Exercise 0.1: Integrating with the table: twelve direct antiderivatives

For each function, find an antiderivative $F$ by reading the table of derivatives from right to left, without substitution, then evaluate the antiderivative taken with $C=0$ at the given point.

One method for all twelve lines: first rewrite the integrand as a SUM of terms the table handles (powers of $x$ , expanding a product, dividing a quotient term by term, a trigonometric identity), then integrate term by term, and finally check by differentiating. The calculator, in radian mode, is only used for the final value.

a) $f(x)=5x^{4}-6x^{2}+4x-3$ , then $F(2)$ .
b) $f(x)=\frac{8}{x^{3}}+\frac{3}{x^{2}}$ , then $F(2)$ .
c) $f(x)=6\sqrt{x}-\frac{2}{\sqrt[3]{x}}$ , then $F(4)$ .
d) $f(x)=3x^{2}-\frac{5}{x}$ , then $F(-2)$ .
e) $f(x)=4e^{x}-7$ , then $F(1)$ .
f) $f(x)=3\cos x-2\sin x$ , then $F\left(\frac{\pi}{3}\right)$ .
g) $f(x)=2\sec^{2}x-\sec x\tan x$ , then $F\left(\frac{\pi}{4}\right)$ .
h) $f(x)=\frac{8}{1+x^{2}}-\frac{5}{\sqrt{1-x^{2}}}$ , then $F\left(\frac{1}{2}\right)$ .
i) $f(x)=x^{2}(x-3)^{2}$ , then $F(2)$ .
j) $f(x)=\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)$ , then $F(4)$ .
k) $f(x)=\frac{2x^{3}+x-6}{x^{2}}$ , then $F(2)$ .
l) $f(x)=\cot^{2}x$ , then $F\left(\frac{\pi}{4}\right)$ .

Type your answers, the page tells you right or wrong 0/12

a)

F(2)

b)

F(2)

c)

F(4)

d)

F(-2)

e)

F(1)

f)

F\left(\frac{\pi}{3}\right)

g)

F\left(\frac{\pi}{4}\right)

h)

F\left(\frac{1}{2}\right)

i)

F(2)

j)

F(4)

k)

F(2)

l)

F\left(\frac{\pi}{4}\right)

Show the solution

Answers

a) $F(x)=x^{5}-2x^{3}+2x^{2}-3x$ ; $F(2)=18$
b) $F(x)=-\frac{4}{x^{2}}-\frac{3}{x}$ ; $F(2)=-\frac{5}{2}$
c) $F(x)=4x^{\frac{3}{2}}-3x^{\frac{2}{3}}$ ; $F(4)\approx 24.440$
d) $F(x)=x^{3}-5\ln\left|x\right|$ ; $F(-2)\approx-11.466$
e) $F(x)=4e^{x}-7x$ ; $F(1)=4e-7\approx 3.873$
f) $F(x)=3\sin x+2\cos x$ ; $F\left(\frac{\pi}{3}\right)=\frac{3\sqrt{3}}{2}+1\approx 3.598$
g) $F(x)=2\tan x-\sec x$ ; $F\left(\frac{\pi}{4}\right)=2-\sqrt{2}\approx 0.586$
h) $F(x)=8\arctan x-5\arcsin x$ ; $F\left(\frac{1}{2}\right)\approx 1.091$
i) $F(x)=\frac{x^{5}}{5}-\frac{3x^{4}}{2}+3x^{3}$ ; $F(2)=\frac{32}{5}$
j) $F(x)=\frac{x^{2}}{2}+\frac{2}{3}x^{\frac{3}{2}}-2x$ ; $F(4)=\frac{16}{3}\approx 5.333$
k) $F(x)=x^{2}+\ln\left|x\right|+\frac{6}{x}$ ; $F(2)=7+\ln 2\approx 7.693$
l) $F(x)=-\cot x-x$ ; $F\left(\frac{\pi}{4}\right)=-1-\frac{\pi}{4}\approx-1.785$

The move that decides everything happens BEFORE integrating: write the integrand as a sum of lines from the table. There is no product rule and no quotient rule for antiderivatives, so a product is expanded, a quotient is divided, a root becomes a power, and $\cot^{2}$ goes through an identity.

a)

Term by term, the exponent goes up by one and you divide by the new exponent: $F(x)=x^{5}-2x^{3}+2x^{2}-3x$ . So $F(2)=32-16+8-6=18$ .

b)

$f(x)=8x^{-3}+3x^{-2}$ , so $F(x)=8\cdot\frac{x^{-2}}{-2}+3\cdot\frac{x^{-1}}{-1}=-\frac{4}{x^{2}}-\frac{3}{x}$ . So $F(2)=-1-\frac{3}{2}=-\frac{5}{2}$ . Pitfall: the exponent $-3$ goes UP to $-2$ , it does not go down to $-4$ , and you divide by $-2$ , hence the minus signs.

c)

$f(x)=6x^{\frac{1}{2}}-2x^{-\frac{1}{3}}$ , so $F(x)=6\cdot\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-2\cdot\frac{x^{\frac{2}{3}}}{\frac{2}{3}}=4x^{\frac{3}{2}}-3x^{\frac{2}{3}}$ . So $F(4)=4\times 8-3\sqrt[3]{16}\approx 32-7.560=24.440$ . Pitfall: dividing by a fractional exponent means multiplying by its reciprocal.

d)

The logarithm line handles the term in $\frac{1}{x}$ , the only power the power rule refuses: $F(x)=x^{3}-5\ln\left|x\right|$ . So $F(-2)=-8-5\ln 2\approx-11.466$ . Pitfall: without the absolute value, $\ln(-2)$ does not exist.

e)

$e^{x}$ is its own antiderivative: $F(x)=4e^{x}-7x$ . So $F(1)=4e-7\approx 3.873$ . Pitfall: treating $e^{x}$ as a power of $x$ and writing $\frac{e^{x+1}}{x+1}$ .

f)

$\int\cos x\,dx=\sin x$ and $\int\sin x\,dx=-\cos x$ : $F(x)=3\sin x+2\cos x$ . So $F\left(\frac{\pi}{3}\right)=\frac{3\sqrt{3}}{2}+1\approx 3.598$ . Pitfall: the sign, $-2\sin x$ has antiderivative $+2\cos x$ .

g)

$\sec^{2}x$ is the derivative of $\tan x$ , and $\sec x\tan x$ that of $\sec x$ : $F(x)=2\tan x-\sec x$ . So $F\left(\frac{\pi}{4}\right)=2-\sqrt{2}\approx 0.586$ .

h)

The two inverse trigonometric lines: $F(x)=8\arctan x-5\arcsin x$ , on $\left(-1, 1\right)$ . So $F\left(\frac{1}{2}\right)=8\arctan\frac{1}{2}-\frac{5\pi}{6}\approx 3.709-2.618=1.091$ .

i)

Expand: $x^{2}(x-3)^{2}=x^{2}\left(x^{2}-6x+9\right)=x^{4}-6x^{3}+9x^{2}$ , so $F(x)=\frac{x^{5}}{5}-\frac{3x^{4}}{2}+3x^{3}$ . So $F(2)=\frac{32}{5}-24+24=\frac{32}{5}=6.4$ . Pitfall: integrating each factor separately and then multiplying, which makes no sense.

j)

Expand: $\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)=x-\sqrt{x}+2\sqrt{x}-2=x+\sqrt{x}-2$ , so $F(x)=\frac{x^{2}}{2}+\frac{2}{3}x^{\frac{3}{2}}-2x$ . So $F(4)=8+\frac{16}{3}-8=\frac{16}{3}\approx 5.333$ .

k)

Divide term by term: $\frac{2x^{3}+x-6}{x^{2}}=2x+\frac{1}{x}-\frac{6}{x^{2}}$ , so $F(x)=x^{2}+\ln\left|x\right|+\frac{6}{x}$ . So $F(2)=4+\ln 2+3\approx 7.693$ . Pitfall: $\int-6x^{-2}dx=-6\cdot\frac{x^{-1}}{-1}=+\frac{6}{x}$ , the two minus signs cancel.

l)

No line handles $\cot^{2}$ , but $1+\cot^{2}x=\csc^{2}x$ gives $\cot^{2}x=\csc^{2}x-1$ , and $\csc^{2}x$ is the negative of the derivative of $\cot x$ : $F(x)=-\cot x-x$ . So $F\left(\frac{\pi}{4}\right)=-1-\frac{\pi}{4}\approx-1.785$ . Pitfall: forgetting the minus sign in $(\cot x)'=-\csc^{2}x$ .

