moving-limit rule
$u(x)=x^{3},\quad u'(x)=3x^{2}$
the moving end named on its own line
$H'(x)=\cos\bigl(x^{3}\bigr)\cdot 3x^{2}$
integrand at the moving end, times the speed of the end
$H'(0)=0\cdot\cos 0=0$
the end is momentarily stationary at $x=0$, so nothing is being accumulated
Handle the two terms separately, and get the second one into a form the power rule recognises.
Hint 2/4
Reverse power rule: an antiderivative of $x^{n}$ is $\dfrac{x^{n+1}}{n+1}$ for $n\neq-1$; and $\sqrt{x}=x^{1/2}$.
Hint 3/4
For $\int_{1}^{4}(2x-3\sqrt{x})\,dx$: the antiderivative is $x^{2}-3\cdot\tfrac{2}{3}x^{3/2}=x^{2}-2x^{3/2}$.
Hint 4/4
Evaluating: $(16-16)-(1-2)=1$.
Show solution
One antiderivative for the whole integrand beats splitting the integral in two, because both terms are powers of $x$ and a single bracket can carry them together.
the power rule cannot act on a root symbol; it acts on an exponent
$F(x)=x^{2}-2x^{3/2}$
term by term, with the $+C$ dropped for a definite integral
Evaluate
$F(4)=16-2\cdot 8=0$
$4^{3/2}=(\sqrt4)^{3}=8$
$F(1)=1-2=-1$
and the double negative in the subtraction is where this is usually lost
$\int_{1}^{4}(2x-3\sqrt{x})\,dx=0-(-1)=1$
top minus bottom
Answer $$1$$
Check
Independent check by splitting the integral: $\int_{1}^{4}2x\,dx=\bigl[x^{2}\bigr]_{1}^{4}=15$ and $\int_{1}^{4}3\sqrt{x}\,dx=\bigl[2x^{3/2}\bigr]_{1}^{4}=16-2=14$, and $15-14=1$.
Turn every root and every fraction into a power before looking for an antiderivative.
4§09.4 — substitution with a constant to repair●●●○○
A definite integral where the inner derivative is present but off by a factor of three. That factor is the only difficulty.
Find the expression that is inside something else, and compare its derivative with the factor sitting outside.
Hint 2/4
Substitution: $\int_{a}^{b} f(g(x))g'(x)\,dx=\int_{g(a)}^{g(b)} f(u)\,du$, and a constant mismatch may be carried outside.
Hint 3/4
For $\int_{0}^{1}x^{2}(x^{3}+2)^{2}\,dx$: take $u=x^{3}+2$, so $du=3x^{2}\,dx$ and $x^{2}\,dx=\tfrac13\,du$; the limits become $u=2$ and $u=3$.
Hint 4/4
The integral is $\tfrac13\int_{2}^{3}u^{2}\,du=\tfrac19(27-8)=\tfrac{19}{9}$.
Show solution
We convert the limits rather than back-substitute: it costs one line and ends the problem in $u$, whereas returning to $x$ means carrying $(x^{3}+2)^{3}$ through the evaluation for no gain.
the constant $\tfrac13$ is a number and may sit outside the integral sign
$=\tfrac19(27-8)=\tfrac{19}{9}$
top minus bottom inside the bracket
Answer $$\frac{19}{9}$$
Check
Independent check by expanding instead: $x^{2}(x^{3}+2)^{2}=x^{8}+4x^{5}+4x^{2}$, whose integral over $[0,1]$ is $\tfrac19+\tfrac46+\tfrac43=\tfrac19+\tfrac{12}{18}+\tfrac{24}{18}$, that is $\tfrac{2}{18}+\tfrac{12}{18}+\tfrac{24}{18}=\tfrac{38}{18}=\tfrac{19}{9}$.
A constant mismatch is repairable. Check for it before deciding a substitution has failed.
5§09.5 — area between a line and a parabola●●○○○
The smallest possible area question, worth doing carefully because every larger one is this with more crossings.
Given
the curves $y=x$ and $y=x^{2}$
the interval $0\le x\le 1$
Find
(a) Say which curve is on top on $(0,1)$, with a test point.
(b) Find the area of the region between them.
Hint 1/4
Evaluate both curves at one convenient number strictly inside the interval, and compare.
Hint 2/4
Area between curves: $A=\int_{a}^{b}[\text{top}-\text{bottom}]\,dx$, with the top decided by a test point.
