Exercise 0.1: Integrate with the table: twelve direct calculations

For each function $f$ , find the general antiderivative using the table of basic antiderivatives and linearity, then the antiderivative $F$ that equals zero at the first point given, and give its value at the second point. Exact value, or to three decimal places if it involves $e$ , $\pi$ or a logarithm.

One method for all twelve rows: first REWRITE the integrand as a sum of terms from the table, roots as fractional powers, fractions as negative powers, a quotient split term by term over the monomial, a product expanded. Then integrate term by term, dividing by the NEW exponent or by the coefficient $k$ of the argument. Finally, the condition $F(a)=0$ fixes the constant, which amounts to computing $G(b)-G(a)$ for any antiderivative $G$ .

a) $f(x)=4x^{3}-6x^{2}+2x-5$ , with $F(0)=0$ : give $F(2)$ .
b) $f(x)=\frac{8}{x^{3}}-\frac{3}{x^{2}}$ , with $F(1)=0$ : give $F(2)$ .
c) $f(x)=\sqrt[3]{x}+\frac{1}{\sqrt[3]{x^{2}}}$ , with $F(1)=0$ : give $F(8)$ .
d) $f(x)=\frac{3x^{4}-x+2}{x^{3}}$ , with $F(1)=0$ : give $F(2)$ .
e) $f(x)=\left(x^{2}+1\right)^{2}$ , with $F(0)=0$ : give $F(1)$ .
f) $f(x)=4e^{2x}-e^{-x/2}$ , with $F(0)=0$ : give $F(\ln 4)$ .
g) $f(x)=4^{x}+\left(\frac{1}{2}\right)^{x}$ , with $F(0)=0$ : give $F(1)$ .
h) $f(x)=\cos(3x)-2\sin\left(\frac{x}{2}\right)$ , with $F(0)=0$ : give $F(\pi)$ .
i) $f(x)=\tan^{2}x+\frac{4}{1+x^{2}}$ , with $F(0)=0$ : give $F(1)$ .
j) $f(x)=\frac{1-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}$ , with $F(0)=0$ : give $F\left(\frac{1}{2}\right)$ .
k) $f(x)=\frac{(x+1)^{2}}{x}$ , with $F(1)=0$ : give $F(e)$ .
l) $f(x)=\frac{1+x\cos x+xe^{x}}{x}$ , with $F(1)=0$ : give $F(2)$ .

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

a)

F(2)

b)

F(2)

c)

F(8)

d)

F(2)

e)

F(1)

f)

F(\ln 4)

g)

F(1)

h)

F(\pi)

i)

F(1)

j)

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

k)

F(e)

l)

F(2)

Show the solution

Answers

a) $F(x)=x^{4}-2x^{3}+x^{2}-5x$ ; $F(2)=-6$
b) $F(x)=-\frac{4}{x^{2}}+\frac{3}{x}+1$ ; $F(2)=\frac{3}{2}$
c) $F(x)=\frac{3}{4}x^{\frac{4}{3}}+3x^{\frac{1}{3}}-\frac{15}{4}$ ; $F(8)=\frac{57}{4}=14.25$
d) $F(x)=\frac{3}{2}x^{2}+\frac{1}{x}-\frac{1}{x^{2}}-\frac{3}{2}$ ; $F(2)=\frac{19}{4}=4.75$
e) $F(x)=\frac{x^{5}}{5}+\frac{2x^{3}}{3}+x$ ; $F(1)=\frac{28}{15}\approx 1.867$
f) $F(x)=2e^{2x}+2e^{-x/2}-4$ ; $F(\ln 4)=29$
g) $F(x)=\frac{4^{x}}{2\ln 2}-\frac{(1/2)^{x}}{\ln 2}+\frac{1}{2\ln 2}$ ; $F(1)=\frac{2}{\ln 2}\approx 2.885$
h) $F(x)=\frac{\sin(3x)}{3}+4\cos\left(\frac{x}{2}\right)-4$ ; $F(\pi)=-4$
i) $F(x)=\tan x-x+4\arctan x$ ; $F(1)=\tan 1-1+\pi\approx 3.699$
j) $F(x)=\arcsin x-x$ ; $F\left(\frac{1}{2}\right)=\frac{\pi}{6}-\frac{1}{2}\approx 0.0236$
k) $F(x)=\frac{x^{2}}{2}+2x+\ln|x|-\frac{5}{2}$ ; $F(e)=\frac{e^{2}}{2}+2e-\frac{3}{2}\approx 7.631$
l) $F(x)=\ln|x|+\sin x+e^{x}-\sin 1-e$ ; $F(2)\approx 5.432$

The move that decides everything happens BEFORE you integrate: rewrite. The table knows neither roots, nor fractions, nor products, nor quotients; it knows $x^{n}$ , $e^{kx}$ , $a^{x}$ , $\sin$ , $\cos$ , $\sec^{2}$ , $\frac{1}{x}$ , $\frac{1}{1+x^{2}}$ and $\frac{1}{\sqrt{1-x^{2}}}$ . Once the sum is written, linearity does the rest, and there is NO product rule and NO quotient rule for antiderivatives.

a)

$F(x)=x^{4}-2x^{3}+x^{2}-5x+C$ , and $F(0)=0$ gives $C=0$ . So $F(2)=16-16+4-10=-6$ . Pitfall: forgetting to divide by the new exponent, $4x^{3}\mapsto 4x^{4}$ instead of $x^{4}$ .

b)

$f(x)=8x^{-3}-3x^{-2}$ , so $F(x)=\frac{8x^{-2}}{-2}-\frac{3x^{-1}}{-1}+C=-\frac{4}{x^{2}}+\frac{3}{x}+C$ . $F(1)=-4+3+C=0$ gives $C=1$ , and $F(2)=-1+\frac{3}{2}+1=\frac{3}{2}$ . Pitfall: the exponent goes UP, $x^{-3}$ becomes $\frac{x^{-2}}{-2}$ and not $\frac{x^{-4}}{-4}$ .

c)

$f(x)=x^{\frac{1}{3}}+x^{-\frac{2}{3}}$ , so $F(x)=\frac{3}{4}x^{\frac{4}{3}}+3x^{\frac{1}{3}}+C$ . $F(1)=\frac{3}{4}+3+C=0$ gives $C=-\frac{15}{4}$ , and $F(8)=\frac{3}{4}\times 16+3\times 2-\frac{15}{4}=18-\frac{15}{4}=\frac{57}{4}=14.25$ . Pitfall: $\frac{1}{\sqrt[3]{x^{2}}}=x^{-\frac{2}{3}}$ , whose antiderivative $\frac{x^{1/3}}{1/3}$ carries a coefficient $3$ .

d)

Split over the monomial: $f(x)=3x-x^{-2}+2x^{-3}$ , so $F(x)=\frac{3}{2}x^{2}+\frac{1}{x}-\frac{1}{x^{2}}+C$ . $F(1)=\frac{3}{2}+1-1+C=0$ gives $C=-\frac{3}{2}$ , and $F(2)=6+\frac{1}{2}-\frac{1}{4}-\frac{3}{2}=\frac{19}{4}=4.75$ . Pitfall: integrating the numerator and the denominator separately, a “quotient rule” that does not exist.

e)

Expand: $\left(x^{2}+1\right)^{2}=x^{4}+2x^{2}+1$ , so $F(x)=\frac{x^{5}}{5}+\frac{2x^{3}}{3}+x+C$ with $C=0$ , and $F(1)=\frac{1}{5}+\frac{2}{3}+1=\frac{28}{15}\approx 1.867$ . Pitfall: answering $\frac{\left(x^{2}+1\right)^{3}}{3}$ , whose derivative is $2x\left(x^{2}+1\right)^{2}$ : the factor $2x$ is missing, so it is not an antiderivative.

f)

$\int e^{kx}dx=\frac{e^{kx}}{k}+C$ , with $k=2$ then $k=-\frac{1}{2}$ : $F(x)=2e^{2x}+2e^{-x/2}+C$ . $F(0)=2+2+C=0$ gives $C=-4$ . At $\ln 4$ : $e^{2\ln 4}=16$ and $e^{-\frac{\ln 4}{2}}=\frac{1}{2}$ , so $F(\ln 4)=32+1-4=29$ . Pitfall: dividing by $-\frac{1}{2}$ multiplies by $-2$ , and the minus sign in front of $e^{-x/2}$ becomes a plus.

g)

$\int a^{x}dx=\frac{a^{x}}{\ln a}+C$ , with $\ln 4=2\ln 2$ and $\ln\frac{1}{2}=-\ln 2$ : $F(x)=\frac{4^{x}}{2\ln 2}-\frac{\left(\frac{1}{2}\right)^{x}}{\ln 2}+C$ . So $F(1)=\frac{4-1}{2\ln 2}-\frac{\frac{1}{2}-1}{\ln 2}=\frac{3}{2\ln 2}+\frac{1}{2\ln 2}=\frac{2}{\ln 2}\approx 2.885$ . Pitfall: $\frac{4^{x+1}}{x+1}$ , the power rule applied to an exponential.

h)

$F(x)=\frac{\sin(3x)}{3}+4\cos\left(\frac{x}{2}\right)+C$ , since $\int\sin(kx)dx=-\frac{\cos(kx)}{k}$ and dividing by $k=\frac{1}{2}$ multiplies by $2$ . $F(0)=0+4+C=0$ gives $C=-4$ , and $F(\pi)=\frac{\sin 3\pi}{3}+4\cos\frac{\pi}{2}-4=-4$ . Pitfall: the sign, $-2\sin\left(\frac{x}{2}\right)$ integrates to $+4\cos\left(\frac{x}{2}\right)$ .

