Corrected practice final exam in integral calculus, 201-SN3-RE (201-NYB)

Product of a polynomial and a logarithm: integration by parts, with $u=\ln x$ (which we cannot integrate directly) and $dv=x^{2}dx$. Then $du=\frac{dx}{x}$ and $v=\frac{x^{3}}{3}$: $\int x^{2}\ln x\,dx=\frac{x^{3}}{3}\ln x-\int\frac{x^{3}}{3}\cdot\frac{dx}{x}=\frac{x^{3}}{3}\ln x-\frac{x^{3}}{9}+C$.

Irreducible quadratic denominator (discriminant $16-52<0$): complete the square. $x^{2}+4x+13=(x+2)^{2}+9$. With $u=x+2$: $\int\frac{du}{u^{2}+9}=\frac{1}{3}\arctan\frac{u}{3}+C=\frac{1}{3}\arctan\dfrac{x+2}{3}+C$.

A proper rational function with a linear factor and an irreducible quadratic factor: partial fractions. Write $\dfrac{3x+5}{(x-1)(x^{2}+4)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^{2}+4}$, hence $A(x^{2}+4)+(Bx+C)(x-1)=3x+5$. At $x=1$: $5A=8$, so $A=\frac{8}{5}$. Coefficient of $x^{2}$: $A+B=0$, so $B=-\frac{8}{5}$. Constant term: $4A-C=5$, so $C=\frac{7}{5}$. (Check with the coefficient of $x$: $-B+C=\frac{8}{5}+\frac{7}{5}=3$.) The integral is $\frac{8}{5}\ln|x-1|-\frac{4}{5}\ln(x^{2}+4)+\frac{7}{10}\arctan\frac{x}{2}+K$, where the constant of integration is called $K$ because $C$ already names a coefficient.

A square root inside an exponent: substitution first, then parts. With $u=\sqrt{x}$, we have $x=u^{2}$ and $dx=2u\,du$: $\int e^{\sqrt{x}}dx=2\int u\,e^{u}du$. Integration by parts ($u$ against $e^{u}$) gives $2(u-1)e^{u}+C$, so $\int e^{\sqrt{x}}dx=2\left(\sqrt{x}-1\right)e^{\sqrt{x}}+C$.

Write $\int_{x^{2}}^{x^{3}}=\int_{0}^{x^{3}}-\int_{0}^{x^{2}}$, then apply the Fundamental Theorem with the chain rule at each limit: $\frac{d}{dx}\int_{x^{2}}^{x^{3}}\sin(t^{2})\,dt=3x^{2}\sin\left(x^{6}\right)-2x\sin\left(x^{4}\right)$. The classic error is to forget the factor coming from the lower limit, or its minus sign.

$x^{3}\cos x$ is odd (odd times even) and the interval is symmetric, so its integral is zero. What remains is $\int_{-2}^{2}x^{2}dx=2\int_{0}^{2}x^{2}dx=2\cdot\frac{8}{3}=\dfrac{16}{3}$. Recognizing the symmetry saves a needless integration by parts.

Power reduction: $\sin^{2}x=\dfrac{1-\cos 2x}{2}$, so $\int_{0}^{\pi/2}\sin^{2}x\,dx=\left[\dfrac{x}{2}-\dfrac{\sin 2x}{4}\right]_{0}^{\pi/2}=\dfrac{\pi}{4}$. Symmetry: the substitution $x\mapsto\frac{\pi}{2}-x$ swaps $\sin^{2}$ and $\cos^{2}$, so the two integrals are equal; and their sum is $\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2}$. Each one is therefore $\dfrac{\pi}{4}$.

By the Fundamental Theorem, $F'(x)=\dfrac{x^{2}-4}{x^{2}+1}$. The denominator is always positive, so the sign is that of $x^{2}-4=(x-2)(x+2)$: $F$ increases on $(-\infty,-2)$, decreases on $(-2,2)$ and increases on $(2,+\infty)$. It therefore has a relative maximum at $x=-2$ and a relative minimum at $x=2$. Finally $F(1)=\int_{1}^{1}=0$: an integral whose two limits coincide is zero, whatever the function.

