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Schaum's Outline of Theory and Problems of Advanced Calculus

Murray R. Spiegel

Chapter 4

Derivatives - all with Video Answers

Educators

Chapter Questions

04:04

Problem 1

If $f(x)$ is continuous in $[a, b]$ prove that
$$
\lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{k=1}^{n} f\left(a+\frac{k(b-a)}{n}\right)=\int_{a}^{b} f(x) d x
$$
Since $f(x)$ is continuous, the limit exists independent of the mode of subdivision (see Problem 35). Choose the subdivision of $[a, b]$ into $n$ equal parts of equal length $\Delta x=(b-a) / n$ (see Fig. 5-1, Page 80). Let $\xi_{k}=a+k(b-a) / n, k=1,2, \ldots, n$. Then
$$
\lim _{n \rightarrow \infty} \sum_{k=1}^{n} f\left(\xi_{k}\right) \Delta x_{k}=\lim _{x \rightarrow \infty} \frac{b-a}{n} \sum_{k=1}^{n} f\left(a+\frac{k(b-a)}{n}\right)=\int_{0}^{b} f(x) d x
$$

Uma Kumari
Uma Kumari
Numerade Educator
00:07

Problem 2

Express $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(\frac{k}{n}\right)$ as a definite integral.
Let $a=0, b=1$ in Problem 1. Then
$$
\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(\frac{k}{n}\right)=\int_{0}^{1} f(x) d x
$$

Fuzail Shakir
Fuzail Shakir
Numerade Educator
10:56

Problem 3

(a) Express $\int_{0}^{1} x^{2} d x$ as a limit of a sum, and use the result to evaluate the given definite integral. (b) Interpret the result geometrically.
(a) If $f(x)=x^{2}$, then $f(k / n)=(k / n)^{2}=k^{3} / n^{2}$. Thus by Problem 2,
$$
\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{k^{3}}{n^{2}}=\int_{0}^{1} x^{2} d x
$$
This can be written, using Problem 29 of Chapter 1 ,
$$
\begin{aligned}
\int_{0}^{1} x^{2} d x &=\lim _{x \rightarrow \infty} \frac{1}{n}\left(\frac{1^{2}}{n^{2}}+\frac{2^{2}}{n^{2}}+\cdots+\frac{n^{2}}{n^{2}}\right)=\lim _{x \rightarrow \infty} \frac{1^{2}+2^{x}+\cdots+n^{2}}{n^{2}} \\
&=\lim _{x \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^{3}}=\lim _{n \rightarrow \infty} \frac{(1+1 / n)(2+1 / n)}{6}=\frac{1}{3}
\end{aligned}
$$
which is the required limit. Note: By using the fundamental theorem of the calculus, we observe that $\int_{0}^{1} x^{3} d x=$ $\left.\left(x^{3} / 3\right)\right|_{0} ^{1}=1^{3} / 3-0^{5} / 3=1 / 3$
(b) The area bounded by the curve $y=x^{2}$, the $x$ axis and the line $x=1$ is equal to $\frac{1}{8}$.

Jacquelyn Trost
Jacquelyn Trost
Numerade Educator
01:09

Problem 4

Evaluate $\lim _{n \rightarrow \infty}\left\{\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}\right\}$
The required limit can be written $$ \begin{aligned} \lim _{n \rightarrow \infty} \frac{1}{n}\left\{\frac{1}{1+1 / n}+\frac{1}{1+2 / n}+\cdots+\frac{1}{1+n / n}\right\} &=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{1+k / n} \\ &=\int_{0}^{1} \frac{d x}{1+x}=\left.\ln (1+x)\right|_{0} ^{1}=\ln 2 \end{aligned} $$
using Problem 2 and the fundamental theorem of the calculus.

Nick Johnson
Nick Johnson
Numerade Educator
01:42

Problem 5

Prove that $\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin \frac{t}{n}+\sin \frac{2 t}{n}+\ldots+\sin \frac{(n-1) t}{n}\right\}=\frac{1-\cos t}{t}$
Let $a=0, b=t, f(x)=\sin x$ in Problem 1. Then
$$
\lim _{n \rightarrow \infty} \frac{t}{n} \sum_{x=1}^{n} \sin \frac{k t}{n}=\int_{0}^{t} \sin x d x=1-\cos t
$$
and so
$$
\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n-1} \sin \frac{k t}{n}=\frac{1-\cos t}{t}
$$
using the fact that $\lim _{n \rightarrow \infty} \frac{\sin t}{n}=0$.

Vishal Parmar
Vishal Parmar
Numerade Educator
07:00

Problem 6

Prove that a countable point set has measure zero.
Let the point set be denoted by $x_{1}, x_{2}, x_{s}, x_{1}, \ldots$ and suppose that intervals of lengths less than $\epsilon / 2, \epsilon / 4, \epsilon / 8, \epsilon / 16, \ldots$ respectively enclose the points, where $e$ is any positive number. Then the sum of the lengths of the intervals is less than $e / 2+e / 4+e / 8+\cdots=e$ (let $a=\epsilon / 2$ and $r=\frac{1}{2}$ in Problem 25(a) of Chapter 3), showing that the set has measure zero.