i)

$\tan^{2}x$ is not in the table, but $\tan^{2}x=\sec^{2}x-1$ is. So $F(x)=\tan x-x+4\arctan x+C$ with $C=0$ , and $F(1)=\tan 1-1+4\times\frac{\pi}{4}=\tan 1-1+\pi\approx 3.699$ . Pitfall: $\frac{\tan^{3}x}{3}$ , whose derivative is $\tan^{2}x\sec^{2}x$ .

j)

Split: $f(x)=\frac{1}{\sqrt{1-x^{2}}}-1$ , so $F(x)=\arcsin x-x+C$ with $C=0$ , and $F\left(\frac{1}{2}\right)=\frac{\pi}{6}-\frac{1}{2}\approx 0.0236$ . Pitfall: $\arcsin\frac{1}{2}=\frac{\pi}{6}$ in radians, not $30$ .

k)

Expand, then split: $\frac{x^{2}+2x+1}{x}=x+2+\frac{1}{x}$ , so $F(x)=\frac{x^{2}}{2}+2x+\ln|x|+C$ . $F(1)=\frac{1}{2}+2+0+C=0$ gives $C=-\frac{5}{2}$ , and $F(e)=\frac{e^{2}}{2}+2e+1-\frac{5}{2}=\frac{e^{2}}{2}+2e-\frac{3}{2}\approx 7.631$ . Pitfall: $\frac{1}{x}=x^{-1}$ is the gap in the power rule, its antiderivative is $\ln|x|$ .

l)

Split over $x$ : $f(x)=\frac{1}{x}+\cos x+e^{x}$ , three families from the table. $F(x)=\ln|x|+\sin x+e^{x}+C$ , $F(1)=\sin 1+e+C=0$ , so $F(2)=\ln 2+\sin 2+e^{2}-\sin 1-e\approx 5.432$ , calculator in radians.

Check: DIFFERENTIATE each antiderivative you find and you must land back on $f$ , in ten seconds per row; this check is what catches the forgotten factor $\frac{1}{k}$ and the sign of the sines. For the value, check that $F(a)=0$ at the starting point, then recompute $F(b)$ on the calculator.

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Exercise 0.2: Integrate by direct substitution: ten direct calculations

For each function $f$ , find an antiderivative by substitution, writing out $u$ and $du$ , then the antiderivative $F$ that equals zero at the first point given, and give its value at the second point. Exact value, or to three decimal places if it involves $e$ or a logarithm.

One method for all ten rows: spot the INNER function $u$ whose derivative is present up to a constant factor; write $du=u'(x)\,dx$ and adjust the constant; integrate in $u$ with the table; return to $x$ . The value asked for is then computed as $G(b)-G(a)$ , where $G$ is the antiderivative found.

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

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

a)

F(2)

b)

F(4)

c)

F(1)

d)

F(2)

e)

F(1)

f)

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

g)

F(e^{4})

h)

F(\ln 3)

i)

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

j)

F(1)

Show the solution

Answers

a) $\frac{(3x-2)^{5}}{15}$ ; $F(2)=\frac{1023}{15}=68.2$
b) $\sqrt{2x+1}$ ; $F(4)=2$
c) $\frac{\left(x^{2}+x+1\right)^{4}}{4}$ ; $F(1)=20$
d) $\ln\left|x^{4}-2x+3\right|$ ; $F(2)=\ln 5\approx 1.609$
e) $\frac{1}{2}e^{x^{2}+2x}$ ; $F(1)=\frac{e^{3}-1}{2}\approx 9.543$
f) $\frac{\sin^{5}x}{5}$ ; $F\left(\frac{\pi}{2}\right)=\frac{1}{5}$
g) $\frac{2}{3}(\ln x)^{\frac{3}{2}}$ ; $F(e^{4})=\frac{16}{3}\approx 5.333$
h) $-\frac{1}{1+e^{x}}$ ; $F(\ln 3)=\frac{1}{4}$
i) $-\frac{1}{2}\ln\left|\cos(2x)\right|$ ; $F\left(\frac{\pi}{6}\right)=\frac{\ln 2}{2}\approx 0.347$
j) $-\frac{3}{2\left(x^{2}+1\right)^{2}}$ ; $F(1)=\frac{9}{8}=1.125$

The move that decides everything: find $u$ by looking for its DERIVATIVE in the integrand. A $2x+1$ next to $x^{2}+x+1$ , a $\frac{1}{x}$ next to $\ln x$ , a $\cos x$ next to $\sin x$ : when the derivative is there, up to a constant, the substitution works; when a factor in $x$ is missing, it does not, and a constant can never be added to make up for an $x$ .

a)

$u=3x-2$ , $du=3\,dx$ : $\int(3x-2)^{4}dx=\frac{1}{3}\int u^{4}du=\frac{(3x-2)^{5}}{15}+C$ . So $F(2)=\frac{4^{5}-1^{5}}{15}=\frac{1023}{15}=68.2$ . Pitfall: forgetting the $\frac{1}{3}$ and answering $\frac{(3x-2)^{5}}{5}$ , three times too large.

b)

$u=2x+1$ , $du=2\,dx$ : $\int(2x+1)^{-\frac{1}{2}}dx=\frac{1}{2}\cdot\frac{u^{\frac{1}{2}}}{\frac{1}{2}}=\sqrt{2x+1}+C$ . So $F(4)=\sqrt{9}-\sqrt{1}=2$ . The $\frac{1}{2}$ from $du$ and the $\frac{1}{2}$ from the exponent cancel exactly.

c)

$u=x^{2}+x+1$ , $du=(2x+1)\,dx$ , present as is: $\int u^{3}du=\frac{\left(x^{2}+x+1\right)^{4}}{4}+C$ . At $0$ , $u=1$ ; at $1$ , $u=3$ . So $F(1)=\frac{81-1}{4}=20$ .

d)

The form $\frac{u'}{u}$ : $u=x^{4}-2x+3$ , $du=(4x^{3}-2)\,dx$ , and $\int\frac{du}{u}=\ln|u|+C$ . Here $u>0$ on $[0, 2]$ (its minimum, at $x=\sqrt[3]{0.5}$ , is about $1.81$ ). At $0$ , $u=3$ ; at $2$ , $u=15$ . So $F(2)=\ln 15-\ln 3=\ln 5\approx 1.609$ . Pitfall: $\frac{u'}{u}$ integrates to $\ln|u|$ , not to $\frac{u^{2}}{2}$ nor to $\ln|u'|$ .

e)

The form $u'e^{u}$ with $u=x^{2}+2x$ , $du=(2x+2)\,dx=2(x+1)\,dx$ : a factor $2$ is missing, a constant, which we adjust. $\int(x+1)e^{x^{2}+2x}dx=\frac{1}{2}e^{x^{2}+2x}+C$ . So $F(1)=\frac{e^{3}-e^{0}}{2}=\frac{e^{3}-1}{2}\approx 9.543$ .

f)

$u=\sin x$ , $du=\cos x\,dx$ : $\int u^{4}du=\frac{\sin^{5}x}{5}+C$ . So $F\left(\frac{\pi}{2}\right)=\frac{1-0}{5}=\frac{1}{5}$ . Pitfall: taking $u=\cos x$ ; its derivative $-\sin x$ only appears to the power $4$ , not on its own.

g)

$u=\ln x$ , $du=\frac{dx}{x}$ : $\int u^{\frac{1}{2}}du=\frac{2}{3}(\ln x)^{\frac{3}{2}}+C$ . At $1$ , $u=0$ ; at $e^{4}$ , $u=4$ . So $F(e^{4})=\frac{2}{3}\times 4^{\frac{3}{2}}=\frac{2}{3}\times 8=\frac{16}{3}\approx 5.333$ .

h)

$u=1+e^{x}$ , $du=e^{x}dx$ : $\int u^{-2}du=-\frac{1}{u}=-\frac{1}{1+e^{x}}+C$ . At $0$ , $u=2$ ; at $\ln 3$ , $u=4$ . So $F(\ln 3)=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}$ . Pitfall: seeing a $\frac{u'}{u}$ form and writing a logarithm; the square in the denominator makes it a power $u^{-2}$ .

i)

$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$ : $u=\cos(2x)$ , $du=-2\sin(2x)\,dx$ , so $\int\tan(2x)\,dx=-\frac{1}{2}\ln\left|\cos(2x)\right|+C$ . At $0$ , $\cos 0=1$ ; at $\frac{\pi}{6}$ , $\cos\frac{\pi}{3}=\frac{1}{2}$ . So $F\left(\frac{\pi}{6}\right)=-\frac{1}{2}\ln\frac{1}{2}=\frac{\ln 2}{2}\approx 0.347$ . Pitfall: two constants at once, the $-1$ from the derivative of cosine and the $2$ from the argument.

j)

$u=x^{2}+1$ , $du=2x\,dx$ , so $6x\,dx=3\,du$ : $\int 3u^{-3}du=-\frac{3}{2u^{2}}=-\frac{3}{2\left(x^{2}+1\right)^{2}}+C$ . At $0$ , $u=1$ ; at $1$ , $u=2$ . So $F(1)=-\frac{3}{8}+\frac{3}{2}=\frac{9}{8}=1.125$ .

Check: differentiate the antiderivative you found with the chain rule, and you must get exactly $f$ back, constant factor included; if a factor $3$ , $\frac{1}{2}$ or $-1$ is missing, the adjustment of $du$ is wrong.

