08Antiderivatives, area, and the definite integral

Try to break the claim rather than to prove it: can you draw a graph that is not flat and still returns zero?

The integral counts area above the axis positively and area below it negatively, so equal amounts of each cancel.

Take $f(x)=x-2$ on $[0,4]$: a triangle of area $2$ below the axis on $[0,2]$ and a triangle of area $2$ above it on $[2,4]$.

One counterexample settles it: the claim is false.

Do not test the four statements on $x^{2}$. Every one of them is true there. Look for the one whose truth does not depend on the shape of $f$ at all.

$L_n=\sum_{i=0}^{n-1}f(x_i)\Delta x$ and $R_n=\sum_{i=1}^{n}f(x_i)\Delta x$: the two lists of heights overlap in everything except their first and last entries.

Subtracting, the terms $f(x_1),\dots,f(x_{n-1})$ appear in both and cancel, leaving $f(x_n)\Delta x-f(x_0)\Delta x$ with $x_0=a$ and $x_n=b$.

$R_n-L_n=(f(b)-f(a))\Delta x$ holds for every $f$ whatsoever.

Ask what the proof of the theorem actually used, and whether it can get from a negative $x$ to a positive one.

A function with zero derivative is constant on an interval. The Mean Value Theorem needs the two points joined by a piece of the domain.

Take $f(x)=1/x^{2}$ and $F(x)=-1/x$. Let $G(x)=-1/x$ for $x<0$ and $G(x)=-1/x+1$ for $x>0$. Then $G'=f$ everywhere on the set, and $F-G$ is $0$ on one piece and $-1$ on the other.

The claim is false: the difference is constant on each piece, but the two constants need not agree.

None of these needs a method. Each term is a line of the differentiation table read from right to left, and the only decision is how to rewrite a term so the table recognises it.

Reversed power rule: $x^{n}\mapsto\frac{x^{n+1}}{n+1}$ for $n\ne-1$; and $\frac1x\mapsto\ln\lvert x\rvert$, $e^{x}\mapsto e^{x}$, $\sin x\mapsto-\cos x$, $\sec x\tan x\mapsto\sec x$, $\sec^{2}x\mapsto\tan x$.

Rewrite $\frac{4}{x^{3}}$ as $4x^{-3}$ first: the new exponent is $-2$ and you divide by it, so the sign flips. The three integrals again: $6x^{2}-4x^{-3}+5$; $e^{x}+\frac3x-2\sin x$; $\sec x\tan x+4\sec^{2}x$.

The three answers are $2x^{3}+\frac{2}{x^{2}}+5x+C$, $e^{x}+3\ln x+2\cos x+C$ and $\sec x+4\tan x+C$.

Two separate decisions live here: what the numbers are, and which of the two sums is the upper bound. Settle the second one from the shape of $f$ before computing anything.

$\Delta x=\frac{b-a}{n}$; $L_n$ uses $x_0,\dots,x_{n-1}$ and $R_n$ uses $x_1,\dots,x_n$; on a decreasing $f$ the left endpoint of each strip is its highest point.

Here $a=0$, $b=2$, $n=4$, so $\Delta x=0.5$ and the points are $0,\ 0.5,\ 1,\ 1.5,\ 2$, at which $f$ is $1,\ 0.8,\ 0.5,\ 0.307692,\ 0.2$.

$L_4=1.303846$, $R_4=0.903846$, so $0.903846\le\int_0^2 f\le 1.303846$, and $M_4=1.108787$.

The whole computation is decided by two objects written before any algebra: the width of a strip, and the $i$-th right endpoint on an interval of length $2$.

$\Delta x=\frac{b-a}{n}$, $x_i=a+i\Delta x$, $R_n=\sum_{i=1}^{n}f(x_i)\Delta x$. A constant term inside the sum contributes $n$ copies.

Here $a=0$, $b=2$: $\Delta x=\frac2n$, $x_i=\frac{2i}{n}$, so $f(x_i)=4-\frac{4i^{2}}{n^{2}}$. The identities again: $\sum 1=n$ and $\sum i^{2}=\frac{n(n+1)(2n+1)}{6}$.

$R_n=8-\frac43\left(2+\frac3n+\frac1{n^{2}}\right)$ and the limit is $\frac{16}{3}$.

Do not try to imagine what $f$, $g$ or $h$ might be. Each part names one property; decide which before writing anything.

Additivity $\int_0^2+\int_2^6=\int_0^6$; linearity for constant multiples and sums; $\int_b^a=-\int_a^b$; and $m\le h\le M$ gives $m(b-a)\le\int_a^b h\le M(b-a)$.

The four given facts again: $\int_0^6 f=10$, $\int_0^2 f=-3$, $\int_2^6 g=8$, and $1\le h\le 5$ on $[0,6]$ whose length is $6$.

(a) $13$; (b) $44$; (c) $-10$; (d) $6\le\int_0^6 h\le 30$.

Each part is two separate questions: a height, and then a place where the graph reaches that height. Do not stop after the first.