Form $\frac{0}{0}$. A first application gives $\dfrac{e^{x}-1}{2x}$, again $\frac{0}{0}$; a second gives $\dfrac{e^{x}}{2}\to\dfrac{1}{2}$. (The Maclaurin expansion $e^{x}=1+x+\frac{x^{2}}{2}+\cdots$ gives the result in one line.)

Form $1^{\infty}$: take the logarithm. $\ln y=3x\ln\left(\dfrac{x+1}{x-2}\right)=3x\ln\left(1+\dfrac{3}{x-2}\right)$. Since $\ln(1+u)\sim u$ as $u\to 0$, we get $\ln y\sim 3x\cdot\dfrac{3}{x-2}\to 9$. So the limit is $e^{9}\approx 8103$.

Hierarchy: any power of the logarithm is negligible compared with any positive power of $x$, so the limit is $0$. With L'Hôpital's rule, four successive applications would be needed: each one lowers the exponent of the logarithm by one, and after four rounds the numerator has no logarithm left at all. The hierarchy gives the result in one line, which is exactly the point of the exercise.

Divide the numerator and the denominator by $x$: $\dfrac{2+\frac{\sin x}{x}}{1+\frac{3}{x}}$. Since $\left|\frac{\sin x}{x}\right|\leq\frac{1}{x}\to 0$ (squeeze theorem), the limit is $\dfrac{2}{1}=2$. L'Hôpital's rule, on the other hand, turns the quotient into $\dfrac{2+\cos x}{1}$, which oscillates between 1 and 3 and has no limit. Conclusion: L'Hôpital's rule is a one-way implication. If the quotient of the derivatives has a limit, then so does the original quotient; but if the quotient of the derivatives has none, nothing can be concluded and you must change methods.

Improper because of the infinite limit. Parts with $u=\ln x$, $dv=x^{-2}dx$: an antiderivative is $-\dfrac{\ln x+1}{x}$. As $x\to\infty$, it tends to 0 (the logarithm loses against $x$); at $x=1$ it equals $-1$. The integral converges and equals $0-(-1)=1$.

Improper because of the singularity at $0$, where $\ln x\to-\infty$. An antiderivative is $x\ln x-x$, and $x\ln x\to 0$ as $x\to 0^{+}$ (growth rates). The integral converges and equals $(1\cdot 0-1)-(0-0)=-1$. An integral can perfectly well converge to a negative value while the function heads to $-\infty$.

Improper because of the infinite limit only: the pole $x=2$ lies outside the interval. Partial fractions: $\dfrac{1}{x^{2}-4}=\dfrac{1}{4}\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right)$, with antiderivative $\dfrac{1}{4}\ln\left|\dfrac{x-2}{x+2}\right|$. As $x\to\infty$, the ratio tends to 1 and the logarithm to 0; at $x=3$ the antiderivative equals $\frac{1}{4}\ln\frac{1}{5}$. The integral converges and equals $\dfrac{1}{4}\ln 5\approx 0.402$. Note that each piece taken separately diverges: it is their difference that converges, so you must find the antiderivative before taking the limit.

First: for $x\geq 1$, $x+e^{x}>e^{x}$, so $0<\dfrac{1}{x+e^{x}}<e^{-x}$. Since $\int_{1}^{\infty}e^{-x}dx=e^{-1}$ converges, the direct comparison test gives convergence. Second: $2+\sin x\geq 1$ for all $x$, so $\dfrac{2+\sin x}{x}\geq\dfrac{1}{x}$. Since $\int_{1}^{\infty}\dfrac{dx}{x}$ diverges, direct comparison gives divergence. The oscillation of the sine changes nothing: the numerator stays bounded away from zero.

On $[0,1]$, $\sqrt{x}\geq x^{2}$. The area is $\int_{0}^{1}\left(\sqrt{x}-x^{2}\right)dx=\left[\frac{2}{3}x^{3/2}-\frac{x^{3}}{3}\right]_{0}^{1}=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$.