Check: DIFFERENTIATE each $F$ you found and you must get exactly $f$ back; on the calculator, the rate $\frac{F(a+0.001)-F(a-0.001)}{0.002}$ must give $f(a)$ to within a thousandth. A discrepancy by a constant factor betrays a forgotten division, a discrepancy in sign a trigonometric line read the wrong way.

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Exercise 0.2: Integrating by substitution: ten direct antiderivatives

For each function, find an antiderivative $F$ by substitution, then evaluate the antiderivative taken with $C=0$ at the given point.

One method for all ten lines: spot the inner function $u$ whose derivative $u'$ is present as a factor, up to a constant; write $du=u'(x)\,dx$ and adjust the constant; integrate in $u$ with a line of the table; go back to $x$ before evaluating.

a) $f(x)=(4x-1)^{3}$ , then $F(1)$ .
b) $f(x)=x\left(x^{2}-1\right)^{4}$ , then $F(2)$ .
c) $f(x)=x^{2}\sqrt{x^{3}+1}$ , then $F(2)$ .
d) $f(x)=\frac{2x+1}{x^{2}+x+3}$ , then $F(1)$ .
e) $f(x)=x\,e^{-x^{2}}$ , then $F(1)$ .
f) $f(x)=\sin(3x+1)$ , then $F(0)$ .
g) $f(x)=\sin^{4}x\cos x$ , then $F\left(\frac{\pi}{2}\right)$ .
h) $f(x)=\tan^{2}(2x)\sec^{2}(2x)$ , then $F\left(\frac{\pi}{8}\right)$ .
i) $f(x)=\frac{x}{1+x^{4}}$ , then $F(1)$ .
j) $f(x)=\frac{1}{\sqrt{1-9x^{2}}}$ , then $F\left(\frac{1}{6}\right)$ .

Type your answers, the page tells you right or wrong 0/10

a)

F(1)

b)

F(2)

c)

F(2)

d)

F(1)

e)

F(1)

f)

F(0)

g)

F\left(\frac{\pi}{2}\right)

h)

F\left(\frac{\pi}{8}\right)

i)

F(1)

j)

F\left(\frac{1}{6}\right)

Show the solution

Answers

a) $F(x)=\frac{(4x-1)^{4}}{16}$ ; $F(1)=\frac{81}{16}$
b) $F(x)=\frac{\left(x^{2}-1\right)^{5}}{10}$ ; $F(2)=\frac{243}{10}$
c) $F(x)=\frac{2}{9}\left(x^{3}+1\right)^{\frac{3}{2}}$ ; $F(2)=6$
d) $F(x)=\ln\left(x^{2}+x+3\right)$ ; $F(1)=\ln 5\approx 1.609$
e) $F(x)=-\frac{1}{2}e^{-x^{2}}$ ; $F(1)=-\frac{1}{2e}\approx-0.184$
f) $F(x)=-\frac{1}{3}\cos(3x+1)$ ; $F(0)=-\frac{\cos 1}{3}\approx-0.180$
g) $F(x)=\frac{\sin^{5}x}{5}$ ; $F\left(\frac{\pi}{2}\right)=\frac{1}{5}$
h) $F(x)=\frac{\tan^{3}(2x)}{6}$ ; $F\left(\frac{\pi}{8}\right)=\frac{1}{6}\approx 0.167$
i) $F(x)=\frac{1}{2}\arctan\left(x^{2}\right)$ ; $F(1)=\frac{\pi}{8}\approx 0.393$
j) $F(x)=\frac{1}{3}\arcsin(3x)$ ; $F\left(\frac{1}{6}\right)=\frac{\pi}{18}\approx 0.175$

The move that decides everything: find $u$ AND its derivative in the integrand. A missing constant is adjusted by dividing; a missing $x$ can never be adjusted. And an answer left in terms of $u$ is not an answer: you go back to $x$ before evaluating.

a)

$u=4x-1$ , $du=4\,dx$ : $\int u^{3}\frac{du}{4}=\frac{u^{4}}{16}$ , so $F(x)=\frac{(4x-1)^{4}}{16}$ and $F(1)=\frac{81}{16}=5.0625$ . Pitfall: $\frac{(4x-1)^{4}}{4}$ , without the factor $\frac{1}{4}$ from the inside.

b)

$u=x^{2}-1$ , $du=2x\,dx$ , so $x\,dx=\frac{du}{2}$ : $\frac{1}{2}\int u^{4}du=\frac{u^{5}}{10}$ , so $F(x)=\frac{\left(x^{2}-1\right)^{5}}{10}$ and $F(2)=\frac{243}{10}=24.3$ .

c)

$u=x^{3}+1$ , $du=3x^{2}dx$ , so $x^{2}dx=\frac{du}{3}$ : $\frac{1}{3}\int u^{\frac{1}{2}}du=\frac{1}{3}\cdot\frac{2}{3}u^{\frac{3}{2}}$ , so $F(x)=\frac{2}{9}\left(x^{3}+1\right)^{\frac{3}{2}}$ and $F(2)=\frac{2}{9}\times 27=6$ .

d)

The numerator IS the derivative of the denominator: $u=x^{2}+x+3$ , $du=(2x+1)dx$ , $\int\frac{du}{u}=\ln\left|u\right|$ . Since $x^{2}+x+3>0$ (discriminant $1-12<0$ ), $F(x)=\ln\left(x^{2}+x+3\right)$ and $F(1)=\ln 5\approx 1.609$ .

e)

$u=-x^{2}$ , $du=-2x\,dx$ , so $x\,dx=-\frac{du}{2}$ : $-\frac{1}{2}\int e^{u}du$ , so $F(x)=-\frac{1}{2}e^{-x^{2}}$ and $F(1)=-\frac{1}{2e}\approx-0.184$ . Pitfall: losing the minus sign that comes from $u'=-2x$ .

f)

$u=3x+1$ , $du=3\,dx$ : $\frac{1}{3}\int\sin u\,du=-\frac{1}{3}\cos u$ , so $F(x)=-\frac{1}{3}\cos(3x+1)$ and $F(0)=-\frac{\cos 1}{3}\approx-0.180$ . Pitfall: the minus sign of $\int\sin u\,du$ AND the $\frac{1}{3}$ .

g)

$u=\sin x$ , $du=\cos x\,dx$ : $\int u^{4}du=\frac{u^{5}}{5}$ , so $F(x)=\frac{\sin^{5}x}{5}$ and $F\left(\frac{\pi}{2}\right)=\frac{1}{5}$ .

h)

$u=\tan(2x)$ , $du=2\sec^{2}(2x)\,dx$ : $\frac{1}{2}\int u^{2}du=\frac{u^{3}}{6}$ , so $F(x)=\frac{\tan^{3}(2x)}{6}$ and, since $\tan\frac{\pi}{4}=1$ , $F\left(\frac{\pi}{8}\right)=\frac{1}{6}\approx 0.167$ . Pitfall: taking $u=\sec(2x)$ , whose derivative $2\sec(2x)\tan(2x)$ is not a factor.

i)

$x^{4}=\left(x^{2}\right)^{2}$ : $u=x^{2}$ , $du=2x\,dx$ , so $\frac{1}{2}\int\frac{du}{1+u^{2}}=\frac{1}{2}\arctan u$ . So $F(x)=\frac{1}{2}\arctan\left(x^{2}\right)$ and $F(1)=\frac{\pi}{8}\approx 0.393$ . Pitfall: looking for a logarithm, when $u'=2x$ is not the derivative of $1+x^{4}$ .

j)

$9x^{2}=(3x)^{2}$ : $u=3x$ , $du=3\,dx$ , so $\frac{1}{3}\int\frac{du}{\sqrt{1-u^{2}}}=\frac{1}{3}\arcsin u$ . So $F(x)=\frac{1}{3}\arcsin(3x)$ and $F\left(\frac{1}{6}\right)=\frac{1}{3}\cdot\frac{\pi}{6}=\frac{\pi}{18}\approx 0.175$ .