Exercise 1: What is an antiderivative?

The figure shows three antiderivatives of $f(x)=2x$ : the parabolas $x^{2}$ , $x^{2}+2$ and $x^{2}-2$ . Each one is obtained from another by a vertical translation, and that is exactly what the constant of integration says.

a) State the definition of an antiderivative, then check that $F(x)=x\sin x+\cos x$ is an antiderivative of $f(x)=x\cos x$ .
b) Show that $F(x)=\sin^{2}x$ and $G(x)=-\cos^{2}x$ are both antiderivatives of $\sin 2x$ . Is this a contradiction? Which theorem settles the question?
c) Find the antiderivative $F$ of $f(x)=3x^{2}-4x+1$ that satisfies $F(2)=5$ . How many such antiderivatives are there?
d) Using the figure: why do the three curves have the same slope at every $x$ -value, and why does exactly one antiderivative pass through a given point of the plane?

Show the solution

Answers

a) $F'(x)=x\cos x$ : $F$ is indeed an antiderivative
b) $F-G=1$ : two antiderivatives differ by a constant, on an interval
c) $F(x)=x^{3}-2x^{2}+x+3$ , and only one
d) Vertical translates: parallel tangent lines, one curve through each point

a)

Definition: $F$ is an antiderivative of $f$ on an interval $I$ if $F$ is differentiable on $I$ and $F'(x)=f(x)$ for every $x\in I$ . Check: $F'(x)=\left(x\sin x\right)'+\left(\cos x\right)'=(\sin x+x\cos x)-\sin x=x\cos x=f(x)$ ✓. The two $\sin x$ terms cancel, and that is the whole point of the form given. Note that we are not LOOKING for the antiderivative here, we are CHECKING it: checking is always easier than finding, which is why every integration answer can be verified by differentiation.

b)

$F'(x)=2\sin x\cos x=\sin 2x$ ✓ and $G'(x)=-2\cos x\cdot(-\sin x)=2\sin x\cos x=\sin 2x$ ✓. No contradiction: $F(x)-G(x)=\sin^{2}x+\cos^{2}x=1$ , a CONSTANT. The theorem that settles the question: if two functions have the same derivative on an interval, their difference is constant there (a consequence of the mean value theorem, since $(F-G)'=0$ ). Two antiderivatives of the same function can therefore differ only by a constant, and that is precisely what the $+C$ encodes. Watch the INTERVAL hypothesis: on a domain in two pieces, such as that of $\frac{1}{x}$ , the “constant” can take one value to the left of zero and another to the right.

c)

$F(x)=x^{3}-2x^{2}+x+C$ . The condition $F(2)=5$ gives $8-8+2+C=5$ , so $C=3$ and $F(x)=x^{3}-2x^{2}+x+3$ . There is exactly ONE: the initial condition selects a single member of the infinite family. This is the central mechanism of differential equations later in the semester: integration provides the family, the initial condition picks out the solution.

d)

The three curves are vertical translates of one another: going from $x^{2}$ to $x^{2}+2$ adds the same amount everywhere, which changes no slope. Their common derivative is $2x$ , so at a given $x$ -value the three tangent lines are parallel. And exactly one antiderivative passes through a point $(x_{0},y_{0})$ of the plane, because the equation $x_{0}^{2}+C=y_{0}$ has the unique solution $C=y_{0}-x_{0}^{2}$ . Geometrically: the family of antiderivatives fills the plane without any two of them crossing, which gives the right picture of the $+C$ as a vertical slider.

Exercise 2: The table of formulas and its single gap

The whole table of basic antiderivatives is the table of derivatives read backwards. Only one formula has a gap, and the exam tests it every semester.

a) Compute $\int\left(x^{5}-3x^{2}+7\right)dx$ , $\int\frac{dx}{x^{4}}$ and $\int\sqrt{x}\,dx$ , first rewriting each integrand as a power of $x$ .
b) Why does the formula $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ fail for $n=-1$ ? What is the integral then, and why the absolute value in the answer?
c) Compute $\int\left(e^{3x}+2^{x}\right)dx$ . Where does the $\ln 2$ in the denominator come from, and where does the factor $\frac{1}{3}$ come from?
d) Compute $\int\left(\sec^{2}x-\frac{4}{1+x^{2}}\right)dx$ and $\int\sin(5x)dx$ . What mistake does the student make who answers $-\cos(5x)+C$ to the second one?

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

a)

\int\frac{dx}{x^{4}}

\int\sqrt{x}\,dx

b)

\int\frac{dx}{x}

c)

Coefficient of

e^{3x}

Antiderivative of

2^{x}

d)

\int\sin(5x)dx

Show the solution

Answers

a) $\frac{x^{6}}{6}-x^{3}+7x+C$ ; $-\frac{1}{3x^{3}}+C$ ; $\frac{2}{3}x^{3/2}+C$
b) $n+1=0$ ; $\int\frac{dx}{x}=\ln|x|+C$
c) $\frac{e^{3x}}{3}+\frac{2^{x}}{\ln 2}+C$
d) $\tan x-4\arctan x+C$ ; $-\frac{\cos 5x}{5}+C$

a)

Rewrite everything as powers before applying the formula. $\int\left(x^{5}-3x^{2}+7\right)dx=\frac{x^{6}}{6}-x^{3}+7x+C$ . Next, $\int\frac{dx}{x^{4}}=\int x^{-4}dx=\frac{x^{-3}}{-3}+C=-\frac{1}{3x^{3}}+C$ : the negative exponent causes no trouble, as long as you remember that $-4+1=-3$ . Finally, $\int\sqrt{x}\,dx=\int x^{1/2}dx=\frac{x^{3/2}}{3/2}+C=\frac{2}{3}x^{3/2}+C$ . Rewriting first is the real technical move: as long as the integrand is not in the form $x^{n}$ , no formula applies.

b)

For $n=-1$ , the denominator $n+1$ is zero: the formula asks for a division by zero, so it makes no sense. This case is handled separately: $\int\frac{dx}{x}=\ln|x|+C$ , because $\left(\ln|x|\right)'=\frac{1}{x}$ . The absolute value is there because $\frac{1}{x}$ also exists for $x<0$ , where $\ln x$ is not defined: for $x<0$ we have $\ln|x|=\ln(-x)$ , whose derivative is $\frac{-1}{-x}=\frac{1}{x}$ ✓. The formula thus covers both sides of the origin. It is the only gap in the table of powers, and it is filled by a function from a completely different family: the power $-1$ integrates to a logarithm, which is far from obvious and deserves to be memorized as an exception.

c)

$\int e^{3x}dx=\frac{e^{3x}}{3}+C$ : when you differentiate $e^{3x}$ , the chain rule brings out a factor 3, so you must divide by 3 to compensate. $\int 2^{x}dx=\frac{2^{x}}{\ln 2}+C$ : differentiating $2^{x}$ gives $2^{x}\ln 2$ , hence the division by $\ln 2$ . Total: $\frac{e^{3x}}{3}+\frac{2^{x}}{\ln 2}+C$ . In both cases the denominator is exactly the stray factor that differentiation would produce: integrating means cancelling in advance what differentiating would create.

d)

$\int\sec^{2}x\,dx=\tan x$ and $\int\frac{dx}{1+x^{2}}=\arctan x$ , so the result is $\tan x-4\arctan x+C$ . For the second one: $\int\sin(5x)dx=-\frac{\cos(5x)}{5}+C$ . The mistake of the student who answers $-\cos(5x)+C$ is forgetting the chain rule: differentiating $-\cos(5x)$ gives $5\sin(5x)$ , five times too much. The checking reflex settles it at once, and the general rule to remember is that for a linear argument $ax+b$ , the antiderivative is divided by $a$ : $\int f(ax+b)dx=\frac{F(ax+b)}{a}+C$ . It is the most useful special case of substitution, the one you end up doing in your head.

Exercise 3: Linearity, and the two rules that do not exist

Integration has ONE algebraic property: linearity. Every semester, a majority of students invent two more, by analogy with differentiation.

a) State the linearity of the integral, then compute $\int\left(4\cos x-\frac{3}{x}+5e^{x}\right)dx$ .
b) A student writes $\int x\,e^{x}dx=\frac{x^{2}}{2}\,e^{x}+C$ . Refute this answer by differentiation, without computing the true antiderivative.
c) Another one writes $\int\frac{x^{2}+3x}{x}dx=\frac{\int(x^{2}+3x)dx}{\int x\,dx}$ . Refute it, then compute the integral correctly.
d) Compute $\int\sqrt{x}\left(x^{2}-1\right)dx$ and $\int\frac{x^{3}+2x-5}{x^{2}}dx$ . What move do these two calculations have in common, and why is it the only one available at this stage?