Oswaldo Jiménez
Oswaldo Jiménez
Numerade Educator
17:12

Problem 7

$(a)$ If $f(x)$ is continuous in $[a, b]$ and $m \leqq f(x) \leqq M$ where $m$ and $M$ are constants, prove that
$$
m(b-a) \leqq \int_{a}^{b} f(x) d x \leqq M(b-a)
$$
(b) Interpret the result of $(a)$ geometrically.
a) We have
$$
m \Delta x_{k} \leq f\left(\xi_{k}\right) \Delta x_{k} \leqq M \Delta x_{k} \quad k=1,2, \ldots, n
$$
Summing from $k=1$ to $n$ and using the fact that
$$
\sum_{k=1}^{n} \Delta x_{k}=\left(x_{1}-a\right)+\left(x_{2}-x_{1}\right)+\cdots+\left(b-x_{s-1}\right)=b-a
$$
it follows that
$$
m(b-a) \leqq \sum_{k=1}^{n} f\left(\xi_{x}\right) \Delta x_{k} \leq M(b-a)
$$
Taking the limit as $n \rightarrow \infty$ and each $\Delta x_{k} \rightarrow 0$ yields the required result.
(b) Assume $f(x) \geqq 0$ and continuous in $[a, b]$ with graph shown in the adjoining Fig. 5-2. It is geometrically evident that

Area $A B C D \leqq$ Area under $y=f(x) \leqq$ Area $A B E F$ i.e.s

A similar interpretation can be made if the restriction $f(x) \geqq 0$ is removed. The result also holds if $f(x)$ is sectionally continuous in $[a, b]$.
Fig.5-2

Regina Hays
Regina Hays
Numerade Educator
05:20

Problem 8

Prove that $\left|\int_{a}^{b} f(x) d x\right| \leqq \int_{a}^{b}|f(x)| d x \quad$ if $a<b$
By inequality 2, Page 3,
Taking the limit as $n \rightarrow \infty$ and each $\Delta x_{k} \rightarrow 0$, we have the required result.

Sam Sohn
Sam Sohn
Numerade Educator
02:03

Problem 9

Prove that $\lim _{n \rightarrow \infty} \int_{0}^{2 \pi} \frac{\sin n x}{x^{2}+n^{2}} d x=0$.
$$
\left|\int_{0}^{2 \pi} \frac{\sin n x}{x^{2}+n^{2}} d x\right| \leqq \int_{0}^{2 \pi}\left|\frac{\sin n x}{x^{3}+n^{2}}\right| d x \leqq \int_{0}^{m \pi} \frac{d x}{n^{2}}=\frac{2 \pi}{n^{2}}
$$
Then $\lim _{x \rightarrow \infty}\left|\int_{0}^{x \pi} \frac{\sin n x}{x^{2}+n^{2}} d x\right|=0$, and so the required result follows.

Nick Johnson
Nick Johnson
Numerade Educator
04:06

Problem 10

(a) If $f(x)$ is continuous in $[a, b]$, prove that there is a point $\xi$ in $(a, b)$ such that
$$
\int_{a}^{b} f(x) d x=(b-a) f(\xi)
$$
(b) Interpret the result of ( $a$ ) geometrically.
(a) Since $f(x)$ is continuous in $[a, b]$, we can find constants $m$ and $M$ such that $m \leqq f(x) \leqq M$. Then by Problem 7,
$$
m \leqq \frac{f_{a}^{b} f(x) d x}{b-a} \leqq M
$$
Since $f(x)$ is continuous it takes on all values between $m$ and $M$ (see Chapter 2, Problems 34, 93). In particular there must be a value $\xi$ such that
$$
f(\xi)=\frac{f_{a}^{b} f(x) d x}{b-a} \quad a<\xi<b
$$
The required result follows on multiplying by $b-a$.
(b) If $f(x) \geqq 0$ with graph as shown in the figure of Problem $7(b)$, we can interpret $\int_{a}^{b} f(x) d x$ as the shaded area under the curve $y=f(x)$. Geometrically this area should equal that of a rectangle with base $b-a$ and height $f(\xi)$ for some value $\xi$ between $a$ and $b$.

Priyanka Sadarangani
Priyanka Sadarangani
Numerade Educator
05:18

Problem 11

If $F(x)=\int_{a}^{x} f(t) d t$ where $f(x)$ is continuous in $[a, b]$, prove that $F^{\prime}(x)=f(x)$
$$
\begin{aligned}
\frac{F(x+h)-F(x)}{h} &=\frac{1}{h}\left\{\int_{a}^{x+\lambda} f(t) d t-\int_{a}^{x} f(t) d t\right\}=\frac{1}{h} \int_{x}^{x+n} f(t) d t \\
&=f(\xi) \quad \xi \text { between } x \text { and } x+h
\end{aligned}
$$
by the first mean value theorem for integrals (Problem 10).
Then if $x$ is any point interior to $[a, b]$,
$$
F^{\prime}(x)=\lim _{n \rightarrow 0} \frac{F(x+h)-F(x)}{h}=\lim _{n \rightarrow 0} f(\xi)=f(x)
$$
since $f$ is continuous.
If $x=a$ or $x=b$, we use right or left hand limits respectively and the result holds in these cases as well.

Bobby Barnes
Bobby Barnes
University of North Texas
03:41

Problem 12

Prove the fundamental theorem of the integral calculus.
By Problem 11, if $F(x)$ is any function whose derivative is $f(x)$, we can write
$$
F(x)=\int_{a}^{x} f(t) d t+c
$$
where $c$ is any constant (see last line of Problem 22, Chapter 4$)$.
Since $F(a)=c$, it follows that $F(b)=\int_{a}^{b} f(t) d t+F(a)$ or $\int_{a}^{b} f(t) d t=F(b)-F(a)$

Muhammad Saleem
Muhammad Saleem
Numerade Educator
04:34

Problem 13

If $f(x)$ is continuous in $\{a, b\}$, prove that $F(x)=\int_{a} f(t) d t$ is continuous in $[a, b]$ If $x$ is any point interior to $[a, b]$, then as in Problem 11,
$\lim _{n \rightarrow 0} F(x+h)-F(x)=\lim _{h \rightarrow 0} h f(\xi)=0$
If $x=a$ and $x=b$, we use right and left hand limits respectively to show that $F(x)$ is continuous at $x=a$ and $x=b$.
Another method:
By Problem 11 and Problem 3 , Chapter 4, it follows that $F^{\prime}(x)$ exists and so $F(x)$ must be continuous.