Check: differentiate each $F$ with the chain rule and you must get $f$ back exactly, the derivative of the inside cancelling the adjustment constant. If the derivative gives $f$ multiplied by a number, the adjustment is missing; if it gives an extra $x$ , the substitution was not allowed.

Exercise 1: The table of derivatives, read from right to left

You have spent the semester filling in a table: a function on the left, its derivative on the right. This chapter asks for nothing new, it asks you to read that same table FROM RIGHT TO LEFT. An antiderivative of $f$ is a function $F$ such that $F'=f$ , so every line of the table of derivatives gives you an antiderivative formula for free.

The power rule $\left(x^{n}\right)'=nx^{n-1}$ reverses into $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ , valid for every $n\neq -1$ . Before integrating anything, rewrite the integrand as POWERS of $x$ : a quotient $\frac{1}{x^{5}}$ and a root $\sqrt[3]{x^{2}}$ cannot be recognized as long as they keep their original form.

a) Compute $\int\left(6x^{2}-8x^{3}+5\right)dx$ , $\int\frac{dx}{x^{5}}$ and $\int\sqrt[3]{x^{2}}\,dx$ , first rewriting each integrand as a power of $x$ .
b) Compute $\int\left(\frac{4}{x}-\frac{4}{x^{2}}\right)dx$ . The two terms look alike: do they come from the same line of the table?
c) Check BY DIFFERENTIATION the third antiderivative in a) and the answer to b). Which differentiation rule do you use each time?
d) A student answers $\int\frac{dx}{x^{5}}=\ln\left|x^{5}\right|+C$ . Refute this in ten seconds, without computing the true antiderivative again.

Type your answers, the page tells you right or wrong 0/6

a)

\int\left(6x^{2}-8x^{3}+5\right)dx

Exponent obtained by integrating

x^{-5}

Coefficient of

x^{5/3}

b)

\int\left(\frac{4}{x}-\frac{4}{x^{2}}\right)dx

Antiderivative of

-\frac{4}{x^{2}}

at

x=2

(with

C=0

)

d)

Derivative of

\ln\left|x^{5}\right|

Show the solution

Answers

a) $2x^{3}-2x^{4}+5x+C$ ; $-\frac{1}{4x^{4}}+C$ ; $\frac{3}{5}x^{5/3}+C$
b) $4\ln\left|x\right|+\frac{4}{x}+C$ , two different lines of the table
c) Both derivatives give back the integrand: power rule, then logarithm
d) False: $\left(\ln\left|x^{5}\right|\right)'=\frac{5}{x}$ and not $\frac{1}{x^{5}}$

a)

First integral: linearity lets you work term by term, and the reversed power rule gives $\int 6x^{2}dx=6\cdot\frac{x^{3}}{3}=2x^{3}$ , $\int -8x^{3}dx=-8\cdot\frac{x^{4}}{4}=-2x^{4}$ and $\int 5\,dx=5x$ . Answer: $2x^{3}-2x^{4}+5x+C$ . Second integral: rewrite $\frac{1}{x^{5}}=x^{-5}$ , then $\int x^{-5}dx=\frac{x^{-4}}{-4}=-\frac{1}{4x^{4}}+C$ . The exponent goes from $-5$ up to $-4$ : it INCREASES by 1, even though it stays negative, and this is where one test in three writes $-6$ . Third integral: $\sqrt[3]{x^{2}}=x^{2/3}$ , so $\int x^{2/3}dx=\frac{x^{5/3}}{5/3}=\frac{3}{5}x^{5/3}+C$ . Dividing by $\frac{5}{3}$ means multiplying by $\frac{3}{5}$ : the fraction flips, it is not copied as is.

b)

The two terms certainly do NOT come from the same line. The first, $\frac{4}{x}=4x^{-1}$ , carries exactly the exponent the power rule forbids, since it would produce a division by $n+1=0$ . The logarithm line handles it: $\int\frac{dx}{x}=\ln\left|x\right|+C$ , so $\int\frac{4}{x}dx=4\ln\left|x\right|+C$ . The second, $-\frac{4}{x^{2}}=-4x^{-2}$ , carries an ordinary exponent: $\int -4x^{-2}dx=-4\cdot\frac{x^{-1}}{-1}=4x^{-1}=\frac{4}{x}$ . Answer: $4\ln\left|x\right|+\frac{4}{x}+C$ . The trap is visual: two quotients that look alike, two formulas with nothing in common. The absolute value in the logarithm is not a frill, it makes the formula valid for $x<0$ as well.

c)

For a): $\left(\frac{3}{5}x^{5/3}\right)'=\frac{3}{5}\cdot\frac{5}{3}x^{2/3}=x^{2/3}=\sqrt[3]{x^{2}}$ , and you get the integrand back: this is the power rule, the one from the first week. For b): $\left(4\ln\left|x\right|+\frac{4}{x}\right)'=\frac{4}{x}-\frac{4}{x^{2}}$ , using $\left(\ln\left|x\right|\right)'=\frac{1}{x}$ and $\left(4x^{-1}\right)'=-4x^{-2}$ ✓. Remember this move: in differential calculus you have no other means of checking, neither a fundamental theorem nor a numerical value to compare with. Differentiating your answer takes ten seconds and catches the great majority of exam errors.

d)

Differentiate the proposed answer: $\left(\ln\left|x^{5}\right|\right)'=\frac{5x^{4}}{x^{5}}=\frac{5}{x}$ , by the chain rule. But the integrand was $\frac{1}{x^{5}}$ . The two do not even look alike, so the answer is wrong, and you did not need to know the right one to say so. The student applied the logarithm line on seeing a quotient, when that line handles ONLY the exponent $-1$ . The confusion typically costs three marks out of ten on the exam, and it is always spotted by the same check.

Exercise 2: The constant C, what differentiation erases

Differentiating destroys information: all the functions $F(x)+C$ have the same derivative. Integrating must therefore give that information back, and nobody can guess which one: hence the $+C$ , which is not a decoration but the trace of a whole family of parallel curves.

The figure shows three members of the family $y=2x^{3/2}+C$ . They have the same slope at every $x$ , hence the same derivative, and only one of them will pass through a prescribed point.

a) Check that $F(x)=\tan^{2}x$ and $G(x)=\sec^{2}x$ are both antiderivatives of $f(x)=2\sec^{2}x\tan x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . What constant separates them?
b) State the theorem guaranteeing that two antiderivatives of the same function on an INTERVAL differ by a constant. Which theorem of the course does it follow from?
c) Find the antiderivative of $f(x)=4x^{3}-\frac{2}{x^{2}}$ that equals $7$ at $x=1$ .
d) Solve $y'=3\sqrt{x}$ with $y(4)=10$ . Which of the three curves in the figure is a solution? Check your answer in two ways.

Show the solution

Answers

a) Both are antiderivatives of $f$ ; $G-F=1$
b) On an INTERVAL, $F'=G'$ implies $F-G$ constant; a consequence of the Mean Value Theorem
c) $F(x)=x^{4}+\frac{2}{x}+4$
d) $y=2x^{3/2}-6$ ; none of the three curves drawn

a)

Differentiate both candidates. $F'(x)=2\tan x\cdot\sec^{2}x$ by the power rule and the chain rule, since $\left(\tan x\right)'=\sec^{2}x$ . $G'(x)=2\sec x\cdot\left(\sec x\tan x\right)=2\sec^{2}x\tan x$ , since $\left(\sec x\right)'=\sec x\tan x$ . Both derivatives equal $f$ , so both functions are indeed antiderivatives of $f$ ✓. Yet they do not look alike. The identity $\sec^{2}x-\tan^{2}x=1$ settles the matter: $G-F=1$ , the constant that separates them is exactly $1$ . Numerical check at $x=\frac{\pi}{4}$ : $F=1$ and $G=2$ ✓. Two answers that look very different can therefore both be right, and this is the most common source of disputes over a test.

b)

Theorem: if $F'=G'$ on an INTERVAL $I$ , then $F-G$ is constant on $I$ . It follows from the Mean Value Theorem, seen in the middle of the semester: set $H=F-G$ , so $H'=0$ everywhere on $I$ , hence for any two points $a<b$ of $I$ there exists $c$ between them such that $H(b)-H(a)=H'(c)(b-a)=0$ , so $H(b)=H(a)$ for every pair of points, that is, $H$ is constant. The word INTERVAL carries all the weight: on $\mathbb{R}\setminus\{0\}$ , which is a union of two intervals, $\ln\left|x\right|+1$ on the left and $\ln\left|x\right|+5$ on the right have the same derivative without differing by a single constant.