Hint 3/4
At $x=\tfrac12$: $y=x$ gives $\tfrac12$ and $y=x^{2}$ gives $\tfrac14$, so the line is on top.
Hint 4/4
So $A=\int_{0}^{1}(x-x^{2})\,dx=\tfrac12-\tfrac13=\tfrac16$.
Show solution
A single top-minus-bottom integral is enough because the curves meet only at the two ends; computing the two areas separately and subtracting them doubles the work and adds a place to lose a sign.
Decide the top curve
$x=\tfrac12:\quad \tfrac12 > \tfrac14$
squaring a number between $0$ and $1$ makes it smaller, which is exactly why the parabola is underneath here
top minus bottom, simplified before antidifferentiating
$=\tfrac12-\tfrac13=\tfrac16$
a common denominator at the last step only
Answer $$A=\frac{1}{6}$$
Check
Independent check by geometry: the region sits inside the triangle under $y=x$ over $[0,1]$, whose area is $\tfrac12$, and the parabola removes $\int_{0}^{1}x^{2}=\tfrac13$ of it. $\tfrac12-\tfrac13=\tfrac16$, and $\tfrac16$ is comfortably less than $\tfrac12$, as it must be.
Top minus bottom, with the top decided by a test point, is the whole of the area method. Everything after that is an ordinary definite integral.
6§09.6 — a separable problem with an interval to find●●●●○
The full pipeline: separate, integrate, use the data point, make $y$ explicit, and say where the answer is valid. The last part is the one that carries marks and gets left out.
Given
$\dfrac{dy}{dx}=\dfrac{x^{2}}{y}$
$y(0)=2$
Find
(a) Identify $g(x)$ and $h(y)$ in the form $\dfrac{dy}{dx}=g(x)h(y)$.
(b) Separate, integrate, and use the initial condition to find the constant.
(c) Solve for $y$ explicitly and give the largest open interval containing $x=0$ on which your formula solves the problem.
Hint 1/4
Check the product form first; only then start moving symbols across the equals sign.
Hint 2/4
Separable method: $\dfrac{dy}{h(y)}=g(x)\,dx$, then antidifferentiate each side in its own variable with one constant.
Hint 3/4
Here $g(x)=x^{2}$ and $h(y)=\dfrac1y$, so the separated form is $y\,dy=x^{2}\,dx$, and $y(0)=2$ gives $\tfrac{4}{2}=0+C$.
Hint 4/4
So $\tfrac{y^{2}}{2}=\tfrac{x^{3}}{3}+2$, that is $y=\sqrt{\tfrac{2}{3}x^{3}+4}$, valid while $\tfrac{2}{3}x^{3}+4>0$.
Show solution
We apply the initial condition while the equation is still implicit, before solving for $y$: the constant is easier to pin down in $\tfrac{y^{2}}{2}$ than inside a square root, and the branch question then has to be answered only once.
Independent check by substitution: $y'=\dfrac{2x^{2}}{2\sqrt{\tfrac23x^{3}+4}}=\dfrac{x^{2}}{y}$, and $y(0)=\sqrt4=2$. Both the equation and the data point are satisfied.
The interval always comes from asking where the formula stops making sense, not from the algebra itself.
7§09.4 — a symmetric interval, term by term●●●○○
Every term here can be integrated the ordinary way, and doing that is not wrong. It is three times the work, and the classification takes about ten seconds.
(a) Say for each of the three terms whether it is even, odd, or neither.
(b) Evaluate the integral exactly.
Hint 1/4
The interval is symmetric about $0$, so before integrating anything, ask what each term does when $x$ is replaced by $-x$.
Hint 2/4
On $[-a,a]$: an odd integrand, $f(-x)=-f(x)$, integrates to $0$; an even one, $f(-x)=f(x)$, integrates to $2\int_{0}^{a}f$. A sum can be split and each piece classified on its own.
Hint 3/4
For $\int_{-1}^{1}\bigl(x^{5}-2x^{2}+x\sqrt{1+x^{2}}\bigr)dx$: $(-x)^{5}=-x^{5}$; $-2(-x)^{2}=-2x^{2}$; $(-x)\sqrt{1+(-x)^{2}}=-x\sqrt{1+x^{2}}$. Only one of the three survives, and it needs $2\int_{0}^{1}(-2x^{2})\,dx$.
Hint 4/4
The two odd terms contribute $0$ and the even term gives $-\tfrac43$, so the integral is $-\tfrac43$.
Show solution
We classify before integrating because two of the three terms then need no antiderivative at all; going straight at the integral means finding an antiderivative for $x\sqrt{1+x^{2}}$, which is a substitution in its own right, only to watch it cancel.