Muhammad Saleem
Muhammad Saleem
Numerade Educator
02:34

Problem 14

Prove the result (14), Page 83, for changing the variable of integration.
Let $F(x)=\int_{a}^{x} f(x) d x$ and $G(t)=\int_{\alpha}^{t} f\{g(t)\} g^{\prime}(t) d t$, where $x=g(t)$.
Then $d F=f(x) d x, \quad d G=f\{g(t)\} g^{\prime}(t) d t$.
Since $d x=g^{\prime}(t) d t$, it follows that $f(x) d x=f\{g(t)\} g^{\prime}(t) d t$ so that $d F(x)=d G(t)$, from which $F(x)=G(t)+c$
Now when $x=a, t=\alpha$ or $F(a)=G(\alpha)+c .$ But $F(a)=G(\alpha)=0$, so that $c=0 .$ Hence $F(x)=G(t)$. Since $x=b$ when $t=\beta$, we have
$\int_{a}^{b} f(x) d x=\int_{\alpha}^{\beta} f\{g(t)\} g^{\prime}(t) d t$

R M
R M
Numerade Educator
04:59

Problem 15

Evaluate:
(a) $\int(x+2) \sin \left(x^{2}+4 x-6\right) d x$
(c) $\int_{-1}^{1} \frac{d x}{\sqrt{(x+2)(3-x)}}$
(e) $\int_{0}^{1 / \sqrt{2}} \frac{x \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} d x$
(b) $\int \frac{\cot (\ln x)}{x} d x$
(c) Method $1:$
$$
\int \frac{d x}{\sqrt{(x+2)(3-x)}}=\int \frac{d x}{\sqrt{6+x-x^{2}}}=\int \frac{d x}{\sqrt{6-\left(x^{2}-x\right)}}=\int \frac{d x}{\sqrt{25 / 4-\left(x-\frac{1}{2}\right)^{2}}}
$$
Letting $x-\frac{1}{2}=u$, this becomes $\int \frac{d u}{\sqrt{25 / 4-u^{2}}}=\sin ^{-1} \frac{u}{5 / 2}+c=\sin ^{-1}\left(\frac{2 x-1}{5}\right)+c$. Then $\int_{-1}^{1} \frac{d x}{\sqrt{(x+2)(3-x)}}=\left.\sin ^{-1}\left(\frac{2 x-1}{5}\right)\right|_{-1} ^{1}=\sin ^{-1}\left(\frac{1}{5}\right)-\sin ^{-1}\left(-\frac{3}{5}\right)$ $=\sin ^{-1} \cdot 2+\sin ^{-1}, 6$
Method 2:
Let $x-\frac{1}{2}=u$ as in Method 1. Now when $x=-1, u=-\frac{3}{2} ;$ and when $x=1, u=\frac{1}{2} .$ Thus by Problem 14
$$
\begin{aligned}
\int_{-1}^{t} \frac{d x}{\sqrt{(x+2)(3-x)}} &=\int_{-1}^{1} \frac{d x}{\sqrt{25 / 4-\left(x-\frac{1}{2}\right)^{2}}}=\int_{-3 / 2}^{1 / 2} \frac{d u}{\sqrt{25 / 4-u^{2}}}=\left.\sin ^{-1} \frac{u}{5 / 2}\right|_{-3 / 2} ^{1.2} \\
&=\sin ^{-1} .2+\sin ^{-1} .6
\end{aligned}
$$
(d) Let $2^{1-s}=u .$ Then $-2^{t-x}(\ln 2) d x=d u$ and $2^{-x} d x=-\frac{d u}{2 \ln 2}$, so that the integral becomes
$$
-\frac{1}{2 \ln 2} \int \tanh u d u=-\frac{1}{2 \ln 2} \ln \cosh 2^{1-x}+c
$$
(e) Let $\sin ^{-1} x^{3}=u$. Then $d u=\frac{1}{\sqrt{1-\left(x^{5}\right)^{2}}} 2 x d x=\frac{2 x d x}{\sqrt{1-x^{4}}}$ and the integral becomes
$$
\frac{1}{2} \int u d u=1 u^{3}+c=\frac{1}{2}\left(\sin ^{-1} x^{3}\right)^{2}+c
$$
Thus $\int_{0}^{1 / \sqrt{x}} \frac{x \sin ^{-1} x^{3}}{\sqrt{1-x^{4}}} d x=1\left(\sin ^{-1} x^{4}\right)||_{0}^{1 / \sqrt{2}}=\frac{1}{\left(\sin ^{-1} \frac{1}{2}\right)^{2}}=\frac{\pi^{3}}{144}$.
$$
\text { (f) } \begin{aligned}
\int \frac{x d x}{\sqrt{x^{2}+x+1}} &=\frac{1}{2} \int \frac{2 x+1-1}{\sqrt{x^{2}+x+1}} d x=\frac{1}{2} \int \frac{2 x+1}{\sqrt{x^{2}+x+1}} d x-\frac{1}{2} \int \frac{d x}{\sqrt{x^{2}+x+1}} \\
&=\frac{1}{2} \int\left(x^{3}+x+1\right)^{-1 / a} d\left(x^{2}+x+1\right)-\frac{1}{2} \int \frac{d x}{\sqrt{\left(x+\frac{1}{2}\right)^{2}+8}} \\
&=\sqrt{x^{3}+x+1}-\frac{1}{2} \ln \left|x+\frac{1}{2}+\sqrt{\left(x+\frac{1}{2}\right)^{2}+1}\right|+c
\end{aligned}
$$
(d) $\int 2^{-x} \tanh 2^{1-x} d x$
(f) $\int \frac{x d x}{\sqrt{x^{2}+x+1}}$
(a) Method 1:
Let $x^{2}+4 x-6=u$. Then $(2 x+4) d x=d u,(x+2) d x=\frac{1}{2} d u$ and the integral becomes
$$
\frac{1}{2} \int \sin u d u=-\frac{1}{2} \cos u+c=-\frac{1}{2} \cos \left(x^{2}+4 x-6\right)+c
$$
Method 2:
$$
\int(x+2) \sin \left(x^{2}+4 x-6\right) d x=\frac{1}{2} \int \sin \left(x^{2}+4 x-6\right) d\left(x^{2}+4 x-6\right)=-\frac{1}{2} \cos \left(x^{2}+4 x-6\right)+c
$$
(b) Let $\ln x=u$. Then $(d x) / x=d u$ and the integral becomes
$$
\int \cot u d u=\ln |\sin u|+c=\ln |\sin (\ln x)|+c
$$