c)

Integrate term by term: $\int 4x^{3}dx=x^{4}$ and $\int -2x^{-2}dx=-2\cdot\frac{x^{-1}}{-1}=\frac{2}{x}$ . The family is $F(x)=x^{4}+\frac{2}{x}+C$ . The condition $F(1)=7$ gives $1+2+C=7$ , so $C=4$ and $F(x)=x^{4}+\frac{2}{x}+4$ . Check by differentiation: $F'(x)=4x^{3}-\frac{2}{x^{2}}$ ✓, and check at the point: $F(1)=1+2+4=7$ ✓. BOTH checks count: the first checks the integration, the second checks the constant, and a wrong constant leaves the first one perfectly green. As an extra check, $F(2)=16+1+4=21$ .

d)

$y'=3\sqrt{x}=3x^{1/2}$ gives $y=3\cdot\frac{x^{3/2}}{3/2}+C=2x^{3/2}+C$ . The condition $y(4)=10$ gives $2\cdot 8+C=10$ , so $C=-6$ and $y=2x^{3/2}-6$ . None of the three curves in the figure is a solution: they have $C=-12$ , $C=-3$ and $C=6$ , and the point $(4, 10)$ lies between the first two, exactly as the figure shows. First check, by differentiation: $\left(2x^{3/2}\right)'=2\cdot\frac{3}{2}x^{1/2}=3\sqrt{x}$ ✓. Second check, with the point: $2\cdot 4^{3/2}-6=16-6=10$ ✓. Extra check: $y(9)=2\cdot 27-6=48$ . An equation of the form $y'=f(x)$ with a condition is the simplest of differential equations, and the whole chapter does nothing but solve it.

Exercise 3: Rewrite first, integrate afterwards

There is no product rule and no quotient rule for integration. When the integrand is a product or a quotient, the only move available is to REWRITE it as a sum of terms the table handles directly: expand, divide, or replace it with a trigonometric identity.

The figure shows the integrand of d), the parabola $y=(2x-3)^{2}$ . This integral can be computed in two perfectly legitimate ways, and the two answers do not look the same.

a) Compute $\int\tan^{2}x\,dx$ . No line of the table handles $\tan^{2}$ : which identity solves the problem?
b) Compute $\int\frac{x^{2}}{x^{2}+1}dx$ . The degree of the numerator equals that of the denominator: what must be done before anything else?
c) Compute $\int\frac{\left(\sqrt{x}+1\right)^{2}}{x}dx$ for $x>0$ .
d) Compute $\int(2x-3)^{2}dx$ in two ways, by expanding and then by reversing the chain rule. Do the two answers contradict each other?

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Answers

a) $\tan x-x+C$ , by $\tan^{2}x=\sec^{2}x-1$
b) $x-\arctan x+C$ , after dividing
c) $x+4\sqrt{x}+\ln x+C$
d) $\frac{4}{3}x^{3}-6x^{2}+9x+C$ and $\frac{(2x-3)^{3}}{6}+C'$ : they differ by $-4.5$ , both are right

a)

The table does not contain $\tan^{2}$ , but it does contain $\sec^{2}$ , which is the derivative of $\tan$ . The identity $1+\tan^{2}x=\sec^{2}x$ gives $\tan^{2}x=\sec^{2}x-1$ , and the integral becomes a sum of two known lines: $\int\left(\sec^{2}x-1\right)dx=\tan x-x+C$ . Check by differentiation: $\left(\tan x-x\right)'=\sec^{2}x-1=\tan^{2}x$ ✓. Numerical check at $x=\frac{\pi}{4}$ : the antiderivative equals $1-\frac{\pi}{4}\approx 0.21$ . The trap in this question is wanting to integrate $\tan^{2}$ as a power, which would give $\frac{\tan^{3}x}{3}$ : this answer is refuted in one differentiation, because the chain rule would bring in a stray $\sec^{2}x$ .

b)

A quotient whose numerator has the same degree as the denominator is divided first. Here the division is immediate: $\frac{x^{2}}{x^{2}+1}=\frac{\left(x^{2}+1\right)-1}{x^{2}+1}=1-\frac{1}{x^{2}+1}$ . Both pieces are then lines of the table, the second being the arctangent line: $\int\frac{x^{2}}{x^{2}+1}dx=x-\arctan x+C$ . Check: $\left(x-\arctan x\right)'=1-\frac{1}{x^{2}+1}=\frac{x^{2}}{x^{2}+1}$ ✓. At $x=1$ the antiderivative equals $1-\frac{\pi}{4}\approx 0.21$ . What you must NOT write is a quotient of integrals: $\int\frac{u}{v}\neq\frac{\int u}{\int v}$ , and differentiating this false result shows it in one line.

c)

Expand the square before anything else: $\left(\sqrt{x}+1\right)^{2}=x+2\sqrt{x}+1$ , then divide each term by $x$ : $\frac{x+2\sqrt{x}+1}{x}=1+2x^{-1/2}+\frac{1}{x}$ . Each piece is now a line of the table: $\int 1\,dx=x$ , $\int 2x^{-1/2}dx=2\cdot\frac{x^{1/2}}{1/2}=4\sqrt{x}$ and $\int\frac{dx}{x}=\ln x$ , the logarithm needing no absolute value since $x>0$ . Answer: $x+4\sqrt{x}+\ln x+C$ . Check: $1+\frac{4}{2\sqrt{x}}+\frac{1}{x}=1+\frac{2}{\sqrt{x}}+\frac{1}{x}$ ✓, which is indeed the expansion obtained. At $x=1$ and with $C=0$ , the antiderivative equals $1+4+0=5$ .

d)

First way, expand: $(2x-3)^{2}=4x^{2}-12x+9$ , so the integral equals $\frac{4}{3}x^{3}-6x^{2}+9x+C$ . Second way, reverse the chain rule: the inside $2x-3$ has derivative $2$ , so $\int(2x-3)^{2}dx=\frac{(2x-3)^{3}}{3\cdot 2}+C'=\frac{(2x-3)^{3}}{6}+C'$ . The two answers do not contradict each other: expanding the second, $\frac{8x^{3}-36x^{2}+54x-27}{6}=\frac{4}{3}x^{3}-6x^{2}+9x-4.5$ . So they differ by the constant $-4.5$ , which is allowed and expected, since two antiderivatives of the same function on an interval differ only by a constant. A marker who sees an unexpected answer differentiates it: that is the only arbiter.

Exercise 4: Transcendental functions, backwards

The second half of the table of derivatives is transcendental: exponential, logarithm, sine and cosine, tangent and secant, arcsine and arctangent. Reversed, it gives as many antiderivative formulas, and this is where SIGN errors pile up.

Remember the asymmetry: $\left(\sin x\right)'=\cos x$ , so $\int\cos x\,dx=\sin x+C$ , but $\left(\cos x\right)'=-\sin x$ , so $\int\sin x\,dx=-\cos x+C$ . A minus sign appears on one side and not on the other.

a) Compute $\int\left(5e^{x}-\frac{3}{\sqrt{1-x^{2}}}\right)dx$ .
b) Compute $\int\left(4\sin x-\sec^{2}x\right)dx$ , then give the value of this antiderivative at $x=0$ with $C=0$ .
c) Compute $\int\left(\frac{6}{1+x^{2}}-\frac{2}{x}\right)dx$ , then check the answer by differentiation.
d) A student answers $\int 4\sin x\,dx=4\cos x+C$ . Differentiate this answer and say which line of the table the student read the wrong way.