Rotation about the $x$-axis: washers, with outer radius $\sqrt{x}$ and inner radius $x^{2}$. $V_{x}=\pi\int_{0}^{1}\left(x-x^{4}\right)dx=\pi\left(\frac{1}{2}-\frac{1}{5}\right)=\frac{3\pi}{10}$. The fatal error would be to write $\pi\int(\sqrt{x}-x^{2})^{2}$: what you need is the difference of the squares, never the square of the difference.

About the $y$-axis, cylindrical shells avoid rewriting the curves in terms of $y$: radius $x$, height $\sqrt{x}-x^{2}$. $V_{y}=2\pi\int_{0}^{1}x\left(\sqrt{x}-x^{2}\right)dx=2\pi\int_{0}^{1}\left(x^{3/2}-x^{3}\right)dx=2\pi\left(\frac{2}{5}-\frac{1}{4}\right)=\frac{3\pi}{10}$. The two volumes are equal, and this is no coincidence: $y=\sqrt{x}$ and $y=x^{2}$ are inverse functions of each other, so the region $R$ is symmetric with respect to the line $y=x$. This reflection swaps the two axes, so it swaps the two solids of revolution, which therefore have the same volume.

About $x=2$, shells again, but the radius becomes the distance to the axis, namely $2-x$ (positive on $[0,1]$): $V=2\pi\int_{0}^{1}(2-x)\left(\sqrt{x}-x^{2}\right)dx=2\pi\left[2\left(\frac{2}{3}-\frac{1}{3}\right)-\left(\frac{2}{5}-\frac{1}{4}\right)\right]=2\pi\left(\frac{2}{3}-\frac{3}{20}\right)=\frac{31\pi}{30}\approx 3.246$.

$y'=x^{1/2}$, so $1+(y')^{2}=1+x$ and $L=\int_{0}^{3}\sqrt{1+x}\,dx=\left[\frac{2}{3}(1+x)^{3/2}\right]_{0}^{3}=\frac{2}{3}(8-1)=\frac{14}{3}\approx 4.67$.

$y=\sqrt{4-x^{2}}$ gives $y'=\dfrac{-x}{\sqrt{4-x^{2}}}$, so $1+(y')^{2}=\dfrac{4}{4-x^{2}}$ and $\sqrt{1+(y')^{2}}=\dfrac{2}{\sqrt{4-x^{2}}}$. The surface area formula gives $S=2\pi\int_{0}^{1}y\sqrt{1+(y')^{2}}\,dx=2\pi\int_{0}^{1}\sqrt{4-x^{2}}\cdot\frac{2}{\sqrt{4-x^{2}}}\,dx=4\pi\int_{0}^{1}dx=4\pi$. The radical disappears completely, and the area equals $2\pi r h$ with $r=2$ and $h=1$: this is Archimedes' theorem, the area of a spherical zone depends only on its height, not on its position on the sphere.

One could let $y$ be the length already hauled up, with $20-y$ metres still hanging, but it is simpler to add up the work slice by slice of cable. So let $y$ be the depth of a slice below the top instead: that slice must rise $y$ metres, and its mass is $5\,dy$ kilograms. Hence $W=\int_{0}^{20}(5\,dy)(9.8)(y)=49\int_{0}^{20}y\,dy=49\cdot\frac{400}{2}=9800\ \text{J}$. In other words, it is as if the total mass of 100 kg rose 10 m, the depth of the centre of mass of the cable.

$\bar{f}=\dfrac{1}{3-0}\int_{0}^{3}\sqrt{1+x}\,dx=\dfrac{1}{3}\cdot\dfrac{14}{3}=\dfrac{14}{9}\approx 1.556$. The Mean Value Theorem for integrals guarantees a $c\in[0,3]$ with $f(c)=\frac{14}{9}$: solve $\sqrt{1+c}=\frac{14}{9}$, which gives $c=\frac{196}{81}-1=\frac{115}{81}\approx 1.42$.