Gaurav Kalra
Gaurav Kalra
Numerade Educator
03:08

Problem 16

Show that $\int_{1}^{2} \frac{d x}{\left(x^{2}-2 x+4\right)^{3 / 2}}=\frac{1}{6}$
Write the integral as $\int_{1}^{2} \frac{d x}{\left[(x-1)^{2}+3\right]^{5 / 2}} .$ Let $x-1=\sqrt{3} \tan u, d x=\sqrt{3} \sec ^{s} u d u$. When $x=1, u=\tan ^{-1} 0=0 ;$ when $x=2, u=\tan ^{-1} 1 / \sqrt{3}=\pi / 6$. Then the integral becomes
$$
\int_{0}^{\pi / 6} \frac{\sqrt{3} \sec ^{2} u d u}{\left[3+3 \tan ^{2} u\right]^{3 / 2}}=\int_{0}^{\pi / 6} \frac{\sqrt{3} \sec ^{2} u d u}{\left[3 \sec ^{2} u\right]^{3 / 2}}=\frac{1}{3} \int_{0}^{\pi / 6} \cos u d u=\left.\frac{1}{3} \sin u\right|_{0} ^{\pi / 6}=\frac{1}{6}
$$

Amy Jiang
Amy Jiang
Numerade Educator
02:17

Problem 17

Determine $\int_{e}^{e z} \frac{d x}{x(\ln x)^{3}}$
Let $\ln x=y,(d x) / x=d y$. When $x=e, y=1$; when $x=e^{x}, y=2$. Then the integral becomes
$$
\int_{1}^{2} \frac{d y}{y^{3}}=\left.\frac{y^{-3}}{-2}\right|_{1} ^{2}=\frac{3}{8}
$$

William Semus
William Semus
Numerade Educator
02:49

Problem 18

Find $\int x^{n} \ln x d x$ if $(a) n \neq-1, \quad($ b $) n=-1$
(a) Use integration by parts, letting $u=\ln x, d v=x^{n} d x$, so that $d u=(d x) / x, v=x^{n+1} /(n+1)$. Then
$$
\begin{aligned}
\int x^{n} \ln x d x &=\int u d v=u v-\int v d u=\frac{x^{n+1}}{n+1} \ln x-\int \frac{x^{n+1}}{n+1} \cdot \frac{d x}{x} \\
&=\frac{x^{n+1}}{n+1} \ln x-\frac{x^{n+1}}{(n+1)^{2}}+c
\end{aligned}
$$
(b) $\int x^{-1} \ln x d x=\int \ln x d(\ln x)=\frac{1}{2}(\ln x)^{2}+c$

Vikash Ranjan
Vikash Ranjan
Numerade Educator
01:02

Problem 19

Find $\int 3^{\sqrt{2 x+1}} d x$
Let $\sqrt{2 x+1}=y, 2 x+1=y^{4}$. Then $d x=y d y$ and the integral becomes $\int 3^{\prime} \cdot y d y$. Integrate by parts, letting $u=y, d v=3^{3} d y ;$ then $d u=d y, v=3 y /(\ln 3)$, and we have
$$
\int 3^{x} \cdot y d y=\int u d v=u v-\int v d u=\frac{y^{\circ} 3^{x}}{\ln 3}-\int \frac{3^{r}}{\ln 3} d y=\frac{y \cdot 3^{\prime}}{\ln 3}-\frac{3^{3}}{(\ln 3)^{2}}+c
$$

Tyler Moulton
Tyler Moulton
Numerade Educator
01:34

Problem 20

Find $\int_{0}^{1} x \ln (x+3) d x$
Let $u=\ln (x+3), d v=x d x .$ Then $d u=\frac{d x}{x+3}, v=\frac{x^{3}}{2} .$ Hence on integrating by parts,
$$
\begin{gathered}
\int x \ln (x+3) d x=\frac{x^{3}}{2} \ln (x+3)-\frac{1}{2} \int \frac{x^{2} d x}{x+3}=\frac{x^{3}}{2} \ln (x+3)-\frac{1}{2} \int\left(x-3+\frac{9}{x+3}\right) d x \\
=\frac{x^{3}}{2} \ln (x+3)-\frac{1}{2}\left\{\frac{x^{\prime}}{2}-3 x+9 \ln (x+3)\right\}+c \\
\int_{0}^{1} x \ln (x+3) d x=\frac{5}{4}-4 \ln 4+\frac{9}{2} \ln 3
\end{gathered}
$$

John Nicolle
John Nicolle
Numerade Educator
01:07

Problem 21

Determine $\int \frac{6-x}{(x-3)(2 x+5)} d x$
Use the method of partial fraction8. Let $\frac{6-x}{(x-3)(2 x+5)}=\frac{A}{x-3}+\frac{B}{2 x+5}$
Method 1:
To determine the constants $A$ and $B$, multiply both sides by $(x-3)(2 x+5)$ to obtain
$$
6-x=A(2 x+5)+B(x-3) \quad \text { or } 6-x=5 A-3 B+(2 A+B) x
$$
Since this is an identity, $5 A-3 B=6,2 A+B=-1$ and $A=3 / 11, B=-17 / 11$. Then
$$
\int \frac{6-x}{(x-3)(2 x+5)} d x=\int \frac{3 / 11}{x-3} d x+\int \frac{-17 / 11}{2 x+5} d x=\frac{3}{11} \ln |x-3|-\frac{17}{22} \ln |2 x+5|+c
$$
Method 2:
Substitute suitable values for $x$ in the identity (1). For example, letting $x=3$ and $x=-5 / 2$ in (1), we find at once $A=3 / 11, B=-17 / 11$