Test each term at −x
$(-x)^{5}=-x^{5}\ \Rightarrow\ \text{odd}$
an odd power is the standard odd function, and the sign comes straight out of the exponent
$-2(-x)^{2}=-2x^{2}\ \Rightarrow\ \text{even}$
the minus sign in front is a constant factor; it does not change the parity, only the height
Independent check, doing all three terms the long way. $\bigl[\tfrac{x^{6}}{6}\bigr]_{-1}^{1}=\tfrac16-\tfrac16=0$. $\bigl[-\tfrac{2x^{3}}{3}\bigr]_{-1}^{1}=-\tfrac23-\tfrac23=-\tfrac43$. And $\tfrac13(1+x^{2})^{3/2}$ is an antiderivative of the third term, taking the same value $\tfrac{2\sqrt2}{3}$ at both endpoints, so that piece is $0$. Same total, no symmetry used.
Parity is decided term by term, not for the sum as a whole. Split first, classify each piece, and integrate only what survives.
C · exam level 4 questions
1§09.1 — the shape of an accumulation function●●●●○
Eight minutes. Everything from the graph-shape material earlier in the term applies to a function defined by an integral, because Part 1 hands you its derivative for nothing.
(a) Find $F'(x)$ and the intervals on which $F$ increases and decreases.
(b) Locate every local extremum of $F$ and give its value.
(c) Compute $F(3)$.
(d) Explain in one sentence how $F$ can be decreasing at $x=1$ while $F(1)$ is negative.
Hint 1/4
Three of the four parts never need a formula for $F$. Get $F'$ first and see how far the sign line alone takes you.
Hint 2/4
Part 1 gives $F'=$ the integrand; then the Increasing/Decreasing Test and the First Derivative Test apply as usual.
Hint 3/4
With $F(x)=\int_{0}^{x}(t^{2}-4)\,dt$: $F'(x)=x^{2}-4=(x-2)(x+2)$, and a formula for $F$ itself is $\tfrac{x^{3}}{3}-4x$.
Hint 4/4
$F$ rises on $(-\infty,-2)$ and $(2,\infty)$, falls on $(-2,2)$, has a local maximum $\tfrac{16}{3}$ at $x=-2$ and a local minimum $-\tfrac{16}{3}$ at $x=2$, and $F(3)=-3$.
Show solution(a) Sign line for F′
$F'(x)=x^{2}-4=(x-2)(x+2)$
Part 1, then factor so the sign is readable
$F'(-3)=5>0,\quad F'(0)=-4<0,\quad F'(3)=5>0$
one test value strictly inside each of the three pieces
the two values are negatives of each other because the integrand is even
(c) One value
$F(3)=9-12=-3$
substitute into the same formula
(d) The interpretation
$F(1)<0\ \text{and}\ F'(1)=-3<0$
$F(1)$ is negative because the integrand was negative on all of $(0,1)$, and $F'(1)$ is negative because the integrand is still negative at $x=1$; the value and the direction answer different questions
Independent check on part (c) without the antiderivative: $F(3)=F(2)+\int_{2}^{3}(t^{2}-4)\,dt$, and the second piece is $\tfrac73$, so $F(3)=-\tfrac{16}{3}+\tfrac73=-\tfrac93=-3$.
Ask which parts of a question genuinely need a formula for $F$. Usually fewer than you think.
2§09.2 — both limits moving, evaluated at a point●●●●○
Exam level, and short. Two terms, two chain factors and one subtraction; the arithmetic at the end is deliberately light.
Write down the four ingredients — the two limits and their two derivatives — before assembling anything.
Hint 2/4
The rule is $\dfrac{d}{dx}\int_{a(x)}^{b(x)}f = f(b(x))b'(x)-f(a(x))a'(x)$.
Hint 3/4
With $f(t)=\sqrt{1+t^{2}}$, $b(x)=x^{2}$, $b'(x)=2x$, $a(x)=2x$, $a'(x)=2$: at $x=1$ we get $b(1)=1$ and $a(1)=2$.
Hint 4/4
So $P'(1)=\sqrt{2}\cdot 2-\sqrt{5}\cdot 2=2\sqrt2-2\sqrt5$.
Show solution
The moving-limit rule is the only route here: $\sqrt{1+t^{2}}$ has no elementary antiderivative, so computing $P$ first and differentiating — the free check used in the worked examples — is unavailable, and the check has to come from somewhere else.