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Answers

a) $5e^{x}-3\arcsin x+C$ on $\left(-1, 1\right)$
b) $-4\cos x-\tan x+C$ ; the antiderivative equals $-4$ at $x=0$
c) $6\arctan x-2\ln\left|x\right|+C$ ; it equals $\frac{3\pi}{2}\approx 4.71$ at $x=1$
d) False: $\left(4\cos x\right)'=-4\sin x$ , the right answer is $-4\cos x+C$

a)

Two lines of the table, taken separately. For the first, $\left(e^{x}\right)'=e^{x}$ is the only function that is its own derivative, so $\int 5e^{x}dx=5e^{x}$ . For the second, $\frac{1}{\sqrt{1-x^{2}}}$ is exactly the derivative of $\arcsin x$ , so $\int-\frac{3}{\sqrt{1-x^{2}}}dx=-3\arcsin x$ . Answer: $5e^{x}-3\arcsin x+C$ , on $\left(-1, 1\right)$ where the radical is defined. Check: $5e^{x}-\frac{3}{\sqrt{1-x^{2}}}$ ✓. The classic trap is to confuse $\frac{1}{\sqrt{1-x^{2}}}$ , which gives an arcsine, with $\frac{1}{1+x^{2}}$ , which gives an arctangent: the radical and the minus sign under the root tell them apart at a glance.

b)

For the first term, $\int\sin x\,dx=-\cos x$ , so $\int 4\sin x\,dx=-4\cos x$ . For the second, $\sec^{2}x$ is the derivative of $\tan x$ , so $\int-\sec^{2}x\,dx=-\tan x$ . Answer: $-4\cos x-\tan x+C$ . Check by differentiation: $4\sin x-\sec^{2}x$ ✓. At $x=0$ with $C=0$ : $-4\cdot 1-0=-4$ . This value is a useful check, because a sign error on the cosine would turn it into $+4$ : a single evaluation at a well-chosen point is often enough to separate a right answer from an answer of the opposite sign.

c)

The first term is the arctangent line: $\int\frac{6}{1+x^{2}}dx=6\arctan x$ . The second is the logarithm line: $\int-\frac{2}{x}dx=-2\ln\left|x\right|$ . Answer: $6\arctan x-2\ln\left|x\right|+C$ . Check by differentiation: $\left(6\arctan x\right)'=\frac{6}{1+x^{2}}$ and $\left(-2\ln\left|x\right|\right)'=-\frac{2}{x}$ , so the sum gives back the integrand ✓. Numerical check at $x=1$ : $6\cdot\frac{\pi}{4}-2\ln 1=\frac{3\pi}{2}\approx 4.71$ . The trap here is to integrate $\frac{6}{1+x^{2}}$ as a logarithm, on the grounds that there is a quotient: but $\ln\left(1+x^{2}\right)$ has derivative $\frac{2x}{1+x^{2}}$ , and the stray $2x$ gives the error away immediately.

d)

Differentiate the proposed answer: $\left(4\cos x\right)'=-4\sin x$ . You get the negative of the integrand, so the answer is wrong, and the right one is $-4\cos x+C$ . The student read the line $\left(\sin x\right)'=\cos x$ and reversed it mechanically, believing that sine gives cosine in both directions. The table is asymmetric: differentiating cosine produces a minus sign, so integrating sine produces one too. This error usually costs half the marks of the question, while the check that catches it takes one line and five seconds. When in doubt about a sign, evaluate at $x=0$ : $\left(-\cos x\right)$ equals $-1$ there and is increasing, which is consistent with an integrand $\sin x$ that is positive just after $0$ .

Exercise 5: The chain rule backwards: seeing u and its derivative

You have differentiated composite functions for six weeks: set the inside $u$ , differentiate the outside, MULTIPLY by $u'$ . Substitution is exactly this move played backwards. You look in the integrand for an inner function $u$ AND its derivative $u'$ as a factor; if both are there, the integral comes down to a line of the table.

The protocol fits in three lines: set $u=\ldots$ , write $du=u'(x)\,dx$ , replace everything, and not a single $x$ may remain; integrate in $u$ , then go BACK to $x$ . An answer left in terms of $u$ is not an answer.

a) Compute $\int 8x\left(x^{2}+3\right)^{3}dx$ , writing out $u$ , $du$ and the return to $x$ .
b) Compute $\int\frac{\cos x}{2+\sin x}dx$ . Why is the absolute value unnecessary in the logarithm?
c) Compute $\int\frac{e^{\sqrt{x}}}{\sqrt{x}}dx$ for $x>0$ .
d) Compute $\int\frac{\arctan x}{1+x^{2}}dx$ , then check by differentiation.

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Answers

a) $u=x^{2}+3$ ; $\left(x^{2}+3\right)^{4}+C$
b) $\ln\left(2+\sin x\right)+C$ ; no absolute value, the argument is at least $1$
c) $2e^{\sqrt{x}}+C$
d) $\frac{\left(\arctan x\right)^{2}}{2}+C$

a)

Set $u=x^{2}+3$ , so $du=2x\,dx$ . The factor $8x\,dx$ is then $4\,du$ , and the integral becomes $\int 4u^{3}du=u^{4}+C$ . Back to $x$ : $\left(x^{2}+3\right)^{4}+C$ . Check by differentiation: $4\left(x^{2}+3\right)^{3}\cdot 2x=8x\left(x^{2}+3\right)^{3}$ ✓, you get exactly the integrand back. Numerical check at $x=1$ with $C=0$ : $4^{4}=256$ . Note that you are not looking for $u'$ exactly, but for $u'$ UP TO A CONSTANT FACTOR: here $8x$ is $4$ times $2x$ , and taking a constant out of the integral is always allowed.

b)

Set $u=2+\sin x$ , so $du=\cos x\,dx$ , and the numerator IS exactly $du$ . The integral becomes $\int\frac{du}{u}=\ln\left|u\right|+C$ , that is $\ln\left(2+\sin x\right)+C$ . The absolute value is unnecessary because $\sin x\in\left[-1, 1\right]$ , so $2+\sin x\in\left[1, 3\right]$ : the argument of the logarithm is always positive, and even bounded below by $1$ . Check: $\left(\ln\left(2+\sin x\right)\right)'=\frac{\cos x}{2+\sin x}$ ✓. Numerical check at $x=0$ : $\ln 2\approx 0.69$ . Writing the absolute value would not be an error, only an unnecessary precaution that you must be able to justify.

c)

Set $u=\sqrt{x}$ , so $du=\frac{dx}{2\sqrt{x}}$ , that is $\frac{dx}{\sqrt{x}}=2\,du$ . The integral becomes $\int e^{u}\cdot 2\,du=2e^{u}+C$ , and going back to $x$ gives $2e^{\sqrt{x}}+C$ . Check by differentiation: $2e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}=\frac{e^{\sqrt{x}}}{\sqrt{x}}$ ✓. Numerical check at $x=1$ : $2e\approx 5.44$ . This is the most instructive case of the set: the derivative of the inside, $\frac{1}{2\sqrt{x}}$ , is hidden in a DENOMINATOR, and you only recognize it by remembering the derivative of the square root. An integrand that contains a function and a strange denominator is almost always a disguised substitution.

d)

Set $u=\arctan x$ , so $du=\frac{dx}{1+x^{2}}$ , and once again the rest of the integrand IS exactly $du$ . The integral becomes $\int u\,du=\frac{u^{2}}{2}+C$ , that is $\frac{\left(\arctan x\right)^{2}}{2}+C$ . Check by differentiation: $\frac{2\arctan x}{2}\cdot\frac{1}{1+x^{2}}=\frac{\arctan x}{1+x^{2}}$ ✓. Numerical check at $x=1$ : $\frac{\left(\pi/4\right)^{2}}{2}\approx 0.31$ . The reflex to build is this one: when $\frac{1}{1+x^{2}}$ is a factor, think of the arctangent not only as a result but as a CANDIDATE $u$ . The same remark holds for $\frac{1}{x}$ , which calls for $u=\ln x$ , and for $\cos x$ , which calls for $u=\sin x$ .

Exercise 6: When a constant factor is missing, and when an x is missing

A constant comes out of the integral, a function of $x$ never does. The whole difficulty of substitution lies on this line: if $u'$ is present up to a CONSTANT factor, you adjust and the integral falls into place; if an $x$ is missing, the integral changes nature completely and often leaves the course.

The figure shows $y=x\sqrt{4-x^{2}}$ , the integrand of a). Its neighbour $y=\sqrt{4-x^{2}}$ , without the factor $x$ , is a semicircle, and it is not handled the same way at all.

a) Compute $\int x\sqrt{4-x^{2}}\,dx$ , stating $u$ and $du$ , then give its value at $x=0$ with $C=0$ .
b) Compute $\int\sec^{2}(7x)\,dx$ . Where does the coefficient in front of the tangent come from?
c) A student writes $\int\sqrt{4-x^{2}}\,dx=-\frac{1}{3}\left(4-x^{2}\right)^{3/2}+C$ . Differentiate this answer: what exactly do you get, and what is missing from the starting integrand?
d) Compute $\int\frac{x}{\sqrt{x^{2}+9}}dx$ and check by differentiation.