Carson Merrill
Carson Merrill
Numerade Educator
05:19

Problem 22

Evaluate $\int \frac{d x}{5+3 \cos x}$ by using the substitution $\tan x / 2=u .$ From Fig. 5-3 we see that
$$
\sin x / 2=\frac{u}{\sqrt{1+u^{2}}}, \quad \cos x / 2=\frac{1}{\sqrt{1+u^{2}}}
$$
Fig. 5-3
Then $\cos x=\cos ^{2} x / 2-\sin ^{2} x / 2=\frac{1-u^{2}}{1+u^{2}}$
Also $d u=\frac{1}{2} \sec ^{3} x / 2 d x$ or $d x=2 \cos ^{2} x / 2 d u=\frac{2 d u}{1+u^{2}}$.
Thus the integral becomes $\int \frac{d u}{u^{2}+4}=\frac{1}{2} \tan ^{-1} u / 2+c=\frac{1}{2} \tan ^{-1}\left(\frac{1}{2} \tan x / 2\right)+c$

Pawan Yadav
Pawan Yadav
Numerade Educator
01:32

Problem 23

Evaluate $\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$
Let $x=\pi-y$. Then
$$
\begin{aligned}
I &=\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x=\int_{0}^{\pi} \frac{(\pi-y) \sin y}{1+\cos ^{2} y} d y=\pi \int_{0}^{\pi} \frac{\sin y}{1+\cos ^{2} y} d y-\int_{0}^{\pi} \frac{y \sin y}{1+\cos ^{2} y} d y \\
&=-\pi \int_{0}^{\pi} \frac{d(\cos y)}{1+\cos ^{2} y}-I=-\left.\pi \tan ^{-1}(\cos y)\right|_{0} ^{\pi}-I=\pi^{2} / 2-1
\end{aligned}
$$
i.e. $\quad I=\pi^{2} / 2-I$ or $I=\pi^{2} / 4$.

Sriram Soundarrajan
Sriram Soundarrajan
Numerade Educator
04:59

Problem 24

Prove that $\int_{0}^{\sin } \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=\frac{\pi}{4}$
Letting $x=\pi / 2-y$, we have
$$
I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=\int_{0}^{\pi / 2} \frac{\sqrt{\cos y}}{\sqrt{\cos y}+\sqrt{\sin y}} d y=\int_{0}^{\pi / 1} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x
$$
Then
$$
\begin{aligned}
I+I &=\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x+\int_{0}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \\
&=\int_{0}^{\pi / 1} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=\int_{0}^{\pi / 2} d x=\frac{\pi}{2}
\end{aligned}
$$
from which $2 I=\pi / 2$ and $I=\pi / 4$.
The same method can be used to prove that for all real values of $m$,
$$
\int_{0}^{\pi / 2} \frac{\sin ^{m} x}{\sin ^{m} x+\cos ^{m} x} d x=\frac{\pi}{4}
$$
Note: This problem and Problem 23 show that some definite integrals can be evaluated without first finding the corresponding indefinite integrals.

Gaurav Kalra
Gaurav Kalra
Numerade Educator
01:02

Problem 25

Evaluate $\int_{0}^{1} \frac{d x}{1+x^{2}}$ approximately, using $(a)$ the trapezoidal rule, (b) Simpson's rule, where the interval $[0,1]$ is divided into $n=4$ equal parts.

Let $f(x)=1 /\left(1+x^{2}\right) .$ Using the notation on Page 85, we find $\Delta x=(b-a) / n=(1-0) / 4=0.25$. Then keeping 4 decimal places, we have: $y_{0}=f(0)=1.0000, y_{1}=f(0.25)=0.9412, \quad y_{2}=f(0.50)=0.8000$, $y_{3}=f(0.75)=0.6400, \quad y_{1}=f(1)=0.5000$
(a) The trapezoidal rule gives
$$
\frac{\Delta x}{2}\left\{y_{0}+2 y_{1}+2 y_{2}+2 y_{3}+y_{4}\right\}=\frac{0.25}{2}\{1.0000+2(0.9412)+2(0.8000)+2(0.6400)+0.500\}=0.7828
$$
(b) Simpson's rule gives
$$
\frac{\Delta x}{3}\left\{y_{0}+4 y_{1}+2 y_{2}+4 y_{3}+y_{4}\right\}=\frac{0.25}{3}\{1.0000+4(0.9412)+2(0.8000)+4(0.6400)+0.5000\}=0.7854
$$
The true value is $\pi / 4 \approx 0.7854$

Tyler Moulton
Tyler Moulton
Numerade Educator
16:28

Problem 26

(a) Evaluate $\int_{0}^{1} e^{x^{2}} d x$ approximately by using Taylor's theorem of the mean and (b) estimate the maximum error.
As in Problem 28, Chapter 4, we find
$$
e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\frac{x^{5} e^{\xi}}{5 !} \quad 0<\xi<x
$$
Then replacing $x$ by $x^{2}$,
$$
e^{x^{2}}=1+x^{2}+\frac{x^{4}}{2 !}+\frac{x^{\beta}}{3 !}+\frac{x^{8}}{4 !}+\frac{x^{10} e^{\xi}}{5 !} \quad 0<\xi<x^{2}
$$
Integrating from 0 to 1 ,
$$
\begin{aligned}
\int_{0}^{1} e^{x^{2}} d x &=\int_{0}^{1}\left(1+x^{2}+\frac{x^{4}}{2 !}+\frac{x^{8}}{3 !}+\frac{x^{8}}{4 !}\right) d x+E \quad\left(\text { where the error } E=\int_{0}^{1} \frac{x^{10}}{5 !} e^{\xi} d x\right) \\
&=1+\frac{1}{3}+\frac{1}{5 \cdot 2 !}+\frac{1}{7 \cdot 3 !}+\frac{1}{9 \cdot 4 !}+E=1.4618+E
\end{aligned}
$$
Now $\quad|E|=\left|\int_{0}^{1} \frac{x^{10}}{5 !} e^{\epsilon} d x\right| \leq \int_{0}^{1}\left|\frac{x^{10}}{5 !} e^{\xi}\right| d x \leqq e \int_{0}^{1} \frac{x^{10}}{5 !} d x=\frac{e}{11 \cdot 5 !}<0.0021$
Thus the maximum error is less than $0.0021$, and so the value of the integral is $1.46$ accurate to two decimal places. By using more terms in Taylor's theorem, better accuracy can be attained.