This is a Riemann sum of $f(x)=\dfrac{1}{1+x}$ on $[0,1]$, with a regular partition and right endpoints. So $u_{n}\to\int_{0}^{1}\dfrac{dx}{1+x}=\ln 2\approx 0.693$.

Form $\infty\cdot 0$. Multiply by the conjugate: $v_{n}=\dfrac{n\left(n^{2}+1-n^{2}\right)}{\sqrt{n^{2}+1}+n}=\dfrac{n}{\sqrt{n^{2}+1}+n}$. Dividing by $n$: $\dfrac{1}{\sqrt{1+1/n^{2}}+1}\to\dfrac{1}{2}$.

Form $1^{\infty}$: $\ln w_{n}=3n\ln\left(1+\frac{2}{n}\right)\sim 3n\cdot\frac{2}{n}=6$, so $w_{n}\to e^{6}\approx 403.4$.

Squeeze. Write $t_{n}=\dfrac{1}{n}\cdot\dfrac{2}{n}\cdot\dfrac{3}{n}\cdots\dfrac{n}{n}$. All the factors after the first are at most 1, so $0<t_{n}\leq\dfrac{1}{n}$. By the squeeze theorem, $t_{n}\to 0$. (The ratio test gives the same result: the ratio of consecutive terms tends to $1/e<1$.)

A recursion $x_{n+1}=g(x_{n})$ with $g(x)=\sqrt{2+x}$. By induction, $0\leq x_{n}\leq 2$: it holds for $x_{0}=0$, and if $0\leq x_{n}\leq 2$ then $\sqrt{2}\leq x_{n+1}\leq 2$. The sequence is increasing because $x_{n+1}^{2}-x_{n}^{2}=2+x_{n}-x_{n}^{2}=(2-x_{n})(1+x_{n})\geq 0$. Increasing and bounded above, it converges (Monotone Convergence Theorem). Its limit $L$ satisfies $L=\sqrt{2+L}$, so $L^{2}-L-2=0$ and $L=2$ (the root $-1$ is rejected because $L\geq 0$). The fixed point is attracting: $|g'(2)|=\frac{1}{2\sqrt{4}}=\frac{1}{4}<1$.

The general term tends to $\dfrac{1}{2}\neq 0$. The divergence test concludes at once: the series diverges. It is the first reflex, before any other test.

Factorials and powers: the ratio test. $\dfrac{a_{n+1}}{a_{n}}=\dfrac{(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^{n}}{n!}=\dfrac{n^{n}}{(n+1)^{n}}=\dfrac{1}{\left(1+\frac{1}{n}\right)^{n}}\to\dfrac{1}{e}\approx 0.368<1$. The series converges (absolutely).

Integral test: $f(x)=\dfrac{1}{x\ln^{2}x}$ is positive, continuous and decreasing on $[2,\infty)$. With $u=\ln x$, $\int_{2}^{\infty}\dfrac{dx}{x\ln^{2}x}=\int_{\ln 2}^{\infty}\dfrac{du}{u^{2}}=\dfrac{1}{\ln 2}\approx 1.443$, which is finite. The series converges. (Compare with $\sum\frac{1}{n\ln n}$, which diverges: the square in the denominator makes all the difference.)

An alternating series. The sequence $\dfrac{1}{\ln(n+1)}$ is positive, decreasing and tends to 0: the alternating series test gives convergence. But the series of absolute values, $\sum\dfrac{1}{\ln(n+1)}$, diverges, because $\ln(n+1)<n$ implies $\dfrac{1}{\ln(n+1)}>\dfrac{1}{n}$ and the harmonic series diverges. The series is therefore conditionally convergent: it converges, but not absolutely.

Limit comparison with the harmonic series: $\dfrac{\sin(1/n)}{1/n}\to 1$, a finite nonzero limit. The two series therefore behave the same way, and since $\sum\frac{1}{n}$ diverges, $\sum\sin\frac{1}{n}$ diverges too. The fact that $\sin\frac{1}{n}\to 0$ proves nothing: it is how fast the terms decrease that decides.