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Answers

a) $-\frac{1}{3}\left(4-x^{2}\right)^{3/2}+C$ ; it equals $-\frac{8}{3}\approx -2.67$ at $x=0$
b) $\frac{1}{7}\tan(7x)+C$ , the $\frac{1}{7}$ offsets the chain rule
c) Its derivative is $x\sqrt{4-x^{2}}$ : an extra factor $x$ comes out, and the integral in c) is not part of 201-SN2-RE
d) $\sqrt{x^{2}+9}+C$ ; it equals $5$ at $x=4$

a)

Set $u=4-x^{2}$ , so $du=-2x\,dx$ , that is $x\,dx=-\frac{1}{2}du$ . The integral becomes $-\frac{1}{2}\int u^{1/2}du=-\frac{1}{2}\cdot\frac{2}{3}u^{3/2}=-\frac{1}{3}u^{3/2}$ , and going back to $x$ gives $-\frac{1}{3}\left(4-x^{2}\right)^{3/2}+C$ . Check by differentiation: $-\frac{1}{3}\cdot\frac{3}{2}\left(4-x^{2}\right)^{1/2}\cdot(-2x)=x\sqrt{4-x^{2}}$ ✓. Value at $x=0$ with $C=0$ : $-\frac{1}{3}\cdot 4^{3/2}=-\frac{8}{3}\approx -2.67$ . The factor $-\frac{1}{2}$ is a CONSTANT, so it passes through the integral without difficulty: this is the comfortable case, where $u'$ is present up to the coefficient.

b)

Set $u=7x$ , so $du=7\,dx$ and $dx=\frac{du}{7}$ . The integral becomes $\frac{1}{7}\int\sec^{2}u\,du=\frac{1}{7}\tan u+C$ , that is $\frac{1}{7}\tan(7x)+C$ . The coefficient $\frac{1}{7}$ comes from the derivative of the inside: when you differentiate $\tan(7x)$ , the chain rule brings out a $7$ , and the $\frac{1}{7}$ is there to cancel it. Check: $\frac{1}{7}\cdot 7\sec^{2}(7x)=\sec^{2}(7x)$ ✓. Numerical check: at $x=\frac{\pi}{28}$ , $\tan\frac{\pi}{4}=1$ and the antiderivative equals $\frac{1}{7}\approx 0.1429$ . The most widespread error is to forget this coefficient and answer $\tan(7x)+C$ , which gives an integrand seven times too large.

c)

Differentiate: $\left(-\frac{1}{3}\left(4-x^{2}\right)^{3/2}\right)'=-\frac{1}{3}\cdot\frac{3}{2}\left(4-x^{2}\right)^{1/2}\cdot(-2x)=x\sqrt{4-x^{2}}$ . So you get the integrand of a) back, not that of c): an EXTRA factor $x$ comes out. This is exactly the error of believing you can manufacture the missing $x$ . But $\int\sqrt{4-x^{2}}\,dx$ is a different matter: substitution does not apply, for lack of $u'$ , and its antiderivative requires a trigonometric substitution, a technique of the integral calculus course that follows. Remember the line: without $u'$ present up to a constant, you do not substitute, you stop.

d)

Set $u=x^{2}+9$ , so $du=2x\,dx$ and $x\,dx=\frac{1}{2}du$ . The integral becomes $\frac{1}{2}\int u^{-1/2}du=\frac{1}{2}\cdot 2u^{1/2}=\sqrt{u}$ , that is $\sqrt{x^{2}+9}+C$ . Check by differentiation: $\left(\sqrt{x^{2}+9}\right)'=\frac{2x}{2\sqrt{x^{2}+9}}=\frac{x}{\sqrt{x^{2}+9}}$ ✓. Numerical check at $x=4$ with $C=0$ : $\sqrt{25}=5$ . Compare with c): here the $x$ in the numerator is precisely what makes the substitution possible, whereas $\int\frac{dx}{\sqrt{x^{2}+9}}$ , without this $x$ , again leaves the course. A single factor $x$ separates a thirty-second integral from an integral of another semester.

Exercise 7: Same numerator, four decisions

Faced with an integral, the question is not which technique to apply, it is which of the three decisions of the chapter to make: read the table directly, REWRITE the integrand, or SUBSTITUTE. The four integrals that follow are built on the same numerator and the same quadratic, and each one calls for a different decision.

The sorting criterion is always the same: is the numerator, up to a constant, the derivative of the denominator? If so, substitute; if not, see whether a rewriting leads back to the table.

a) Compute $\int\frac{x+1}{x^{2}}dx$ .
b) Compute $\int\frac{x+1}{x^{2}+2x+5}dx$ . Why is the absolute value superfluous here?
c) Compute $\int(x+1)e^{x^{2}+2x}dx$ .
d) Compute $\int\frac{dx}{x^{2}+2x+5}$ , first writing the denominator in vertex form (by completing the square). Which of the four integrals needed NO substitution?

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Answers

a) $\ln\left|x\right|-\frac{1}{x}+C$
b) $\frac{1}{2}\ln\left(x^{2}+2x+5\right)+C$ ; the quadratic is at least $4$ , so no absolute value
c) $\frac{1}{2}e^{x^{2}+2x}+C$
d) $\frac{1}{2}\arctan\frac{x+1}{2}+C$ ; only the integral in a) needed no substitution

a)

The denominator is $x^{2}$ , whose derivative is $2x$ : the numerator $x+1$ is not a constant multiple of it, so substitution is ruled out. REWRITE: $\frac{x+1}{x^{2}}=\frac{1}{x}+\frac{1}{x^{2}}=x^{-1}+x^{-2}$ . Two lines of the table follow: $\int x^{-1}dx=\ln\left|x\right|$ and $\int x^{-2}dx=-\frac{1}{x}$ . Answer: $\ln\left|x\right|-\frac{1}{x}+C$ . Check by differentiation: $\frac{1}{x}+\frac{1}{x^{2}}$ ✓, which is indeed the starting rewriting. Value at $x=1$ with $C=0$ : $0-1=-1$ . This is the only one of the four integrals settled without setting any $u$ .

b)

The denominator $x^{2}+2x+5$ has derivative $2x+2=2(x+1)$ : the numerator is therefore $u'$ up to a factor $\frac{1}{2}$ , and substitution is called for. Set $u=x^{2}+2x+5$ , $du=(2x+2)dx$ , so $(x+1)dx=\frac{1}{2}du$ . The integral becomes $\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln\left|u\right|+C$ , that is $\frac{1}{2}\ln\left(x^{2}+2x+5\right)+C$ . The absolute value is superfluous because $x^{2}+2x+5=(x+1)^{2}+4\geq 4>0$ for every $x$ : the quadratic has discriminant $4-20=-16<0$ , it never vanishes. Check: $\frac{1}{2}\cdot\frac{2x+2}{x^{2}+2x+5}=\frac{x+1}{x^{2}+2x+5}$ ✓. At $x=0$ : $\frac{1}{2}\ln 5\approx 0.80$ .

c)

Here the exponent is $x^{2}+2x$ , whose derivative is $2x+2=2(x+1)$ , and the factor in front of the exponential is precisely $x+1$ : same structure as in b), same substitution. Set $u=x^{2}+2x$ , $du=(2x+2)dx$ , so $(x+1)dx=\frac{1}{2}du$ , and the integral becomes $\frac{1}{2}\int e^{u}du=\frac{1}{2}e^{u}+C$ , that is $\frac{1}{2}e^{x^{2}+2x}+C$ . Check by differentiation: $\frac{1}{2}e^{x^{2}+2x}\cdot(2x+2)=(x+1)e^{x^{2}+2x}$ ✓. Value at $x=0$ with $C=0$ : $\frac{1}{2}e^{0}=0.5$ . The methodological point is that $u$ is not chosen from the look of the integrand, but from whose derivative is available.

d)

The numerator is $1$ : it is not the derivative of the denominator, so the previous substitution does not work. Complete the square, $x^{2}+2x+5=(x+1)^{2}+4$ , then aim for the arctangent line, $\int\frac{du}{u^{2}+1}$ . Set $u=\frac{x+1}{2}$ , so $x+1=2u$ and $dx=2\,du$ : the integral becomes $\int\frac{2\,du}{4u^{2}+4}=\frac{1}{2}\int\frac{du}{u^{2}+1}=\frac{1}{2}\arctan u+C$ , that is $\frac{1}{2}\arctan\frac{x+1}{2}+C$ . Check: $\frac{1}{2}\cdot\frac{1}{1+\left(\frac{x+1}{2}\right)^{2}}\cdot\frac{1}{2}=\frac{1}{4+(x+1)^{2}}=\frac{1}{x^{2}+2x+5}$ ✓. At $x=1$ : $\frac{1}{2}\arctan 1=\frac{\pi}{8}\approx 0.39$ . The one that needed no substitution is the integral in a).