Chris Trentman
Chris Trentman
Numerade Educator
01:10

Problem 27

Find the $(a)$ area and $(b)$ moment of inertia about the $y$ axis of the region in the $x y$ plane bounded by $y=4-x^{2}$ and the $x$ axis.
(a) Subdivide the region into rectangles as in the figure on Page 80. A typical rectangle is shown in the adjoining Fig. 5-4. Then
(b) Assuming unit density, the moment of inertia about the $y$ axis of the typical rectangle shown above is
Fig. 5-4 $\xi_{k}^{2} f\left(\xi_{k}\right) \Delta x_{k}$. Then
$$
\begin{aligned}
\text { Required moment of inertia } &=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \xi_{k}^{2} f\left(\xi_{k}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \xi_{k}^{2}\left(4-\xi_{k}^{2}\right) \Delta x_{k} \\
&=\int_{-2}^{2} x^{2}\left(4-x^{2}\right) d x=\frac{128}{15}
\end{aligned}
$$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
04:40

Problem 28

Find the length of arc of the parabola $y=x^{2}$ from $x=0$ to $x=1$.
$$
\begin{aligned}
\text { Required arc length } &=\int_{0}^{1} \sqrt{1+(d y / d x)^{8}} d x=\int_{0}^{1} \sqrt{1+(2 x)^{2}} d x \\
&=\int_{0}^{1} \sqrt{1+4 x^{2}} d x=\frac{1}{2} \int_{0}^{2} \sqrt{1+u^{2}} d u \\
&=\left.\frac{1}{2}\left\{\frac{1}{2} u \sqrt{1+u^{2}}+\frac{1}{2} \ln \left(u+\sqrt{1+u^{2}}\right)\right\}\right|_{0} ^{2}=\frac{1}{2} \sqrt{5}+1 \ln (2+\sqrt{5})
\end{aligned}
$$

Linda Hand
Linda Hand
Numerade Educator
03:18

Problem 29

Find the volume generated by revolving the region of Problem 27 about the $x$ axis.
$$
\text { Required volume }=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \pi y_{k}^{2} \Delta x_{k}=\pi \int_{-2}^{2}\left(4-x^{2}\right)^{2} d x=512 \pi / 15
$$

Stephen Hobbs
Stephen Hobbs
Numerade Educator
01:03

Problem 30

If $f(x)$ and $g(x)$ are continuous in $[a, b]$, prove Schwarz's inequality for integrals:
$$
\left(\int_{a}^{b} f(x) g(x) d x\right)^{2} \leqq \int_{a}^{b}\{f(x)\}^{2} d x \int_{a}^{0}\{g(x)\}^{2} d x
$$
We have
$$
\int_{a}^{b}\{f(x)+\lambda g(x)\}^{2} d x=\int_{a}^{b}\{f(x)\}^{2} d x+2 \lambda \int_{a}^{b} f(x) g(x) d x+\lambda^{2} \int_{a}^{b}\{g(x)\}^{2} d x \geq 0
$$
for all real values of $\lambda$. Hence by Problem 13 of Chapter 1, using (1) with
$$
A^{2}=\int_{a}^{b}\{g(x)\}^{2} d x, \quad B^{2}=\int_{a}^{0}\{f(x)\}^{2} d x, \quad C=\int_{a}^{b} f(x) g(x) d x
$$
we find $C^{4} \leqq A^{2} B^{2}$, which gives the required result.

Aman Gupta
Aman Gupta
Numerade Educator
08:43

Problem 31

Prove the second mean value theorem of equation (8), Page 82, under the assumption of the existence and continuity of $g^{\prime}(x)$ in $[a, b]$ in addition to the other assumptions.
Let $F(x)=\int_{a}^{x} f(t) d t .$ Then on integrating by parts,
$$
\begin{aligned}
\int_{a}^{b} f(x) g(x) d x &=\int_{a}^{b} g(x) F^{\prime}(x) d x \\
&=\left.g(x) F(x)\right|_{0} ^{\circ}-\int_{0}^{b} g^{\prime}(x) F(x) d x \\
&=g(b) F(b)-\int_{a}^{b} g^{\prime}(x) F(x) d x
\end{aligned}
$$
Case 1: $g(x)$ is monotonic inereasing, i.e. $g^{\prime}(x) \geq 0$.
Then by the generalized first mean value theorem for integrals (see Page 82) we have,
$$
\int_{a}^{0} g^{\prime}(x) F(x) d x=F(\xi) \int_{a}^{0} g^{\prime}(x) d x=F(\xi)[g(b)-g(a)]
$$
where $\xi$ is in $(a, b)$ so that
$$
\begin{aligned}
\int_{a}^{b} f(x) g(x) d x &=g(b) F(b)-F(\xi)[g(b)-g(a)] \\
&=g(a) F(\xi)+g(b)[F(b)-F(\xi)] \\
&=g(a) \int_{a}^{t} f(x) d x+g(b) \int_{i}^{b} f(x) d x
\end{aligned}
$$
Case $2: g(x)$ is monotonic decreasing, i.e. $g^{\prime}(x) \leqslant 0$.
The proof is similar to that in Case 1 .