$f_{\text{avg}}=\frac{1}{b-a}\int_a^b f$, and the Mean Value Theorem for integrals guarantees a $c$ in $[a,b]$ with $f(c)=f_{\text{avg}}$ when $f$ is continuous.

For (i): $\int_0^2 3t^{2}\,dt=8$ and $b-a=2$. For (ii): the quarter disc of radius $5$ has area $\frac14\pi(5)^{2}=\frac{25\pi}{4}$ and $b-a=5$.

(a) $v_{\text{avg}}=4$ and $c=\frac{2}{\sqrt3}\approx1.155$. (b) $h_{\text{avg}}=\frac{5\pi}{4}\approx3.927$ and $c=5\sqrt{1-\frac{\pi^{2}}{16}}\approx3.095$.

There is no function to integrate. Decide first how many intervals five readings cut the ten minutes into, and which reading you would use on each interval to be certain of undershooting.

A rate summed over time is a total: $\sum r(t_i^{*})\Delta t$. On a rate that never falls, the reading at the start of each interval is the smallest it gets there, and the reading at the end the largest. The gap between the two sums is $\left(r(b)-r(a)\right)\Delta t$.

The readings again, at $t=0,2,4,6,8,10$: $12,15,19,22,24,25$ litres per minute, with $\Delta t=2$ minutes and five intervals. The tank starts with $40$ litres.

$184\le V\le 210$ litres entered, so the tank holds between $224$ and $250$ litres at $t=10$; the bracket is $26$ litres wide and needs $\Delta t\le\frac{5}{13}$ minutes, about $23$ seconds, to fall to $5$.

The property turns a floor and a ceiling on the function into two rectangles. Ask what the width of those rectangles is here.

If $m\le f\le M$ on $[a,b]$ then $m(b-a)\le\int_a^b f\le M(b-a)$.

Here $m=\frac14$, $M=1$ and $b-a=4-1=3$, so the two rectangles have areas $\frac14\cdot3$ and $1\cdot3$.

$\tfrac34\le I\le 3$.

Draw the graph before doing anything else, and mark the one place it crosses the axis. Then treat parts (a) and (b) as two different questions rather than one.

Additivity splits the interval; a rectangle has area base times height and a triangle half of that; $f_{\text{avg}}=\frac{1}{b-a}\int_a^b f$; and area below the axis enters the integral negatively but the total area positively.

The pieces again: $p=2$ on $[0,2]$; $p=6-2x$ on $[2,4]$, crossing zero at $x=3$; $p=-2$ on $[4,6]$. That gives a rectangle of area $4$, two triangles of area $1$ each, and a rectangle of area $4$.

(a) $0$; (b) $10$; (c) $0$; (d) $c=3$ only; (e) $5$.

Part (a) and part (b) are different kinds of question. One asks for a largest value of something that varies; the other asks for the size of a fixed region.

For (a): write the area as a function of one variable and set its derivative to zero. For (b): the region is symmetric about the $y$-axis, and an integral over a symmetric interval splits into two equal halves.

The rectangle has width $2x$ and height $4-x^{2}$, so $A(x)=2x(4-x^{2})=8x-2x^{3}$ on $0

$A'(x)=8-6x^{2}=0$ at $x=\frac{2}{\sqrt3}$, giving largest area $\frac{32\sqrt3}{9}\approx6.158$; the region has area $\frac{32}{3}\approx10.667$; the ratio is $\frac{\sqrt3}{3}\approx0.577$.

Part (b) is not a new problem. Look at what part (a) produced and ask which question that line already answers.

The chain rule gives $\frac{d}{dx}\left[g(x)\right]^{3}=3\left[g(x)\right]^{2}g'(x)$; and an antiderivative statement is a derivative statement read from right to left. Constant multiples come out of an integral.

With $F(x)=(x^{2}+1)^{3}$: $F'(x)=3(x^{2}+1)^{2}(2x)=6x(x^{2}+1)^{2}$, which is exactly the integrand of (b). For (c), the integrand is one sixth of that.

(a) $6x(x^{2}+1)^{2}$; (b) $(x^{2}+1)^{3}+C$; (c) $\tfrac16(x^{2}+1)^{3}+C$.

Neither part asks you to find anything. Both ask whether something has to exist, so the answer is a theorem and its hypotheses, not a computation.

The Intermediate Value Theorem: a continuous function on $[a,b]$ takes every value between $f(a)$ and $f(b)$. The Mean Value Theorem for integrals: a continuous $f$ takes the value $f_{\text{avg}}=\frac{1}{b-a}\int_a^b f$ somewhere in $[a,b]$.

Here $f(1)=-3$ and $f(5)=7$, so $0$ lies between them; and $f_{\text{avg}}=\frac{1}{5-1}(8)=2$, which is the number part (b) asks about.

(a) Yes, by the Intermediate Value Theorem. (b) Yes, by the Mean Value Theorem for integrals, because $2$ is exactly the average value. (c) No — continuity is what that theorem needs.