Exercise 8: Five statements to correct

Each of the five statements below was taken from a student's test. For each one, say whether it is true or false, correct it, and justify the correction WITH A DIFFERENTIATION or with a numerical counterexample.

1) “ $\int f(x)g(x)\,dx=\left(\int f(x)dx\right)\left(\int g(x)dx\right)$ , just like the derivative of a product.”
2) “The constant $C$ is a detail of notation: $F(x)=x^{2}$ and $F(x)=x^{2}+7$ are the same antiderivative.”
3) “ $\int\frac{dx}{x}=\ln x+C$ on all of $\mathbb{R}\setminus\{0\}$ .”
4) “ $\int(3x+2)^{5}dx=\frac{(3x+2)^{6}}{6}+C$ , by the power rule.”
5) “Differentiating then integrating, or integrating then differentiating, comes to the same thing: you get $f$ back in both cases.”

Type your answers, the page tells you right or wrong 0/6

1)

\int x^{2}dx

at

x=3

(with

C=0

)

\left(\frac{x^{2}}{2}\right)^{2}

at

x=3

2)

x^{2}

and

x^{2}+7

are

3)

Correct statement

4)

Correct denominator

5)

What gets lost

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Answers

1) False: $\frac{x^{3}}{3}$ equals $9$ at $x=3$ , $\frac{x^{4}}{4}$ equals $20.25$
2) False: two different antiderivatives of the same function
3) False: $\ln\left|x\right|+C$ , and one constant PER INTERVAL
4) False: $\frac{(3x+2)^{6}}{18}+C$
5) False in one direction: $\int f'(x)dx=f(x)+C$ , the constant is lost

1)

FALSE, and doubly so. First, there is no product rule for integration; second, the product rule for differentiation does not have this form either, since $\left(fg\right)'=f'g+fg'$ . Numerical counterexample with $f(x)=g(x)=x$ : the left-hand side is $\int x^{2}dx=\frac{x^{3}}{3}$ , which equals $9$ at $x=3$ with $C=0$ ; the right-hand side is $\left(\frac{x^{2}}{2}\right)^{2}=\frac{x^{4}}{4}$ , which equals $20.25$ at $x=3$ . The two numbers differ, so the statement falls. Correction: faced with a product, you must REWRITE, or recognize a substitution, or wait for the integral calculus course and integration by parts.

2)

FALSE. These two functions have the same DERIVATIVE, $2x$ , but they are not the same function: their graphs are two distinct curves, shifted vertically by $7$ , and $F(0)$ equals $0$ in one case and $7$ in the other. Correction: they are two different antiderivatives of the same function. The expression $x^{2}+C$ does not denote a function but a FAMILY of functions, and an extra condition, a prescribed point, selects exactly one of them. Forgetting the $+C$ in a general answer costs one mark; getting $C$ wrong in an initial value problem costs the whole problem.

3)

FALSE because of $\ln x$ , which is defined only for $x>0$ , and false in its scope as well. Correction in two steps. First the absolute value: $\int\frac{dx}{x}=\ln\left|x\right|+C$ , and differentiating confirms that for $x<0$ , $\left(\ln(-x)\right)'=\frac{-1}{-x}=\frac{1}{x}$ ✓. Then the scope: $\mathbb{R}\setminus\{0\}$ is not an INTERVAL but the union of two intervals, and the theorem on antiderivatives applies only on an interval. The general antiderivative there therefore calls for TWO constants, one per interval. In practice, you work on the interval where the problem lives and a single constant is enough.

4)

FALSE, a factor $\frac{1}{3}$ is missing. The check shows it in one line: $\left(\frac{(3x+2)^{6}}{6}\right)'=\frac{6(3x+2)^{5}\cdot 3}{6}=3(3x+2)^{5}$ , that is three times the integrand. Correction: $\int(3x+2)^{5}dx=\frac{(3x+2)^{6}}{18}+C$ , the correct denominator being $18=6\times 3$ . This is the most frequent error of the chapter, and it comes from applying the power rule to a composite function without offsetting the derivative of the inside. The substitution $u=3x+2$ , $du=3\,dx$ , makes it impossible: the $\frac{1}{3}$ appears by itself as soon as you write $dx=\frac{du}{3}$ .

5)

FALSE in one direction out of two. Integrating then differentiating gives back exactly $f$ : $\frac{d}{dx}\left(\int f(x)dx\right)=f(x)$ , the constant disappears when you differentiate. But differentiating then integrating does not give back $f$ : $\int f'(x)dx=f(x)+C$ , and nothing says this $C$ is zero. Example: start from $f(x)=x^{2}+7$ , differentiate to get $2x$ , integrate to get $x^{2}+C$ . The information $7$ is lost, and no computation recovers it. Correction: differentiation ERASES the constant, and integration cannot guess which one to put back. That is precisely why a serious antiderivative problem always supplies an initial condition.

Exercise 9: Problem: two constants, and a condition that is not at t = 0

A cart leaves a platform and accelerates along a horizontal rail. A sensor gives its acceleration, $a(t)=\frac{6}{\sqrt{t+1}}$ in metres per second squared, for $t\geq 0$ in seconds. The radar on the platform measured $v(3)=30$ metres per second, and the origin of positions is taken at the platform, so $x(0)=0$ .

Working back from $a$ to $x$ takes TWO integrations, hence two constants, hence two conditions. Here the first condition is not given at time zero, which means it cannot be read directly off the formula.

a) Find $v(t)$ . Which piece of data fixes the first constant, and why can it not be applied before integrating?
b) Find $x(t)$ , then the position of the cart at $t=3$ seconds.
c) Check your $x(t)$ by two successive differentiations, and check both conditions.
d) A student sets both constants equal to zero “since the cart starts from the platform”. What does he find at $t=3$ , and by how many metres is he off?
e) If you knew only $a(t)$ and $x(0)$ , could you find $x(t)$ ? Explain by counting the constants.

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Answers

a) $v(t)=12\sqrt{t+1}+6$ , with $C_{1}=6$ fixed by $v(3)=30$
b) $x(t)=8(t+1)^{3/2}+6t-8$ ; $x(3)=74$ m
c) $x'=v$ and $v'=a$ ✓, with $x(0)=0$ and $v(3)=30$ ✓
d) He finds $64$ m, that is $10$ m short, and his $v(3)=24$ contradicts the radar
e) No: two integrations, two constants, so two conditions are needed

a)

Write $a(t)=6(t+1)^{-1/2}$ and integrate with the reversed power rule: $\int 6(t+1)^{-1/2}dt=6\cdot\frac{(t+1)^{1/2}}{1/2}=12\sqrt{t+1}$ , the inside $t+1$ having derivative $1$ , which adds no coefficient. So $v(t)=12\sqrt{t+1}+C_{1}$ . The data $v(3)=30$ fixes $C_{1}$ : $12\sqrt{4}+C_{1}=24+C_{1}=30$ , so $C_{1}=6$ and $v(t)=12\sqrt{t+1}+6$ . The condition cannot be applied before integrating, because $C_{1}$ only exists after the integration: the condition says nothing about $a$ , it gives a value of $v$ , and $v$ is known only once the antiderivative is written. Check: $v(0)=12+6=18$ metres per second.

b)