List the ingredients
$b(x)=x^{2},\ b'(x)=2x;\qquad a(x)=2x,\ a'(x)=2$
both limits move here, so both terms of the rule survive
$b(1)=1$ and $a(1)=2$, so the two roots are $\sqrt{2}$ and $\sqrt{5}$
Answer $$P'(1)=2\sqrt2-2\sqrt5\approx-1.64$$
Check
Two independent checks. Structural: replace $\sqrt{1+t^{2}}$ by the constant $1$, so that $P(x)=x^{2}-2x$ exactly and $P'(1)=0$; the formula gives $1\cdot 2-1\cdot 2=0$ as well. Sign: at $x=1$ both ends move right at speed $2$, but the lower end sits where the integrand is $\sqrt5$ and the upper end where it is only $\sqrt2$, so more is being taken out than put in and the rate must be negative.
Test a rule on a case whose answer you already know — replacing a messy integrand by the constant $1$ is the cheapest way there is to catch a sign error.
3§09.5 — a region whose limits you must find yourself●●●●○
Exam level. No interval is given, the curves supply it, and the parabola dips below the axis inside the region — which changes nothing, and that is the point being tested.
Given
the curves $y=x^{2}-2x$ and $y=x$
Find
(a) Find the two crossings.
(b) Say which curve is on top between them, using a test point.
(c) Find the area enclosed.
Hint 1/4
With no interval given, the region is whatever the two curves trap between them, so the crossings are the limits.
Hint 2/4
$A=\int_{a}^{b}[\text{top}-\text{bottom}]\,dx$, where $a$ and $b$ are consecutive crossings and the top is settled by a test point.
Hint 3/4
Setting $x^{2}-2x=x$ gives $x^{2}-3x=0$, so $x=0$ and $x=3$; at $x=1$ the line gives $1$ and the parabola gives $-1$.
Hint 4/4
So $A=\int_{0}^{3}\bigl[x-(x^{2}-2x)\bigr]dx=\int_{0}^{3}(3x-x^{2})\,dx=\tfrac92$.
Show solution
We find the crossings first because the question gives no interval; integrating in $y$ instead would mean solving the parabola for $x$ and splitting the region, which buys nothing when the curves are already functions of $x$.
(a) Crossings
$x^{2}-2x=x \iff x^{2}-3x=0 \iff x(x-3)=0$
factor rather than use the quadratic formula; the constant term is missing, which makes factoring immediate
$x=0,\quad x=3$
these are the limits of integration
(b) Which curve is on top
$x=1:\quad x=1 \ \text{against}\ x^{2}-2x=-1$
one test value strictly inside; the line wins by $2$ there
Independent check with the parabolic-arch fact: the height function $3x-x^{2}$ has roots $0$ and $3$ and maximum $\tfrac94$ at $x=\tfrac32$, and two thirds of the bounding rectangle is $\tfrac23\cdot 3\cdot\tfrac94=\tfrac92$.
Nothing in the strip formula refers to the axis, so a region that straddles it needs no extra work.
4§09.6 — how far a solution reaches●●●●○
Exam level, and the part of the separable question that is most often left blank. The formula is handed to you; only its domain is in question.
Given
$\dfrac{dy}{dx}=xy^{2}$ with $y(0)=1$
the solution is $y=\dfrac{2}{2-x^{2}}$
Find
What is the largest open interval on which this solves the initial value problem?
Hint 1/4
Ask where the given formula stops being defined, and then which side of those places the initial point is on.
Hint 2/4
The solution of an initial value problem lives on the largest open interval containing $x_{0}$ on which the formula exists and satisfies the equation.
Hint 3/4
Here $y=\dfrac{2}{2-x^{2}}$ and $x_{0}=0$; the denominator vanishes when $x^{2}=2$.
Hint 4/4
The bad points are $x=\pm\sqrt2$, and $0$ lies between them, so the interval is $(-\sqrt2,\sqrt2)$.
Show solution
The interval is decided by the denominator of the given formula alone, so re-solving the differential equation is wasted work; only the points where the formula dies and the location of the initial point matter.