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Answers

a) $4\sin x-3\ln|x|+5e^{x}+C$
b) The derivative gives $xe^{x}+\frac{x^{2}}{2}e^{x}$ : wrong
c) $\frac{x^{2}}{2}+3x+C$
d) $\frac{2}{7}x^{7/2}-\frac{2}{3}x^{3/2}+C$ ; $\frac{x^{2}}{2}+2\ln|x|+\frac{5}{x}+C$

a)

Linearity: $\int\left(af(x)+bg(x)\right)dx=a\int f(x)dx+b\int g(x)dx$ for all constants $a$ and $b$ . It allows two things, and only two: splitting a SUM, and pulling out a CONSTANT. Application: $\int\left(4\cos x-\frac{3}{x}+5e^{x}\right)dx=4\sin x-3\ln|x|+5e^{x}+C$ . A single constant $C$ is enough for the whole expression, since the sum of three arbitrary constants is still an arbitrary constant.

b)

It is enough to differentiate the proposed answer: $\left(\frac{x^{2}}{2}e^{x}\right)'=xe^{x}+\frac{x^{2}}{2}e^{x}$ , by the product rule. We wanted $xe^{x}$ and we get $xe^{x}+\frac{x^{2}}{2}e^{x}$ : the stray term $\frac{x^{2}}{2}e^{x}$ is nonzero for every $x\neq 0$ , so the answer is wrong. The lesson goes beyond the exercise: the integral of a product IS NOT the product of the integrals, because differentiating a product produces two terms, not one. There is no product rule for integration; there is integration by parts, which makes up for that absence, and which would give $(x-1)e^{x}+C$ here.

c)

The same method of refutation with a simple case: $\int\frac{x}{x}dx=\int 1\,dx=x+C$ , whereas the proposed formula would give $\frac{x^{2}/2}{x^{2}/2}=1$ , a constant. A nonconstant function cannot equal a constant, so the rule is false. The correct calculation goes through algebraic SIMPLIFICATION before any integration: $\frac{x^{2}+3x}{x}=x+3$ for $x\neq 0$ , hence $\int\frac{x^{2}+3x}{x}dx=\frac{x^{2}}{2}+3x+C$ .

d)

First calculation: distribute before integrating, $\sqrt{x}(x^{2}-1)=x^{5/2}-x^{1/2}$ , so $\int=\frac{2}{7}x^{7/2}-\frac{2}{3}x^{3/2}+C$ . Second calculation: split the fraction term by term, $\frac{x^{3}+2x-5}{x^{2}}=x+\frac{2}{x}-\frac{5}{x^{2}}$ , so $\int=\frac{x^{2}}{2}+2\ln|x|+\frac{5}{x}+C$ . The common move is to REWRITE the integrand as a sum of terms from the table before integrating anything. It is the only one available because linearity can only handle sums: as long as a product or a quotient is still standing, no rule of this chapter applies. Hence the opening strategy for every integral in this chapter: distribute, split, simplify, and only then see whether any work is left.

Exercise 4: Absolute value and disjoint domains

Two situations where the “ $+C$ ” hides a subtlety: when the integrand has a corner, and when the domain is in two pieces.

a) Compute $\int|x|\,dx$ by treating $x\geq 0$ and $x<0$ separately, then join the two pieces.
b) Show that the antiderivative obtained can be written $F(x)=\frac{x|x|}{2}+C$ and check that it is differentiable AT 0.
c) On $\mathbb{R}\setminus\{0\}$ , is $\ln|x|+C$ the general antiderivative of the function $\frac{1}{x}$ ? Justify precisely.
d) Which hypothesis of the theorem “two antiderivatives differ by a constant” is at stake in part c)?

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Answers

a) $\pm\frac{x^{2}}{2}+C$ , a single constant by continuity
b) $F(x)=\frac{x|x|}{2}+C$ , $F'(0)=0$
c) No: two independent constants, one per piece
d) The interval hypothesis

a)

For $x\geq 0$ , $|x|=x$ , so an antiderivative is $\frac{x^{2}}{2}+C_{1}$ . For $x<0$ , $|x|=-x$ , so an antiderivative is $-\frac{x^{2}}{2}+C_{2}$ . These two pieces must JOIN UP: an antiderivative is differentiable, hence continuous, and continuity at 0 forces $\frac{0}{2}+C_{1}=-\frac{0}{2}+C_{2}$ , that is, $C_{1}=C_{2}$ . So there is only one constant, not two: $F(x)=\frac{x^{2}}{2}+C$ for $x\geq 0$ , $F(x)=-\frac{x^{2}}{2}+C$ for $x<0$ .

b)

The two pieces combine into a single formula: if $x\geq 0$ , $\frac{x|x|}{2}=\frac{x\cdot x}{2}=\frac{x^{2}}{2}$ ✓; if $x<0$ , $\frac{x|x|}{2}=\frac{x(-x)}{2}=-\frac{x^{2}}{2}$ ✓. So $F(x)=\frac{x|x|}{2}+C$ . Differentiability at 0: the difference quotient is $\frac{F(h)-F(0)}{h}=\frac{h|h|/2}{h}=\frac{|h|}{2}$ , which tends to 0 as $h\to 0$ from BOTH sides. So $F'(0)=0=|0|$ ✓. The antiderivative is indeed differentiable everywhere, including at the corner of the integrand: integration SMOOTHS, whereas differentiation breaks. This is a general observation, and it is why a function that is merely continuous still has differentiable antiderivatives.

c)

NO, not as the GENERAL antiderivative. It is true that $\left(\ln|x|\right)'=\frac{1}{x}$ on all of $\mathbb{R}\setminus\{0\}$ . But the domain is in TWO disjoint pieces, $\left(-\infty, 0\right)$ and $\left(0, +\infty\right)$ , and nothing forces the constant to be the same on both. The general antiderivative is therefore $F(x)=\ln|x|+C_{1}$ for $x>0$ and $F(x)=\ln|x|+C_{2}$ for $x<0$ , with $C_{1}$ and $C_{2}$ INDEPENDENT. For example, the function equal to $\ln x$ for $x>0$ and to $\ln(-x)+7$ for $x<0$ does have derivative $\frac{1}{x}$ everywhere, and it cannot be written $\ln|x|+C$ . Unlike part a), no continuity condition links the two constants, since the pieces do not touch.

d)

The hypothesis at stake is that of an INTERVAL. The theorem reads: if $F'=G'$ ON AN INTERVAL, then $F-G$ is constant there. Its proof relies on the mean value theorem, which requires being able to connect two points while staying in the domain. On $\mathbb{R}\setminus\{0\}$ , which is not an interval, you cannot go from a negative point to a positive point without crossing 0, where the function does not exist: the argument breaks down, and the conclusion falls with it. In practice in CEGEP, we still write $\int\frac{dx}{x}=\ln|x|+C$ because we almost always work on only one of the two pieces at a time; but as soon as a definite integral crosses 0, that expression becomes wrong, and the integral is in fact divergent. That is exactly the pitfall of the chapter on improper integrals.

Exercise 5: Recognize the form: the table of derivatives backwards

Three patterns come back in half of all exam integrals. Spotting them in a few seconds beats any heavy technique.

a) State the three forms $\int\frac{f'}{f}$ , $\int f'f^{n}$ and $\int f'e^{f}$ , with their antiderivatives.
b) Compute $\int\frac{2x}{x^{2}+1}dx$ , $\int\frac{e^{x}}{e^{x}+3}dx$ and $\int\frac{dx}{x\ln x}$ , identifying $f$ each time.
c) Compute $\int 3x^{2}\left(x^{3}+1\right)^{4}dx$ and $\int\cos x\,e^{\sin x}dx$ .
d) Compute $\int\frac{x}{x^{2}+1}dx$ : the form is not exactly the one in a). What adjustment is needed, and why is it allowed?

Type your answers, the page tells you right or wrong 0/7

a)

\int f'e^{f}

b)

\int\frac{e^{x}}{e^{x}+3}dx

\int\frac{dx}{x\ln x}

c)

\int 3x^{2}(x^{3}+1)^{4}dx

\int\cos x\,e^{\sin x}dx

d)

Constant in front of

\ln(x^{2}+1)

The adjustment is allowed because the missing factor is

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Answers

a) $\ln|f|+C$ ; $\frac{f^{n+1}}{n+1}+C$ ; $e^{f}+C$
b) $\ln(x^{2}+1)$ ; $\ln(e^{x}+3)$ ; $\ln|\ln x|$ , plus $C$
c) $\frac{(x^{3}+1)^{5}}{5}+C$ ; $e^{\sin x}+C$
d) $\frac{1}{2}\ln(x^{2}+1)+C$ : pull out the constant $\frac{1}{2}$

a)

The three forms, all coming from the chain rule read backwards. $\int\frac{f'(x)}{f(x)}dx=\ln\left|f(x)\right|+C$ . $\int f'(x)\left[f(x)\right]^{n}dx=\frac{\left[f(x)\right]^{n+1}}{n+1}+C$ for $n\neq-1$ . $\int f'(x)e^{f(x)}dx=e^{f(x)}+C$ . In all three cases the signature is the same: a composite function MULTIPLIED by the derivative of its inside.

b)

First case: $f=x^{2}+1$ , $f'=2x$ , exactly in the numerator, so $\int\frac{2x}{x^{2}+1}dx=\ln\left(x^{2}+1\right)+C$ (the absolute value is unnecessary, since $x^{2}+1$ is always positive). Second: $f=e^{x}+3$ , $f'=e^{x}$ , present, so $\int\frac{e^{x}}{e^{x}+3}dx=\ln\left(e^{x}+3\right)+C$ . Third: rewrite $\frac{1}{x\ln x}=\frac{1/x}{\ln x}$ , so $f=\ln x$ and $f'=\frac{1}{x}$ , hence $\int\frac{dx}{x\ln x}=\ln\left|\ln x\right|+C$ . A logarithm of a logarithm, perfectly legitimate.

c)

First: $f=x^{3}+1$ , $f'=3x^{2}$ , exactly as a factor, with $n=4$ : $\int 3x^{2}\left(x^{3}+1\right)^{4}dx=\frac{\left(x^{3}+1\right)^{5}}{5}+C$ . Second: $f=\sin x$ , $f'=\cos x$ , present: $\int\cos x\,e^{\sin x}dx=e^{\sin x}+C$ . Systematic check by differentiation: $\left(e^{\sin x}\right)'=\cos x\,e^{\sin x}$ ✓.

d)

The derivative of $f=x^{2}+1$ is $2x$ , but the numerator only contains $x$ : the factor 2 is missing. We ADJUST by writing $\frac{x}{x^{2}+1}=\frac{1}{2}\cdot\frac{2x}{x^{2}+1}$ , hence $\int\frac{x}{x^{2}+1}dx=\frac{1}{2}\ln\left(x^{2}+1\right)+C$ . The adjustment is allowed because $\frac{1}{2}$ is a CONSTANT, and linearity lets you pull constants out of the integral. It would be forbidden if a factor in $x$ were missing: you can never write $\int x\,g(x)dx=x\int g(x)dx$ . All the flexibility of the method lies in this distinction, and it is what separates a correct substitution from a faulty argument.