Ratio test on $\left|\dfrac{x^{n}}{n}\right|$: $\dfrac{|x|^{n+1}}{n+1}\cdot\dfrac{n}{|x|^{n}}=|x|\dfrac{n}{n+1}\to|x|$. The series converges absolutely for $|x|<1$: the radius is $R=1$. At $x=1$ we get the harmonic series $\sum\frac{1}{n}$, which diverges. At $x=-1$ we get the alternating harmonic series, which converges by the alternating series test. The interval of convergence is therefore $[-1,1)$.

Inside its interval of convergence, a power series can be differentiated term by term, and the differentiated series has the same radius. Here $f'(x)=\sum_{n\geq 1}x^{n-1}=\dfrac{1}{1-x}$ for $|x|<1$ (geometric series). So $f(x)=-\ln(1-x)+K$, and $f(0)=0$ forces $K=0$: $f(x)=-\ln(1-x),\qquad -1\leq x<1$.

Evaluate at $x=-1$, which is legitimate because the series converges there and its sum is continuous there (Abel's theorem): $\sum_{n\geq 1}\frac{(-1)^{n}}{n}=f(-1)=-\ln 2,\quad\text{so}\quad\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}=\ln 2\approx 0.693$.

Evaluate at $x=\frac{1}{2}$, well inside the interval: $\sum_{n\geq 1}\frac{1}{n\,2^{n}}=f\!\left(\tfrac{1}{2}\right)=-\ln\!\left(1-\tfrac{1}{2}\right)=\ln 2$. Both sums equal $\ln 2$, but the second converges geometrically fast (10 terms already give 4 decimal places) while the first needs hundreds of thousands of terms. Same sum, two speeds beyond comparison.

The derivatives at 0: $f(0)=0$, $f'(x)=\frac{1}{1+x}$ gives $f'(0)=1$, $f''(x)=\frac{-1}{(1+x)^{2}}$ gives $f''(0)=-1$, $f'''(x)=\frac{2}{(1+x)^{3}}$ gives $f'''(0)=2$. So $P_{3}(x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3},\qquad R_{3}(x)=\frac{f^{(4)}(c)}{4!}x^{4}=\frac{-6}{(1+c)^{4}}\cdot\frac{x^{4}}{24}$ for some $c$ between 0 and $x$.

$P_{3}(0.2)=0.2-0.02+\dfrac{0.008}{3}=0.1826667$. For the bound: on $[0,0.2]$, $\left|f^{(4)}(c)\right|=\dfrac{6}{(1+c)^{4}}$ is largest at $c=0$, where it equals 6. So $|R_{3}|\leq\frac{6}{24}(0.2)^{4}=\frac{0.0016}{4}=4.0\times 10^{-4}$. The exact value is $\ln(1.2)=0.1823216$, so the actual error is $3.45\times 10^{-4}$: it is indeed below the bound, and of the same order of magnitude, which shows that the bound is tight.

Replace $\ln(1+x)$ by its expansion: $\ln(1+x)-x+\frac{x^{2}}{2}=\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots\right)-x+\frac{x^{2}}{2}=\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots$ Dividing by $x^{3}$: $\dfrac{1}{3}-\dfrac{x}{4}+\cdots\to\dfrac{1}{3}$. L'Hôpital's rule would give the same result, but only after three successive differentiations of an increasingly heavy quotient.

Start from $e^{u}=\sum\frac{u^{n}}{n!}$ with $u=-x^{2}$, then integrate term by term (legitimate, the radius is infinite): $\int_{0}^{a}e^{-x^{2}}dx=\sum_{n\geq 0}\frac{(-1)^{n}a^{2n+1}}{n!\,(2n+1)}$. For $a=0.5$, the first five terms are $0.5-0.0416667+0.0031250-0.0001860+0.0000090$, whose sum is $0.4612814$. The series is alternating with decreasing terms, so the error is bounded by the first neglected term: $\dfrac{0.5^{11}}{5!\cdot 11}=3.7\times 10^{-7}$. We can therefore state $0.461281$ with six reliable decimal places (the exact value is $0.4612810$).