Find where the formula dies
$2-x^{2}=0 \iff x=\pm\sqrt2$
a zero denominator, so the function is undefined at exactly two points
Take the piece containing the initial point
$0\in(-\sqrt2,\sqrt2)$
the real line is cut into three pieces by the two bad points, and the solution is the one on the middle piece
$\text{interval}=(-\sqrt2,\sqrt2)$
the formula also defines functions on the two outer pieces, but those are different solutions with different initial data, not continuations of this one
Answer $$(-\sqrt2,\ \sqrt2)$$
Check
Independent check on the size of the answer: as $x$ approaches $\sqrt2$ from below, $y$ grows without bound, and the equation $y'=xy^{2}$ makes the growth faster the larger $y$ gets. A solution that blows up in finite $x$ is exactly what a $y^{2}$ on the right-hand side produces.
A solution formula solves the problem only on the interval that contains the initial point. Find where the formula dies before reporting anything.
D · interleaved 5 questions
1§09 — mixed practice 1●●●●○
Nothing announces which tool this wants. Read it, decide, then work. The deciding is the part worth practising.
(b) Find the absolute maximum of $F$ on $[0,3]$ and where it occurs.
(c) Find the absolute minimum on the same interval.
Hint 1/4
An absolute extremum on a closed interval is found by comparing the critical values with the two endpoint values, so the first job is to get a derivative.
Hint 2/4
Part 1 supplies $F'$; then the closed-interval method compares $F$ at every critical number with $F$ at $a$ and at $b$.
Hint 3/4
With $F(x)=\int_{0}^{x}(4-t^{2})\,dt$: $F'(x)=4-x^{2}$, zero at $x=2$ inside $[0,3]$, and $F(x)=4x-\tfrac{x^{3}}{3}$.
Hint 4/4
$F(0)=0$, $F(2)=\tfrac{16}{3}$, $F(3)=3$, so the maximum is $\tfrac{16}{3}$ at $x=2$ and the minimum is $0$ at $x=0$.
Show solution
We reach for Part 1 rather than for a formula for $F$ first, because the closed-interval method only needs $F'$ to find the candidates; the formula for $F$ is written later and only where values have to be compared.
Critical numbers
$F'(x)=4-x^{2}=0 \iff x=\pm 2$
Part 1; $F'$ exists everywhere, so zeros of $F'$ are the only candidates
$x=2\ \text{is in}\ [0,3];\ x=-2\ \text{is not}$
candidates outside the interval are discarded, not classified
Compare three values
$F(x)=4x-\tfrac{x^{3}}{3}$
a formula is needed here because values are being compared
Independent check without any formula for $F$: the integrand $4-t^{2}$ is positive on $[0,2)$ and negative on $(2,3]$, so $F$ climbs until $x=2$ and then falls. The largest accumulation must therefore be at $x=2$, and the smallest at the start, where nothing has been accumulated yet.
When the integrand changes sign, the accumulation function turns around at exactly that point.
2§09 — mixed practice 2●●●●○
This one looks like a limit question from the start of the term. It is, and it is also a one-line question if you notice what the quotient is.
Give the integral a name and look at the shape of the whole expression. It is a difference quotient in disguise.
Hint 2/4
Definition of the derivative: $G'(0)=\lim_{x\to0}\dfrac{G(x)-G(0)}{x-0}$; and Part 1 says $G'(x)$ is the integrand at $x$.
Hint 3/4
Let $G(x)=\int_{0}^{x}\sqrt{1+t^{3}}\,dt$. Then $G(0)=0$, so the expression is exactly $\dfrac{G(x)-G(0)}{x-0}$, and $G'(x)=\sqrt{1+x^{3}}$.
Hint 4/4
The limit is $G'(0)=\sqrt{1+0}=1$.
Show solution
We read the expression as a difference quotient instead of expanding the integrand, because $\sqrt{1+t^{3}}$ has no convenient antiderivative and any attempt to compute the integral first stalls immediately.
this is the difference quotient of $G$ at $0$, not merely something that looks like one
Take the limit
$\lim_{x\to0}\frac{G(x)-G(0)}{x-0}=G'(0)$
the definition of the derivative, run in the direction it is usually not read
$G'(x)=\sqrt{1+x^{3}} \Rightarrow G'(0)=1$
Part 1 supplies $G'$, and substituting $0$ finishes it
Answer $$1$$
Check
Independent check by estimating directly: for small $x$ the integrand is close to $\sqrt{1+0}=1$ on the whole tiny interval, so the integral is close to $1\cdot x$ and the quotient is close to $1$. The two arguments use different facts and agree.
A $0/0$ limit with an integral on top is almost always a derivative in disguise. Look for it before reaching for anything heavier.
3§09 — mixed practice 3●●●●○
A moving boundary and a question about speed. Decide what is a function of what before writing any derivative.