Exercise 6: Direct substitution: recognizing the hidden derivative

Substitution is the only tool in this chapter that goes beyond the table. Its signature is always the same: a function and its derivative live side by side in the integrand.

a) Compute $\int 2x\left(x^{2}+5\right)^{7}dx$ , writing out $u$ , $du$ and the return to $x$ .
b) Compute $\int\frac{3x^{2}}{x^{3}+4}dx$ , then $\int\frac{\ln x}{x}dx$ . What structure do these two integrals share?
c) Compute $\int x\,e^{x^{2}}dx$ . Compare with $\int e^{x^{2}}dx$ : what difference does the factor $x$ make?
d) Compute $\int\cos^{3}x\sin x\,dx$ , then state in one sentence the test that decides whether a substitution will work.

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Answers

a) $\frac{(x^{2}+5)^{8}}{8}+C$
b) $\ln|x^{3}+4|+C$ ; $\frac{(\ln x)^{2}}{2}+C$
c) $\frac{e^{x^{2}}}{2}+C$ ; $\int e^{x^{2}}dx$ is not elementary
d) $-\frac{\cos^{4}x}{4}+C$ ; no $x$ may remain

a)

Set $u=x^{2}+5$ , so $du=2x\,dx$ : the factor $2x\,dx$ in the integral is exactly $du$ . The integral becomes $\int u^{7}du=\frac{u^{8}}{8}+C$ , and returning to $x$ gives $\frac{(x^{2}+5)^{8}}{8}+C$ . The return is COMPULSORY for an indefinite integral: an answer left in $u$ does not answer the question, which was about a function of $x$ . Check by differentiation: $\frac{8(x^{2}+5)^{7}\cdot 2x}{8}=2x(x^{2}+5)^{7}$ ✓.

b)

First case: $u=x^{3}+4$ , $du=3x^{2}dx$ , the integral becomes $\int\frac{du}{u}=\ln|u|+C=\ln\left|x^{3}+4\right|+C$ . Second case: $u=\ln x$ , $du=\frac{dx}{x}$ , the integral becomes $\int u\,du=\frac{u^{2}}{2}+C=\frac{(\ln x)^{2}}{2}+C$ . The common structure is $\int f(g(x))\,g'(x)\,dx$ : a composite function multiplied by the derivative of its inside. It is the chain rule read backwards, and it is the only thing substitution knows how to undo. Note along the way the form $\int\frac{g'}{g}dx=\ln|g|+C$ , which comes up so often that it deserves to be recognized directly.

c)

With $u=x^{2}$ , $du=2x\,dx$ , so $x\,dx=\frac{du}{2}$ : $\int xe^{x^{2}}dx=\frac{1}{2}\int e^{u}du=\frac{e^{x^{2}}}{2}+C$ . On the other hand, $\int e^{x^{2}}dx$ has NO elementary antiderivative: no finite combination of the usual functions gives it, and this is not a matter of skill but a theorem. So the factor $x$ changes everything: it provides the $du$ that makes the substitution possible. It is the most instructive comparison in the chapter, because it shows that integration is not a guaranteed procedure the way differentiation is. An elementary function always has an elementary derivative; it does not always have an elementary antiderivative.

d)

Set $u=\cos x$ , $du=-\sin x\,dx$ , so $\sin x\,dx=-du$ : $\int\cos^{3}x\sin x\,dx=-\int u^{3}du=-\frac{u^{4}}{4}+C=-\frac{\cos^{4}x}{4}+C$ . The test, in one sentence: the substitution $u=g(x)$ works if, after replacing $g(x)$ by $u$ and $g'(x)dx$ by $du$ , there is NO $x$ LEFT in the integral. That is the complete working criterion, and it can be checked in ten seconds before committing to the calculation. If a lone $x$ remains, there are two ways out: express it in terms of $u$ (the back-substitution of the next exercise), or change technique.

Exercise 7: When substitution resists: constants and back-substitution

Two complications separate exam substitution from textbook substitution: the derivative present up to a constant factor, and the leftover $x$ that has to be re-expressed in terms of $u$ .

a) Compute $\int x^{2}\left(x^{3}-2\right)^{5}dx$ : the derivative of the inside is not exactly there, so how do you adjust, and why are you allowed to?
b) Compute $\int\frac{dx}{\left(4x+3\right)^{2}}$ and $\int\sqrt{7-2x}\,dx$ (linear arguments).
c) Compute $\int x\sqrt{x+1}\,dx$ . Here the substitution $u=x+1$ leaves an $x$ standing: what do you do, and why is this move legitimate?
d) A student writes $\int x^{2}\sqrt{x^{3}+1}\,dx=\frac{x^{3}}{3}\cdot\frac{2}{3}\left(x^{3}+1\right)^{3/2}+C$ , “pulling out” the $x^{2}$ the way you pull out a constant. Diagnose the error precisely.

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Answers

a) $\frac{(x^{3}-2)^{6}}{18}+C$
b) $-\frac{1}{4(4x+3)}+C$ ; $-\frac{(7-2x)^{3/2}}{3}+C$
c) $\frac{2}{5}(x+1)^{5/2}-\frac{2}{3}(x+1)^{3/2}+C$
d) A factor in $x$ does not come out: $\frac{2}{9}(x^{3}+1)^{3/2}+C$

a)

Set $u=x^{3}-2$ , so $du=3x^{2}dx$ . The integral contains $x^{2}dx$ , that is, $\frac{du}{3}$ : we write $\int x^{2}(x^{3}-2)^{5}dx=\frac{1}{3}\int u^{5}du=\frac{u^{6}}{18}+C=\frac{(x^{3}-2)^{6}}{18}+C$ . We are allowed to because $\frac{1}{3}$ is a CONSTANT, and linearity lets you pull constants out. It would be forbidden with a factor in $x$ : $\int x\,f(x)dx\neq x\int f(x)dx$ , ever. This is the decisive distinction of the chapter, and it separates a correct substitution from the mistake in part d).

b)

Linear arguments, so divide by the coefficient of $x$ . First case: $u=4x+3$ , $du=4dx$ , $\int\frac{dx}{(4x+3)^{2}}=\frac{1}{4}\int u^{-2}du=-\frac{1}{4u}+C=-\frac{1}{4(4x+3)}+C$ . Second case: $u=7-2x$ , $du=-2dx$ , $\int\sqrt{7-2x}\,dx=-\frac{1}{2}\int\sqrt{u}\,du=-\frac{1}{3}u^{3/2}+C=-\frac{(7-2x)^{3/2}}{3}+C$ . The negative sign comes from the coefficient $-2$ and is very often forgotten: checking by differentiation catches it every time.

c)

With $u=x+1$ we have $du=dx$ and $\sqrt{x+1}=\sqrt{u}$ , but the factor $x$ remains. Express it in terms of $u$ : since $u=x+1$ , $x=u-1$ . The integral becomes $\int(u-1)\sqrt{u}\,du=\int\left(u^{3/2}-u^{1/2}\right)du=\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}+C$ , that is, $\frac{2}{5}(x+1)^{5/2}-\frac{2}{3}(x+1)^{3/2}+C$ . The move is legitimate because a change of variable is a COMPLETE SUBSTITUTION: everything that depends on $x$ must be rewritten in terms of $u$ , including the $x$ 's that were not part of the original plan. The relation $u=x+1$ is invertible, so it gives $x=u-1$ for free. This back-substitution is the move the exam rewards most reliably, because it requires understanding the substitution rather than just recognizing it.

d)

The error is treating $x^{2}$ as a constant, “integrating it on its own” as $\frac{x^{3}}{3}$ while integrating the rest separately. No rule allows that: the integral of a product is not the product of the integrals, and pulling a factor out of the integral is only allowed if it is CONSTANT. The immediate refutation is by differentiation: the proposed answer needs the product rule and gives back $\frac{2}{3}x^{2}\left(x^{3}+1\right)^{3/2}+x^{5}\sqrt{x^{3}+1}$ , which is not the integrand $x^{2}\sqrt{x^{3}+1}$ : it is wrong. The correct calculation is an ordinary substitution: $u=x^{3}+1$ , $du=3x^{2}dx$ , hence $\frac{1}{3}\int\sqrt{u}\,du=\frac{2}{9}\left(x^{3}+1\right)^{3/2}+C$ . The irony: the student had the right $x^{2}$ right there, and it was exactly the $du$ needed.