Muhammad Saleem
Muhammad Saleem
Numerade Educator
11:30

Problem 32

(a) If $f^{(n+1)}(x)$ is continuous in $[a, b]$, prove that for $x$ in $[a, b]$,
$$
f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)(x-a)^{2}}{2 !}+\ldots+\frac{f^{(*)}(a)(x-a)^{*}}{n !}+\frac{1}{n !} \int_{a}^{\pi}(x-t)^{*} f^{(x+n)}(t) d t
$$
(b) Use (a) to obtain the Lagrange and Cauchy forms of the remainder in Taylor's theorem (see Page 61).
(a) We use mathematical induction (see Chapter 1). The result holds for $n=0$ since
$$
f(x)=f(a)+\int_{a}^{n} f^{\prime}(t) d t=f(a)+\left.f(t)\right|_{0} ^{n}=f(a)+f(x)-f(a)
$$
Assume that it holds for $n=k$. Then integrating by parts, using
$$
\frac{(z-t)^{v}}{k !} d t=d v, f^{\alpha+1)}(t)=u \text { so that } v=-\frac{(x-t)^{k+1}}{(k+1) !}, d u=f^{(k+1)}(t) d t
$$
we find
$\frac{1}{k !} \int_{a}^{x}(x-t)^{k} f^{(k+1)}(t) d t=-\left.\frac{f^{(k+1)}(t)(x-t)^{k+1}}{(k+1) !}\right|_{0} ^{*}+\frac{1}{(k+1) !} \int_{a}^{x}(x-t)^{k+1} f^{k+1)}(t) d t$
$=\frac{f^{(k+1)}(a)(x-a)^{n+1}}{(k+1) !}+\frac{1}{(k+1) !} \int_{a}^{x}(x-t)^{k+1} f^{(k+1)}(t) d t$
which shows that the result holds for $n=k+1 .$ Thus it holds for all integers $n \geq 0$.
(b) We have by the generalized flrst mean value theorem for integrals (see Page 82),
$$
\int_{0}^{\infty} F(t) G(t) d t=F(\xi) \int_{a}^{*} G(t) d t \quad \& \text { between } a \text { and } x
$$
Letting $F(t)=f^{(*+1)}(t), G(t)=\frac{(x-t)^{*}}{n !}$, we obtain
$$
\frac{1}{n !} \int_{a}^{*}(x-t)^{*} f^{(n+1)}(t) d t=\frac{f^{(n+1)}(\xi)}{n !} \int_{a}^{x}(x-t)^{*} d t=\frac{f^{(*+1)}(\xi)(x-a)^{n+1}}{(n+1) !}
$$
giving the Lagrange form of the remainder, [equation (80), Page 61], with $b$ replaced by $x$.
Letting $F(t)=\frac{f^{(*+1)}(t)(x-t)^{n}}{n !}, G(t)=1$, we have
$$
\frac{1}{n !} \int_{0}^{*}(x-t)^{*} f^{(*+1)}(t) d t=\frac{f^{(*+n}(\xi)(x-6)^{*}}{n !} \int_{0}^{*} 1 d t=\frac{f^{(x+1)}(k)(x-\xi)^{*}(x-a)}{n !}
$$
giving the Cauehy form of the remainder, [equation ( $(21)$, Page 61], with $b$ replaced by $x$.

Jack Chen
Jack Chen
Numerade Educator
03:50

Problem 33

Prove that $\lim _{M \rightarrow \infty} \int_{0}^{\infty} \frac{d x}{x^{4}+4}=\frac{\pi}{8}$
We have $x^{4}+4=x^{4}+4 x^{2}+4-4 x^{4}=\left(x^{4}+2\right)^{2}-(2 x)^{\prime}=\left(x^{3}+2+2 x\right)\left(x^{2}+2-2 x\right)$ According to the method of partial fractions, assume
$$
\frac{1}{x^{4}+4}=\frac{A x+B}{x^{2}+2 x+2}+\frac{C x+D}{x^{2}-2 x+2}
$$
Then $1=(A+C) x^{3}+(B-2 A+2 C+D) x^{3}+(2 A-2 B+2 C+2 D) x+2 B+2 D$ so that $A+C=0, B-2 A+2 C+D=0,2 A-2 B+2 C+2 D=0,2 B+2 D=1$
Solving simultaneously, $A=\frac{1}{8}, B=1, C=-\frac{1}{8}, D=4$. Thus
$$
\begin{aligned}
\int \frac{d x}{x^{4}+4} &=\frac{1}{8} \int \frac{x+2}{x^{3}+2 x+2} d x-\frac{1}{8} \int \frac{x-2}{x^{2}-2 x+2} d x \\
&=\frac{1}{8} \int \frac{x+1}{(x+1)^{2}+1} d x+\frac{1}{8} \int \frac{d x}{(x+1)^{\prime}+1}-\frac{1}{8} \int \frac{x-1}{(x-1)^{2}+1} d x+\frac{1}{8} \int \frac{d x}{(x-1)^{2}+1} \\
&=\frac{1}{16} \ln \left(x^{3}+2 x+2\right)+\frac{1}{8} \tan ^{-1}(x+1)-\frac{1}{16} \ln \left(x^{2}-2 x+2\right)+\frac{1}{8} \tan ^{-1}(x-1)+C
\end{aligned}
$$
Then
$$
\lim _{\boldsymbol{w} \rightarrow \infty} \int_{0}^{\mathrm{s}} \frac{d x}{x^{4}+4}=\lim _{x \rightarrow \infty}\left\{\frac{1}{16} \ln \left(\frac{M^{2}+2 M+2}{M^{2}-2 M+2}\right)+\frac{1}{8} \tan ^{-1}(M+1)+\frac{1}{8} \tan ^{-1}(M-1)\right\}=\frac{\pi}{8}
$$
We denote this limit by $\int_{0}^{\infty} \frac{d x}{x^{4}+4}$, called an improper integral of the firat kind. Such integrals are considered further in Chapter 12. See also Problem $78 .$