Given
$A(x)=\displaystyle\int_{0}^{x}\sqrt{s}\,ds$ is the area under $y=\sqrt{s}$ from $0$ to $x$
the right-hand edge $x$ moves right at $3$ units per second
Find
How fast is the area growing at the instant when $x=4$?
Hint 1/4
Two rates are involved and only one of them is given. Write down which quantity depends on which before differentiating anything.
Hint 2/4
Chain rule for related rates: $\dfrac{dA}{dt}=\dfrac{dA}{dx}\cdot\dfrac{dx}{dt}$, and Part 1 supplies $\dfrac{dA}{dx}$.
Hint 3/4
Here $\dfrac{dA}{dx}=\sqrt{x}$ and $\dfrac{dx}{dt}=3$; at the instant in question $x=4$, so $\sqrt{x}=2$.
Hint 4/4
So $\dfrac{dA}{dt}=2\cdot 3=6$ square units per second.
Show solution
We differentiate the area with respect to the edge and chain it to time, rather than writing $A$ as a function of $t$ first: Part 1 hands over $dA/dx$ with no work, while composing the two functions by hand adds an algebra step that can only introduce errors.
Get dA/dx for free
$\frac{dA}{dx}=\sqrt{x}$
Part 1, with $s$ as the dummy variable so that nothing collides with the time variable $t$
the area depends on the edge and the edge depends on time, which is exactly the chain rule's shape
$x=4 \Rightarrow \frac{dA}{dt}=2\cdot 3=6$
substitute the instant only at the end, after the general rate is written
Answer $$\frac{dA}{dt}=6\ \text{square units per second}$$
Check
Independent check by units and by size: the strip being added each second is about $3$ wide and about $\sqrt4=2$ tall, giving about $6$ square units, and width times height does carry the units of area per second.
The height of the curve at the moving edge is the rate at which area is being made. That sentence is Part 1 in words.
4§09 — mixed practice 4●●●●○
No tool is named, and the interval is not given either. The curve has to tell you where the region starts and stops before any integral can be written.
Given
the curve $y=x\sqrt{4-x^{2}}$
the $x$-axis
the part of the picture with $x\ge 0$
Find
(a) Find the two values of $x\ge 0$ at which the curve meets the $x$-axis, and say why the curve exists only for $\lvert x\rvert\le 2$.
(b) Find the exact area of the region enclosed by the curve and the $x$-axis between those two values.
Hint 1/4
Nothing here says which of the tools is wanted. Notice first that no limits of integration are given: the region is described by curves, so the crossings have to be found before anything is integrated.
Hint 2/4
Area between an upper and a lower boundary is $\int_{a}^{b}\bigl[\text{top}-\text{bottom}\bigr]dx$, and here the bottom is the $x$-axis. For an integrand of the form (factor)$\cdot\sqrt{\text{inner}}$, check whether the factor is the derivative of the inner function up to a constant.
Hint 3/4
For $y=x\sqrt{4-x^{2}}$: the root needs $4-x^{2}\ge 0$, and $y=0$ when $x=0$ or $4-x^{2}=0$. The area is $\int_{0}^{2}x\sqrt{4-x^{2}}\,dx$; with $u=4-x^{2}$ we get $du=-2x\,dx$, so $x\,dx=-\tfrac12\,du$, and the limits become $u=4$ and $u=0$.
Hint 4/4
The crossings are $x=0$ and $x=2$, and the area is $\tfrac83$.
Show solution
We look for a substitution rather than an antiderivative of $x\sqrt{4-x^{2}}$ by inspection, because the factor $x$ outside is the derivative of $4-x^{2}$ up to the constant $-2$; expanding or squaring is not available here, and guessing costs more than the two lines the substitution takes.
Where the curve lives and where it meets the axis
$4-x^{2}\ge 0 \iff -2\le x\le 2$
a square root is defined only where its inside is not negative, so this is the widest the picture can be
flipping the limits absorbs the minus sign, which is cheaper than carrying it through the evaluation
Answer $$A=\frac{8}{3}\approx 2.67$$
Check
Independent check by size: the curve is inside the rectangle $[0,2]\times[0,2]$, because $x\le 2$ and $\sqrt{4-x^{2}}\le 2$, so the area cannot exceed $4$; and the tallest point is $x=\sqrt2$, where $y=2$, so the region is a fat hump rather than a sliver. A value of $\tfrac83\approx 2.67$, two thirds of the rectangle, is the right order. A second, algebraic check: $-\tfrac13(4-x^{2})^{3/2}$ differentiates to $x\sqrt{4-x^{2}}$, and evaluating it from $0$ to $2$ gives $0-(-\tfrac83)=\tfrac83$.