Exercise 8: Synthesis: from antiderivative to motion

The indefinite integral is not an exercise in symbolic computation: it is the reconstruction of a quantity from its rate of change. Physics makes the most direct use of it.

a) A particle has acceleration $a(t)=6t-4$ (in m/s²), with $v(0)=2$ m/s and $s(0)=3$ m. Find $v(t)$ , then $s(t)$ , justifying each constant.
b) The rate of change of a population is $P'(t)=200e^{0.1t}$ individuals per year, and $P(0)=5000$ . Find $P(t)$ , then the population after 10 years.
c) Compute $\int\frac{2x+3}{x^{2}+3x+1}dx$ and $\int\frac{dx}{x\left(\ln x\right)^{2}}$ : what common structure governs these two integrals?
d) Faced with an unfamiliar integral from this chapter, in what order should you ask the questions? Illustrate each step with one of the integrals already met.

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Answers

a) $v=3t^{2}-4t+2$ ; $s=t^{3}-2t^{2}+2t+3$
b) $P(t)=2000e^{0.1t}+3000$ ; $P(10)\approx 8437$
c) $\ln|x^{2}+3x+1|+C$ ; $-\frac{1}{\ln x}+C$
d) Simplify, table, substitution, back-substitution, then differentiate

a)

Acceleration is the derivative of velocity, so velocity is an antiderivative of acceleration: $v(t)=3t^{2}-4t+C_{1}$ . The condition $v(0)=2$ gives $C_{1}=2$ , hence $v(t)=3t^{2}-4t+2$ . Velocity is in turn the derivative of position: $s(t)=t^{3}-2t^{2}+2t+C_{2}$ , and $s(0)=3$ gives $C_{2}=3$ , hence $s(t)=t^{3}-2t^{2}+2t+3$ . Each integration produces its own constant, and each constant calls for its own initial condition: two integrations, two conditions, which is exactly why mechanics needs both the initial position AND the initial velocity to predict a motion. Check: $s'(t)=3t^{2}-4t+2=v(t)$ ✓ and $v'(t)=6t-4=a(t)$ ✓. The figure shows the consequence: the velocity never vanishes (the discriminant of the quadratic is 16 − 24, which is negative) and stays positive, so the position keeps increasing.

b)

$P(t)=\int 200e^{0.1t}dt=\frac{200}{0.1}e^{0.1t}+C=2000e^{0.1t}+C$ . The condition $P(0)=5000$ gives $2000+C=5000$ , so $C=3000$ and $P(t)=2000e^{0.1t}+3000$ . After 10 years: $P(10)=2000e^{1}+3000\approx 2000(2.71828)+3000\approx 8437$ individuals. Note that the constant $3000$ is not a cosmetic detail: it represents the part of the population that does not take part in the exponential growth, and forgetting it would give about $5437$ individuals instead of $8437$ , more than a third less.

c)

Both are of the form $\int\frac{g'(x)}{\left(g(x)\right)^{n}}dx$ . First case: $g=x^{2}+3x+1$ , $g'=2x+3$ is exactly in the numerator, so $\int\frac{2x+3}{x^{2}+3x+1}dx=\ln\left|x^{2}+3x+1\right|+C$ (the case $n=1$ , which gives the logarithm). Second case: rewrite $\frac{1}{x(\ln x)^{2}}=\frac{1}{(\ln x)^{2}}\cdot\frac{1}{x}$ , with $g=\ln x$ and $g'=\frac{1}{x}$ present: $\int u^{-2}du=-\frac{1}{u}+C=-\frac{1}{\ln x}+C$ (the case $n=2$ , which gives a power). The common structure is therefore the same substitution, and only the value of $n$ decides whether the answer is a logarithm or a power: it is the gap at $n=-1$ in the table once again, met in another disguise.

d)

The efficient order: 1) Can I SIMPLIFY the integrand algebraically (distribute, split a fraction, cancel)? Example: $\int\frac{x^{2}+3x}{x}dx$ , which becomes $\int(x+3)dx$ . 2) Is it a DIRECT form from the table, possibly with a linear argument? Example: $\int\sin(5x)dx$ , where dividing by 5 is enough. 3) Do I see a function AND its derivative, up to a constant? Then SUBSTITUTION. Example: $\int x^{2}(x^{3}-2)^{5}dx$ . 4) Does the substitution leave an $x$ standing? Then express it in terms of $u$ . Example: $\int x\sqrt{x+1}\,dx$ . 5) None of this works? The integral belongs to the next chapter (parts, partial fractions, trigonometry), or has no elementary antiderivative, like $\int e^{x^{2}}dx$ . And in every case: DIFFERENTIATE the answer before handing in the paper.

Exercise 9: Antiderivatives that do not exist

Every elementary function has an elementary derivative. The converse is FALSE, and that is a theorem, not an admission of ignorance.

a) What exactly does “elementary function” mean? Give three common integrands whose antiderivative is not elementary.
b) Compare $\int xe^{x^{2}}dx$ and $\int e^{x^{2}}dx$ : the first can be computed, the second cannot. What changes?
c) Does the function $e^{-x^{2}}$ have antiderivatives? Reconcile your answer with part a).
d) What do you actually do when faced with $\int_{0}^{1}e^{-x^{2}}dx$ ? Name the two routes from the course and say which one comes with an error guarantee.

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Answers

a) Finitely many sums, products, quotients, compositions; $e^{-x^{2}}$ , $\frac{\sin x}{x}$ , $\sqrt{1+x^{3}}$
b) The factor $x$ provides the $du$
c) Yes: $\int_{0}^{x}e^{-t^{2}}dt$ exists, without being elementary
d) Series or Simpson, $\approx 0.7468$ , both with an error bound

a)

An ELEMENTARY function is one obtained from polynomials, exponentials, logarithms, trigonometric functions and their inverses by a FINITE number of sums, products, quotients and compositions. Three common integrands with no elementary antiderivative: $e^{-x^{2}}$ (the normal distribution in statistics), $\frac{\sin x}{x}$ (the sinc function, in optics and signal processing), and $\sqrt{1+x^{3}}$ (the arc length of certain curves). You can add $\frac{1}{\ln x}$ and $\frac{e^{x}}{x}$ . Liouville's theorem proves this impossibility: it is not that nobody has found it yet, it is that there is nothing to find.

b)

What changes is the presence of the factor $x$ . For the first one: $u=x^{2}$ gives $du=2x\,dx$ , so the $x\,dx$ present equals $\frac{du}{2}$ and the integral becomes $\frac{1}{2}\int e^{u}du=\frac{e^{x^{2}}}{2}+C$ . The substitution works BECAUSE the derivative of the inside is there, up to a constant. For the second one, that factor is missing: no substitution can absorb the $dx$ , and we saw in part a) that no other technique will succeed either. A seemingly harmless factor $x$ thus makes all the difference between a thirty-second integral and an impossible one: it is the most instructive comparison in the chapter.

c)

YES, it does. There is no contradiction, because these are two DIFFERENT statements. “Having an antiderivative” and “having an ELEMENTARY antiderivative” are not the same thing. The function $e^{-x^{2}}$ is continuous on $\mathbb{R}$ , and the fundamental theorem of calculus then guarantees that the function $F(x)=\int_{0}^{x}e^{-t^{2}}dt$ is differentiable with derivative $e^{-x^{2}}$ : it really is an antiderivative, it EXISTS. It simply cannot be written with the usual symbols. Mathematicians have in fact given it a name, the error function $\operatorname{erf}$ , just as $\ln$ was once the name given to the antiderivative of $\frac{1}{x}$ . A “new” function is often just an antiderivative that finally got a name.

d)

Two routes. First: SERIES EXPANSION. Replace $e^{-x^{2}}$ by its Maclaurin series $1-x^{2}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+\cdots$ , and integrate term by term, which gives $1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\cdots\approx 0.7468$ . Since the series is alternating, the error is bounded by the first term left out: the guarantee is explicit. Second: NUMERICAL INTEGRATION, by Simpson's rule for example, which gives $0.746855$ with $n=4$ against an exact value of $0.746824$ ; here too an error bound is available, $\frac{M(b-a)^{5}}{180n^{4}}$ . BOTH routes therefore come with an error guarantee, which is precisely what makes them acceptable in science: an approximation without a bound is worthless. The series is more convenient near its centre of expansion; the numerical method does not care about that, and even applies to data with no formula.

Exercise 10: The method of undetermined coefficients

When you GUESS the form of the antiderivative, all that remains is to adjust coefficients by matching. The technique sometimes saves several integrations by parts.

a) Find $a$ and $b$ so that $F(x)=(ax+b)e^{x}$ is an antiderivative of $xe^{x}$ .
b) Find $a$ , $b$ , $c$ so that $F(x)=\left(ax^{2}+bx+c\right)e^{x}$ is an antiderivative of $x^{2}e^{x}$ .
c) Guess the form, then find the coefficients of an antiderivative of $e^{x}\sin x$ .
d) What is the advantage of this method over integration by parts, and what is its limitation?