Integrate $v$ : $\int 12(t+1)^{1/2}dt=12\cdot\frac{(t+1)^{3/2}}{3/2}=8(t+1)^{3/2}$ , and $\int 6\,dt=6t$ . So $x(t)=8(t+1)^{3/2}+6t+C_{2}$ . The condition $x(0)=0$ gives $8\cdot 1+0+C_{2}=0$ , so $C_{2}=-8$ and $x(t)=8(t+1)^{3/2}+6t-8$ . Position at $t=3$ : $8\cdot 4^{3/2}+18-8=8\cdot 8+10=74$ metres. Notice that $C_{2}$ is not zero even though the initial position is: the term $8(t+1)^{3/2}$ already equals $8$ at $t=0$ , and that is what the constant must offset. Confusing “zero initial position” with “zero constant” is the error in d).

c)

First differentiation: $x'(t)=8\cdot\frac{3}{2}(t+1)^{1/2}+6=12\sqrt{t+1}+6=v(t)$ ✓. Second differentiation: $v'(t)=12\cdot\frac{1}{2\sqrt{t+1}}=\frac{6}{\sqrt{t+1}}=a(t)$ ✓. You get the starting acceleration back, so both integrations are right. That leaves the conditions, which are a SEPARATE check: $x(0)=8+0-8=0$ ✓ and $v(3)=12\cdot 2+6=30$ ✓. The two checks are independent, and that is what makes them useful: an antiderivative error makes the first one fail, a constant error makes only the second one fail.

d)

With $C_{1}=C_{2}=0$ , he gets $v(t)=12\sqrt{t+1}$ and $x(t)=8(t+1)^{3/2}$ , so $x(3)=8\cdot 8=64$ metres instead of $74$ : he is off by $10$ metres. His error can be detected without even knowing the right answer, since his $v(3)=12\cdot 2=24$ contradicts the radar measurement, which said $30$ . This is the check to make every time: once the constants are set, substitute EACH piece of data from the problem into the formula obtained. A wrong constant leaves the check by differentiation perfectly green, which is why it so often goes unnoticed.

e)

No. Each integration introduces a constant, so working back from $a$ to $x$ introduces two, and exactly TWO conditions are needed to fix them. With the single piece of data $x(0)=0$ , you get $x(t)=8(t+1)^{3/2}+C_{1}t+C_{2}$ with $C_{2}=-8-0$ , that is $x(t)=8(t+1)^{3/2}+C_{1}t-8$ : a family of trajectories indexed by $C_{1}$ , that is, by the unknown initial velocity. Physically, this is obvious: an accelerometer cannot tell whether the cart was already moving. The same remark applies to a phone, which measures its acceleration without ever being able to deduce on its own where it is or how fast it is going.

Exercise 10: Problem: the drone, a substitution and a forgotten C

A drone takes off from the ground and climbs vertically. Its rate of climb is $v(t)=\frac{20t}{t^{2}+4}$ in metres per second, for $t\geq 0$ in seconds, and its altitude at takeoff is zero, $z(0)=0$ .

The figure shows the shape of this velocity. It starts at zero, goes through a maximum, then decreases slowly: the drone climbs more and more slowly, which does not mean that it stops climbing.

a) Find $z(t)$ by a substitution that you will state, then fix the constant.
b) At what altitude is the drone at $t=6$ seconds?
c) At what instant is the rate of climb at its maximum, and what is it then?
d) Check $z(t)$ by differentiation. Does the drone eventually reach a limiting altitude?
e) A student forgets the constant and announces $z(6)=10\ln 40$ . What altitude does he announce, by how much is he off, and what does he unwittingly claim about the moment of takeoff?

Show the solution

Answers

a) $u=t^{2}+4$ ; $z(t)=10\ln\frac{t^{2}+4}{4}$ , with $C=-10\ln 4\approx -13.86$
b) $z(6)=10\ln 10\approx 23.03$ m
c) $t=2$ s, with $v(2)=5$ m/s
d) $z'=v$ ✓; no limiting altitude, $z(t)\to+\infty$
e) He announces $36.89$ m, that is $13.86$ m too much, and he places the takeoff at an altitude of $13.86$ m

a)

The denominator $t^{2}+4$ has derivative $2t$ , and the numerator $20t$ is its constant multiple $10\times 2t$ : substitution is called for. Set $u=t^{2}+4$ , so $du=2t\,dt$ and $20t\,dt=10\,du$ . The integral becomes $10\int\frac{du}{u}=10\ln\left|u\right|+C$ , that is $10\ln\left(t^{2}+4\right)+C$ , the absolute value being unnecessary since $t^{2}+4\geq 4>0$ . The condition $z(0)=0$ gives $10\ln 4+C=0$ , so $C=-10\ln 4$ and $z(t)=10\ln\left(t^{2}+4\right)-10\ln 4=10\ln\frac{t^{2}+4}{4}$ . Writing the answer in this condensed form is a real gain: the initial condition can be read in it, since the logarithm equals $\ln 1=0$ at $t=0$ .

b)

$z(6)=10\ln\frac{36+4}{4}=10\ln 10\approx 23.03$ metres. You can check the order of magnitude without a calculator: the velocity exceeds $4$ metres per second roughly between the first and the fourth second, then drops back towards $2$ , so about twenty metres in six seconds is plausible. This plausibility check is worth as much as the computation: an answer of $200$ metres or $2$ metres should have raised a flag. Notice also the pleasant coincidence, $\frac{40}{4}=10$ , which produces a natural logarithm of ten.

c)

The velocity is at its maximum where $v'(t)=0$ . Differentiate with the quotient rule: $v'(t)=20\cdot\frac{\left(t^{2}+4\right)-t\cdot 2t}{\left(t^{2}+4\right)^{2}}=\frac{20\left(4-t^{2}\right)}{\left(t^{2}+4\right)^{2}}$ . The denominator is always positive, so the sign is that of $4-t^{2}$ : positive for $t<2$ , negative for $t>2$ . The velocity increases then decreases, so $t=2$ seconds is a MAXIMUM, and it is absolute on $\left[0, +\infty\right)$ since it is the only critical number. Value: $v(2)=\frac{40}{8}=5$ metres per second. This question is pure differential calculus, asked in the middle of an integration problem: the two halves of the course answer each other.

d)

Check: $z'(t)=10\cdot\frac{2t}{t^{2}+4}=\frac{20t}{t^{2}+4}=v(t)$ ✓, differentiating $\ln\left(t^{2}+4\right)$ with the chain rule, the $-10\ln 4$ being a constant with zero derivative. There is NO limiting altitude: $t^{2}+4\to+\infty$ , so $z(t)\to+\infty$ . The drone climbs forever, more and more slowly, but without ever levelling off. A velocity that tends to zero does not imply a bounded altitude, and this is a common misreading. Order of magnitude: it would take about $109$ seconds to pass $80$ metres, the logarithm being formidably slow.

e)

He announces $10\ln 40\approx 36.89$ metres, instead of $23.03$ : he is off by $10\ln 4\approx 13.86$ metres, too high. And he unwittingly claims that the drone was already at an altitude of $10\ln 4\approx 13.86$ metres AT THE MOMENT OF TAKEOFF, since his formula gives $z(0)=10\ln 4$ . This is how a forgotten $C$ is spotted: you evaluate your own answer at the instant where the problem gives the condition, and you see whether it says what the problem says. The check takes five seconds, and it saves the five marks this kind of problem puts on the constant.

The rate of climb is at its maximum at

t=2

s, the only critical number of

v

on

\left[0, +\infty\right)

, where it equals

v(2)=5

m/s.

Corrected exercises: elementary integration (201-SN2-RE)

Course recap

Calculations: antiderivatives read from the table

Exercise 0.1: Integrating with the table: twelve direct antiderivatives

Calculations: antiderivatives by substitution

Exercise 0.2: Integrating by substitution: ten direct antiderivatives

Part A: the basics (/50)

Exercise 1: The table of derivatives, read from right to left

Exercise 2: The constant C, what differentiation erases

Exercise 3: Rewrite first, integrate afterwards

Exercise 4: Transcendental functions, backwards

Exercise 5: The chain rule backwards: seeing u and its derivative

Part B: problems and reasoning (/50)

Exercise 6: When a constant factor is missing, and when an x is missing

Exercise 7: Same numerator, four decisions

Exercise 8: Five statements to correct

Exercise 9: Problem: two constants, and a condition that is not at t = 0

Exercise 10: Problem: the drone, a substitution and a forgotten C

See also

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