When a question gives curves but no limits, the limits are part of the answer: find them first, decide the top boundary second, and only then look for the integration technique.
5§09 — mixed practice 5●●●●●
A rate is described in words, an initial value is given, and nothing says which of the tools this is. Read it, decide, then work.
Given
$\dfrac{dy}{dx}=\dfrac{x\sqrt{1+x^{2}}}{y}$
$y(0)=2$
Find
(a) Find $y$ explicitly as a function of $x$.
(b) Give the largest open interval containing $x=0$ on which your formula solves the problem.
(c) Check your formula by differentiating it, without solving the equation again.
Hint 1/4
There is an unknown function and its derivative in the same equation, so nothing can be evaluated directly. Ask first whether the right-hand side splits into a part in $x$ times a part in $y$.
Hint 2/4
If $\dfrac{dy}{dx}=g(x)h(y)$ then $\displaystyle\int\frac{dy}{h(y)}=\int g(x)\,dx$, and the integral in $x$ may itself need a substitution before it can be done.
Hint 3/4
Here $g(x)=x\sqrt{1+x^{2}}$ and $h(y)=\tfrac1y$, with $y(0)=2$. Separating gives $y\,dy=x\sqrt{1+x^{2}}\,dx$, and the right side asks for $u=1+x^{2}$, $x\,dx=\tfrac12\,du$.
Hint 4/4
Integrating gives $\tfrac{y^{2}}{2}=\tfrac13(1+x^{2})^{3/2}+C$ with $C=\tfrac53$, so $y=\sqrt{\tfrac23(1+x^{2})^{3/2}+\tfrac{10}{3}}$ on all of $(-\infty,\infty)$.
Show solution
We separate before doing anything to the right-hand side, because the substitution $u=1+x^{2}$ is only worth setting up once, on the side where it belongs; rearranging into an explicit $y$ first is impossible here, since $y$ is still unknown.
Independent check by implicit differentiation, which never uses the solving we just did: from $\tfrac{y^{2}}{2}=\tfrac13(1+x^{2})^{3/2}+\tfrac53$, differentiating both sides in $x$ gives $y\,y'=\tfrac13\cdot\tfrac32(1+x^{2})^{1/2}\cdot 2x=x\sqrt{1+x^{2}}$, so $y'=x\sqrt{1+x^{2}}/y$, the original equation. And $y(0)=\sqrt{\tfrac23+\tfrac{10}{3}}=\sqrt4=2$.
A separable equation can hide a second technique on the $x$-side. Separate first and find out what that side needs afterwards; deciding both at once is where the algebra goes wrong.
Mistake ledger (15 entries)
⚠ Differentiating the integrand instead of evaluating it
The word derivative is in the question, the integrand is the only visible function, and the reflex fires before the theorem is read.
⚠ Using the same letter for the dummy variable and the limit
Both slots want a variable and $x$ is the one on the brain. The result is an expression in which $x$ has two incompatible jobs and no meaning.
wrong$F(x)=\int_{a}^{x} f(x)\,dx$
right$F(x)=\int_{a}^{x} f(t)\,dt$
⚠ Dropping the derivative of the moving limit
Part 1 is remembered as "the answer is the integrand", and with a bare $x$ on top that is true. The extra factor only appears when the top is something else, so it is exactly the case that is never rehearsed.
⚠ Applying Part 2 across a point where the integrand is not defined
The antiderivative formula keeps working symbolically on both sides of the bad point, so nothing on the page looks broken until the answer has the wrong sign.
⚠ Deciding which curve is on top by looking at the formulas
A cubic "grows faster" than a line, which is true far out and false on $[-1,1]$. The general fact gets applied to a particular interval where it does not hold.
Both sides were antidifferentiated, and each antiderivative genuinely has a family constant. Keeping both makes the initial condition look like one equation in two unknowns.
The algebra ends with a formula, the formula looks like an answer, and nothing on the page reminds you that it stops being a solution at the first blow-up.
⚠ Using the symmetry shortcut on an interval that is not symmetric
The odd-or-even test is about the integrand, and once it comes out clean it feels like the answer. The condition that actually carries the shortcut sits on the integral sign: the two limits must be $-a$ and $a$.
$h(y)\neq0$ on the piece being found; roots of $h$ give constant solutions separately
Check yourself
Close the page and write, from memory: both halves of the theorem with their hypotheses, the moving-limit formula including its minus sign, the two things a definite substitution has to convert, and the one condition the area formula needs before it may be used on an interval.