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Answers

a) $a=1$ , $b=-1$ : $(x-1)e^{x}+C$
b) $(x^{2}-2x+2)e^{x}+C$
c) $\frac{e^{x}(\sin x-\cos x)}{2}+C$
d) Fast and self-checking; you have to guess the form

a)

Differentiate the assumed form: $F'(x)=ae^{x}+(ax+b)e^{x}=e^{x}\left(ax+a+b\right)$ . We want $F'(x)=xe^{x}=e^{x}\cdot x$ . Dividing by $e^{x}$ , which is never zero, leaves $ax+(a+b)=x$ . By MATCHING coefficients: the coefficient of $x$ gives $a=1$ ; the constant term gives $a+b=0$ , so $b=-1$ . Hence $\int xe^{x}dx=(x-1)e^{x}+C$ . Check: $\left[(x-1)e^{x}\right]'=e^{x}+(x-1)e^{x}=xe^{x}$ ✓.

b)

$F'(x)=(2ax+b)e^{x}+\left(ax^{2}+bx+c\right)e^{x}=e^{x}\left[ax^{2}+(2a+b)x+(b+c)\right]$ . We want $x^{2}e^{x}$ , so after dividing by $e^{x}$ : $ax^{2}+(2a+b)x+(b+c)=x^{2}$ . Matching: $a=1$ ; then $2a+b=0$ gives $b=-2$ ; then $b+c=0$ gives $c=2$ . Hence $\int x^{2}e^{x}dx=\left(x^{2}-2x+2\right)e^{x}+C$ , which agrees with the result of two successive integrations by parts ✓. A single linear system has replaced two rounds of parts.

c)

Differentiating a product $e^{x}\times$ trigonometric function always regenerates both functions $\sin$ and $\cos$ : so we guess the form $F(x)=e^{x}\left(a\sin x+b\cos x\right)$ . Differentiate: $F'(x)=e^{x}\left(a\sin x+b\cos x\right)+e^{x}\left(a\cos x-b\sin x\right)=e^{x}\left[(a-b)\sin x+(a+b)\cos x\right]$ . We want $e^{x}\sin x$ , so $a-b=1$ and $a+b=0$ . The second gives $b=-a$ , and substituting: $2a=1$ , so $a=\frac{1}{2}$ and $b=-\frac{1}{2}$ . Conclusion: $\int e^{x}\sin x\,dx=\frac{e^{x}\left(\sin x-\cos x\right)}{2}+C$ . This is the so-called CYCLIC case, which parts solve in two rounds plus an equation in $I$ ; here a single matching is enough.

d)

ADVANTAGE: the method replaces several successive integrations with a single differentiation followed by a linear system, which is faster and above all less exposed to sign errors, very common in repeated parts. It also makes checking immediate, since the answer has been differentiated along the way. LIMITATION: you have to GUESS the right form at the start, which is only possible for well-identified families, essentially polynomial times exponential, polynomial times sine or cosine, and exponential times trigonometric function. Faced with $\int\ln x\,dx$ or $\int\frac{dx}{x^{2}+1}$ , no obvious form comes to mind and the method cannot get started. It therefore complements integration by parts without replacing it, just as the method of undetermined coefficients will complement variation of parameters in the chapter on differential equations.

Exercise 11: Other bases: 10 to the x, the common log and the three e's

The table gives $\int e^{x}dx=e^{x}+C$ and $\int\frac{dx}{x}=\ln|x|+C$ . As soon as the base is no longer $e$ , a natural logarithm appears in the denominator, and that is the one people forget.

a) Compute $\int 10^{x}dx$ and give the decimal value of the coefficient obtained in front of $10^{x}$ .
b) Compute $\int 3^{2x}dx$ in two ways: by the substitution $u=2x$ , then by writing $3^{2x}=9^{x}$ . Show that the two answers agree.
c) Compute $\int x\,5^{x^{2}}dx$ .
d) Compute $\int\frac{\log_{10}x}{x}dx$ , starting by writing $\log_{10}x=\frac{\ln x}{\ln 10}$ .
e) Compute $\int x^{e}dx$ , $\int e^{x}dx$ and $\int e^{e}dx$ . Why do these three integrals, which look alike, fall under three different rules?

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Answers

a) $\frac{10^{x}}{\ln 10}+C$ , coefficient $\approx 0.4343$
b) $\frac{3^{2x}}{2\ln 3}+C=\frac{9^{x}}{\ln 9}+C$
c) $\frac{5^{x^{2}}}{2\ln 5}+C$
d) $\frac{(\ln x)^{2}}{2\ln 10}+C$
e) $\frac{x^{e+1}}{e+1}$ , $e^{x}$ , $e^{e}x$ : power, exponential, constant

a)

Since $\left(10^{x}\right)'=10^{x}\ln 10$ , we compensate: $\int 10^{x}dx=\frac{10^{x}}{\ln 10}+C$ . The coefficient is $\frac{1}{\ln 10}\approx 0.4343$ . Check: $\left(\frac{10^{x}}{\ln 10}\right)'=\frac{10^{x}\ln 10}{\ln 10}=10^{x}$ ✓.

b)

Substitution: $u=2x$ , $du=2\,dx$ , so $\int 3^{2x}dx=\frac{1}{2}\int 3^{u}du=\frac{3^{2x}}{2\ln 3}+C$ . Rewriting: $3^{2x}=\left(3^{2}\right)^{x}=9^{x}$ , so $\int 9^{x}dx=\frac{9^{x}}{\ln 9}+C$ . Now $\ln 9=\ln 3^{2}=2\ln 3$ and $9^{x}=3^{2x}$ : the two answers are identical, with coefficient $\frac{1}{2\ln 3}\approx 0.4551$ .

c)

$u=x^{2}$ , $du=2x\,dx$ , so $x\,dx=\frac{du}{2}$ : $\int x\,5^{x^{2}}dx=\frac{1}{2}\int 5^{u}du=\frac{5^{x^{2}}}{2\ln 5}+C$ . The coefficient is $\frac{1}{2\ln 5}\approx 0.3107$ . Two factors pile up in the denominator: the $2$ from the substitution and the $\ln 5$ from the base.

d)

$\int\frac{\log_{10}x}{x}dx=\frac{1}{\ln 10}\int\frac{\ln x}{x}dx$ . With $u=\ln x$ , $du=\frac{dx}{x}$ : $\frac{1}{\ln 10}\cdot\frac{(\ln x)^{2}}{2}+C=\frac{(\ln x)^{2}}{2\ln 10}+C$ , with coefficient $\frac{1}{2\ln 10}\approx 0.2171$ . Without the change of base, it is tempting to write $\frac{(\log_{10}x)^{2}}{2}$ , which is wrong: its derivative is $\frac{\log_{10}x}{x\ln 10}$ , a factor $\ln 10$ is missing.

e)

$\int x^{e}dx=\frac{x^{e+1}}{e+1}+C$ : it is a POWER of $x$ with a constant exponent, so the power rule applies. $\int e^{x}dx=e^{x}+C$ : it is an EXPONENTIAL, the variable is in the exponent. $\int e^{e}dx=e^{e}x+C$ : it is a CONSTANT, about $15.15$ , which integrates to $e^{e}x$ . The rule depends on where the variable sits, never on what the symbols look like.

Exercise 12: Recovering a function from its second derivative and a tangent line

A function $f$ satisfies $f''(x)=6x-12$ for every real $x$ , and the line with equation $y=3x-2$ is tangent to its graph at the point with $x$ -coordinate $1$ .

a) Write $f'(x)$ with a constant $C_{1}$ . Which piece of information in the statement fixes $C_{1}$ ? Compute it.
b) Write $f(x)$ with a constant $C_{2}$ . Which second piece of information fixes $C_{2}$ ? Compute it.
c) Compute $f(2)$ , then check that $f(x)=(x-2)^{3}+2$ .
d) At what other point is the tangent line to the graph parallel to the line $y=3x-2$ ? Give its coordinates and the equation of this tangent line.
e) A student writes “ $f'(1)=3x-2$ evaluated at $1$ , so $f'(1)=1$ ”. What is the student confusing?

Show the solution

Answers

a) $f'(x)=3x^{2}-12x+12$
b) $f(x)=x^{3}-6x^{2}+12x-6$
c) $f(2)=2$ ; $f(x)=(x-2)^{3}+2$
d) $(3, 3)$ , tangent line $y=3x-6$
e) The $y$ -value $1$ is not the slope $3$

a)

$f'(x)=3x^{2}-12x+C_{1}$ . The tangent line at $x=1$ has slope $3$ , so $f'(1)=3$ : $3-12+C_{1}=3$ , hence $C_{1}=12$ and $f'(x)=3x^{2}-12x+12$ . The SLOPE of the tangent line fixes the constant of the derivative.

b)

$f(x)=x^{3}-6x^{2}+12x+C_{2}$ . The point of tangency lies both on the graph and on the tangent line: $f(1)=3\times 1-2=1$ . So $1-6+12+C_{2}=1$ , hence $C_{2}=-6$ and $f(x)=x^{3}-6x^{2}+12x-6$ . A tangent line provides TWO conditions, a slope and a point, exactly what two integrations need.

c)

$f(2)=8-24+24-6=2$ . And $(x-2)^{3}+2=x^{3}-6x^{2}+12x-8+2=x^{3}-6x^{2}+12x-6$ ✓. The graph is that of $x^{3}$ shifted $2$ to the right and $2$ up: $(2, 2)$ is its centre of symmetry and its inflection point, where $f''(x)=6x-12$ is zero.

d)

We want $f'(x)=3$ : $3x^{2}-12x+12=3$ , that is, $x^{2}-4x+3=0$ , with roots $x=1$ and $x=3$ . The other point is $(3, f(3))=(3, 3)$ , since $f(3)=1+2=3$ . The tangent line there has equation $y=3(x-3)+3=3x-6$ , parallel to $y=3x-2$ without being the same line. The two points of tangency are symmetric about the centre $(2, 2)$ , as the symmetry of the graph suggested.

e)

The student confuses the $y$ -VALUE of the line at $x=1$ , which is $1$ and gives $f(1)$ , with its SLOPE, which is $3$ and gives $f'(1)$ . With $f'(1)=1$ , the student would find $C_{1}=10$ and everything after it would be wrong. The figure in the solution shows the graph and its two tangent lines of slope $3$ .

Exercise 13: Five antiderivatives taken from a test paper

Here are five answers taken from the same exam paper. For each one, say whether it is true or false by DIFFERENTIATING the proposed answer, and correct it if it is false.

a) $\int\ln x\,dx=\frac{1}{x}+C$ .
b) $\int\frac{dx}{2x+1}=\ln|2x+1|+C$ .
c) $\int\sec x\tan x\,dx=\sec x+C$ .
d) $\int e^{-x}dx=e^{-x}+C$ .
e) $\int\frac{dx}{(1+x)^{2}}=-\frac{1}{1+x}+C$ .

Show the solution

Answers

a) False: $x\ln x-x+C$
b) False: $\frac{1}{2}\ln|2x+1|+C$
c) True
d) False: $-e^{-x}+C$
e) True

a)

FALSE. $\left(\frac{1}{x}\right)'=-\frac{1}{x^{2}}$ , not $\ln x$ : the paper gave the DERIVATIVE of $\ln x$ instead of an antiderivative. The correct answer is $x\ln x-x+C$ , which we check: $\left(x\ln x-x\right)'=\ln x+1-1=\ln x$ ✓. It is obtained by parts in the next chapter.

b)

FALSE. $\left(\ln|2x+1|\right)'=\frac{2}{2x+1}$ : twice too much. The chain rule brings out the factor $2$ of the linear argument, so you must divide: $\int\frac{dx}{2x+1}=\frac{1}{2}\ln|2x+1|+C$ .

c)

TRUE. $\left(\sec x\right)'=\left(\frac{1}{\cos x}\right)'=\frac{\sin x}{\cos^{2}x}=\sec x\tan x$ ✓. It is a row of the table of derivatives, read backwards.

d)

FALSE. $\left(e^{-x}\right)'=-e^{-x}$ : the sign is wrong. The correct answer is $-e^{-x}+C$ , since the argument $-x$ has coefficient $-1$ .

e)

TRUE. $\left(-\frac{1}{1+x}\right)'=-\left(-\frac{1}{(1+x)^{2}}\right)=\frac{1}{(1+x)^{2}}$ ✓. With $u=1+x$ : $\int u^{-2}du=-u^{-1}+C$ . Verdict on the paper: two correct answers out of five, and the three wrong ones all fell to differentiation, in under a minute.

Exercise 14: Problem: the deflection of a cantilever beam

A horizontal beam of length $L=3$ m is built into a wall at $x=0$ and free at the other end. Under a distributed load, its deflection curve $y(x)$ (in metres, negative downward) satisfies $y''(x)=-k\left(L-x\right)^{2}$ with $k=0.0004$ m $^{-3}$ . The built-in end forces $y(0)=0$ and $y'(0)=0$ : the beam leaves the wall horizontally.

The figure below shows the situation; it is not to scale.

a) Compute $y'(x)$ in terms of $k$ , $L$ and $x$ . Watch the sign produced by the argument $L-x$ .
b) Compute $y(x)$ in terms of $k$ , $L$ and $x$ .
c) Show that the deflection at the free end is $y(L)=-\frac{kL^{4}}{4}$ , then compute it in millimetres.
d) Compute the slope $y'(L)$ at the end of the beam.
e) The beam is replaced by a beam with the same cross-section that is twice as long, under the same load per metre. By what factor is the deflection multiplied?

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Answers

a) $y'(x)=\frac{k(L-x)^{3}}{3}-\frac{kL^{3}}{3}$
b) $y(x)=-\frac{k(L-x)^{4}}{12}-\frac{kL^{3}x}{3}+\frac{kL^{4}}{12}$
c) $y(L)=-\frac{kL^{4}}{4}=-8.1$ mm
d) $y'(L)=-0.0036$
e) Deflection in $L^{4}$ : multiplied by $16$

a)

With $u=L-x$ , $du=-dx$ : $\int-k(L-x)^{2}dx=k\int u^{2}du=\frac{k(L-x)^{3}}{3}+C_{1}$ . The minus sign of $du=-dx$ has changed the sign of the antiderivative. The condition $y'(0)=0$ gives $\frac{kL^{3}}{3}+C_{1}=0$ , so $y'(x)=\frac{k(L-x)^{3}}{3}-\frac{kL^{3}}{3}$ .

b)

$\int\frac{k(L-x)^{3}}{3}dx=-\frac{k(L-x)^{4}}{12}$ , by the same substitution. Hence $y(x)=-\frac{k(L-x)^{4}}{12}-\frac{kL^{3}}{3}x+C_{2}$ , and $y(0)=0$ gives $-\frac{kL^{4}}{12}+C_{2}=0$ : $y(x)=-\frac{k(L-x)^{4}}{12}-\frac{kL^{3}x}{3}+\frac{kL^{4}}{12}$ .

c)

At $x=L$ , the first term vanishes: $y(L)=-\frac{kL^{4}}{3}+\frac{kL^{4}}{12}=-\frac{4kL^{4}-kL^{4}}{12}=-\frac{kL^{4}}{4}$ . Numerically: $-\frac{0.0004\times 81}{4}=-0.0081$ m, that is, a deflection of $8.1$ mm downward.

d)

$y'(L)=0-\frac{kL^{3}}{3}=-\frac{0.0004\times 27}{3}=-0.0036$ . The beam drops $3.6$ mm per metre at its end, where it is most steeply inclined.

e)

The deflection is proportional to $L^{4}$ : doubling the length multiplies it by $2^{4}=16$ . That is why a balcony twice as deep does not need a beam twice as strong, but far more than that. The figure in the solution plots the deflection curve in millimetres.

Exercise 15: Problem: the sales of a game launched without advertising

A small studio launches a game with no advertising campaign: sales come from word of mouth, which starts slowly and then runs out of steam. The sales rate is modelled by $S'(t)=5000\,t\,e^{-t^{2}/2}$ copies per week, with $t$ in weeks after release, and $S(0)=0$ .

a) Using a substitution that you state, find $S(t)$ .
b) How many copies are sold during the first two weeks?
c) What value does $S(t)$ approach as $t\to+\infty$ ? Interpret.
d) After how many weeks will the studio have sold $90$ percent of that total?
e) At what time is the sales rate at its maximum, and how many copies per week are being sold then?

Show the solution

Answers

a) $S(t)=5000\left(1-e^{-t^{2}/2}\right)$
b) $\approx 4323$ copies
c) $5000$ copies in total
d) $t=\sqrt{2\ln 10}\approx 2.15$ weeks
e) At $t=1$ week, $\approx 3033$ per week

a)

Set $u=-\frac{t^{2}}{2}$ , so $du=-t\,dt$ and $t\,dt=-du$ : $\int 5000\,t\,e^{-t^{2}/2}dt=-5000\int e^{u}du=-5000e^{-t^{2}/2}+C$ . The condition $S(0)=0$ gives $-5000+C=0$ , so $C=5000$ and $S(t)=5000\left(1-e^{-t^{2}/2}\right)$ .

b)

$S(2)=5000\left(1-e^{-2}\right)\approx 5000\times 0.8647\approx 4323$ copies.

c)

$e^{-t^{2}/2}\to 0$ , so $S(t)\to 5000$ . The model predicts a total of $5000$ copies: that is exactly the constant of integration, which here measures the market that word of mouth will eventually reach.

d)

$5000\left(1-e^{-t^{2}/2}\right)=4500$ gives $e^{-t^{2}/2}=0.1$ , that is, $\frac{t^{2}}{2}=\ln 10$ and $t=\sqrt{2\ln 10}\approx 2.15$ weeks. Most of the sales therefore happen in just over two weeks.

e)

Differentiate the rate: $\left(t\,e^{-t^{2}/2}\right)'=e^{-t^{2}/2}-t^{2}e^{-t^{2}/2}=\left(1-t^{2}\right)e^{-t^{2}/2}$ , positive before $t=1$ and negative after. The rate is at its maximum at $t=1$ week and equals $5000e^{-1/2}\approx 3033$ copies per week. The figure in the solution shows both curves: the rate peaks at the inflection point of the cumulative sales.

Corrected exercises: antiderivatives and the indefinite integral (201-SN3-RE)

Course recap

Calculations: antiderivatives to compute with the table

Exercise 0.1: Integrate with the table: twelve direct calculations

Calculations: antiderivatives to compute by substitution

Exercise 0.2: Integrate by direct substitution: ten direct calculations

Part A: The antiderivative, the table and the forms to recognize (/50)

Exercise 1: What is an antiderivative?

Exercise 2: The table of formulas and its single gap

Exercise 3: Linearity, and the two rules that do not exist

Exercise 4: Absolute value and disjoint domains

Exercise 5: Recognize the form: the table of derivatives backwards

Part B: Substitution, the limits of the method and undetermined coefficients (/50)

Exercise 6: Direct substitution: recognizing the hidden derivative

Exercise 7: When substitution resists: constants and back-substitution

Exercise 8: Synthesis: from antiderivative to motion

Exercise 9: Antiderivatives that do not exist

Exercise 10: The method of undetermined coefficients

Part C: the classics (/50)

Exercise 11: Other bases: 10 to the x, the common log and the three e's

Exercise 12: Recovering a function from its second derivative and a tangent line

Exercise 13: Five antiderivatives taken from a test paper

Exercise 14: Problem: the deflection of a cantilever beam

Exercise 15: Problem: the sales of a game launched without advertising

See also

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