Differentiate $\int_{1}^{x}\sqrt{2+t^{6}}\,dt$ and say what would go wrong if the integrand had a jump inside the interval.
c-ftc1
Write $\dfrac{d}{dx}\int_{x^{2}}^{x^{3}}f(t)\,dt$ from memory, with both chain factors and the correct sign.
c-moving-limits
State the hypothesis of Part 2 and give a specific integral where it fails and the formula produces a wrong number.
c-ftc2
Say in one sentence how to tell, before starting, whether a substitution will finish an integral or merely relabel it.
c-substitution
Explain why $\int_{a}^{b}(f-g)\,dx$ can be zero while the region between the curves has positive area.
c-area-between
Solve $y'=2xy^{2}$ with $y(0)=1$ and give the interval, without looking anything up.
c-separable
Glossary (17 terms)
Fundamental Theorem of Calculus, Part 1Analizin Temel Teoremi, birinci kısım
On an interval where $f$ is continuous, the accumulation function $F(x)=\int_{a}^{x} f(t)\,dt$ is differentiable and $F'=f$.
Fundamental Theorem of Calculus, Part 2Analizin Temel Teoremi, ikinci kısım
If $f$ is continuous on $[a,b]$ and $F'=f$ there, then $\int_{a}^{b} f=F(b)-F(a)$, for any such $F$.
accumulation functionbiriken alan fonksiyonu
The function $F(x)=\int_{a}^{x} f(t)\,dt$, which collects the signed area under $f$ from the fixed left end $a$ up to $x$.
antiderivativeters türev
A function $F$ with $F'=f$ on the interval under consideration; two antiderivatives of the same $f$ differ by a constant.
indefinite integralbelirsiz integral
The whole family of antiderivatives of $f$, written $\int f(x)\,dx=F(x)+C$; a family of functions rather than a number.
definite integralbelirli integral
The number $\int_{a}^{b} f(x)\,dx$ defined as a limit of Riemann sums and computable by the evaluation theorem when $f$ is continuous.
dummy variablesözde değişken
The letter used for the variable of integration inside an integral; it has no meaning outside and may be renamed freely.
substitution ruledeğişken değiştirme kuralı
The chain rule read backwards: $\int f(g(x))g'(x)\,dx=\int f(u)\,du$ with $u=g(x)$, and both limits converted in the definite case.
net changenet değişim
The quantity $F(b)-F(a)$, read as the total change in $F$ produced by accumulating its rate $F'$ from $a$ to $b$.
displacementyer değiştirme
The signed integral of velocity over a time interval; it records where the object ended up relative to where it started.
area between curveseğriler arası alan
The integral of top minus bottom over each piece where one curve stays above the other, with the pieces added.
vertical stripdikey şerit
A thin column of width $dx$ used when a region is integrated with respect to $x$; its height is the top curve minus the bottom curve.
horizontal stripyatay şerit
A thin row of height $dy$ used when a region is integrated with respect to $y$; its width is the right curve minus the left curve.
separable equationayrılabilir denklem
A first-order equation that can be written $dy/dx=g(x)h(y)$, so that each variable can be collected on its own side and integrated.
initial conditionbaşlangıç koşulu
One data point $y(x_{0})=y_{0}$ that the solution must pass through; it fixes the single constant of integration.
equilibrium solutiondenge çözümü
A constant solution $y\equiv y_{0}$ arising at a root of $h$; it is exactly what dividing by $h(y)$ removes from view.
interval of existencetanımlılık aralığı
The largest open interval containing the initial point on which the solution formula is defined and satisfies the equation.
What comes next
§10 · Volumes by disks and washers
Here the strip between two curves was a length and adding the lengths gave an area. Next the same strip is spun around a line, so each one becomes a disc with a thickness, and adding those gives a volume. The setting-up is identical — sketch, find the crossings, decide the direction of the strips — and only the thing being added changes.
Sources
James Stewart, Calculus, Ninth Edition — sections 4.4, 4.5, 5.1 and 9.3 The theorems are stated as this book states them; the worked functions here are different ones.
Course syllabus, week 9: Integrals 4.4, 4.5, 9.3, 5.1 The assessment weights quoted on the card come from the same syllabus and from nowhere else.
The parabolic-arch fact used as an independent check on three area answers The area under a parabolic arch is two thirds of the rectangle that encloses it; used here only to verify answers reached by integration.