Stanley Enemuo
Stanley Enemuo
Numerade Educator
View

Problem 34

Evaluate $\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sin t^{3} d t}{x^{4}}$
The conditions of L'Hospital's rule are satisfied, so that the required limit is
$$
\lim _{s \rightarrow 0} \frac{\frac{d}{d x} \int_{0}^{x} \sin t^{3} d t}{\frac{d}{d x}\left(x^{4}\right)}=\lim _{x \rightarrow 0} \frac{\sin x^{3}}{4 x^{8}}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(\sin x^{3}\right)}{\frac{d}{d x}\left(4 x^{5}\right)}=\lim _{s \rightarrow 0} \frac{3 x^{3} \cos x^{3}}{12 x^{2}}=\frac{1}{4}
$$

Suzanne W.
Suzanne W.
Numerade Educator
01:06

Problem 35

Prove that if $f(x)$ is continuous in $[a, b]$ then $\int_{a}^{0} f(x) d x$ exists.
Let $\sigma=\sum_{k=1}^{n} f\left(\xi_{k}\right) \Delta x_{k}$, using the notation of Page 80. Since $f(x)$ is continuous we can find numbers $M_{k}$ and $m_{k}$ representing the l.u.b. and g.1.b. of $f(x)$ in the interval $\left[x_{k-1}, x_{k}\right]$, i.e. such that $m_{k} \leqq f(x) \leqq M_{k} .$ We then have
$$
m(b-a) \leqq g=\sum_{k=1}^{n} m_{k} \Delta x_{k} \leqq \sigma \leqq \sum_{k=1}^{n} M_{k} \Delta x_{k}=S \leqq M(b-a)
$$
where $m$ and $M$ are the g.l.b. and l.u.b. of $f(x)$ in $[a, b] .$ The sums 8 and $S$ are sometimes called the lower and upper sums respectively.

Now choose a second mode of subdivision of $[a, b]$ and consider the corresponding lower and upper sums denoted by $\boldsymbol{s}^{\prime}$ and $S^{\prime}$ respectively. We must have
$$
8^{\prime} \leqq S \quad \text { and } \quad S^{\prime} \geqq 8
$$
To prove this we choose a third mode of subdivision obtained by using the division points of both the first and second modes of subdivision and consider the corresponding lower and upper sums, denoted by $t$ and $\boldsymbol{T}$ respectively. By Problem 89 , we have
$$
\boldsymbol{g} \leqq t \leqq \boldsymbol{T} \leqq S^{\prime} \quad \text { and } \quad \boldsymbol{8}^{\prime} \leqq t \leqq T \leqq S
$$
which proves (2).
From ( $($ ) it is also clear that as the number of subdivisiong is increased, the upper sums are monotonic decreasing and the lower sums are monotonic increasing. Since according to (1) these sums are also bounded, it follows that they have limiting values which we shall call $\delta$ and $\underline{S}$ respectively. By Problem $90, \overline{8} \leqq \underline{S} .$ In order to prove that the integral exists, we must show that $\nabla=\underline{S}$

Since $f(x)$ is continuous in the closed interval $[a, b]$, it is uniformly continuous. Then given any $>0$, we can take each $\Delta x_{k}$ so small that $M_{k}-m_{k}<e /(b-a) .$ It follows that
$$
S-g=\sum_{n=1}^{n}\left(M_{k}-m_{k}\right) \Delta x_{k}<\frac{e}{b-a} \sum_{k=1}^{n} \Delta x_{k}=e
$$
Now $S-8=(S-\underline{S})+(S-\mathbf{8})+(\overline{8}-8)$ and it follows that each term in parentheses is positive and so is less than e by (4). In particular, since $\underline{S}-8$ is a definite number it must be zero, i.e. $\underline{S}=\overline{8}$. Thus the limits of the upper and lower sums are equal and the proof is complete.

Carson Merrill
Carson Merrill
Numerade Educator
01:52

Problem 36

(a) Express $\int_{0}^{1} x^{3} d x$ as a limit of a sum. (b) Use the result of $(a)$ to evaluate the given definite integral. (c) Interpret the result geometrically. Ans. (b) $t$

Ma. Theresa Alin
Ma. Theresa Alin
Numerade Educator
01:31

Problem 37

Using the definition, evaluate
(a) $\int_{0}^{2}(3 x+1) d x$,
(b) $\int_{s}^{6}\left(x^{2}-4 x\right) d x$

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
01:21

Problem 38

$$
\text { Prove that } \lim _{n \rightarrow \infty}\left\{\frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2 !}+\cdots+\frac{n}{n^{2}+n^{2}}\right\}=\frac{\pi}{4}
$$

Nick Johnson
Nick Johnson
Numerade Educator
01:12

Problem 39

$$
\text { Prove that } \lim _{n \rightarrow \infty}\left\{\frac{1^{p}+2^{p}+3^{p}+\cdots+n^{p}}{n^{p+1}}=\frac{1}{p+1}\right\} \text { if } p>-1 \text {. }
$$

Nick Johnson
Nick Johnson
Numerade Educator
03:11

Problem 40

$$
\text { Using the definition, prove that } \int_{a} e^{x} d x=e^{b}-e^{a}
$$

Rukhmani Jain
Rukhmani Jain
Numerade Educator
02:32

Problem 41

$$
\text { Work Problem } 5 \text { directly, using Problem } 94 \text { of Chapter } 1 .
$$

Sanjeev Kumar
Sanjeev Kumar
Numerade Educator
02:06

Problem 42

$$
\text { Prove that } \lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{n^{2}+1^{2}}}+\frac{1}{\sqrt{n^{2}+2^{2}}}+\cdots+\frac{1}{\sqrt{n^{2}+n^{2}}}\right\}=\ln (1+\sqrt{2})
$$

Bobby Barnes
Bobby Barnes
University of North Texas
01:00

Problem 43

$$
\text { Prove that } \lim _{n \rightarrow \infty} \sum_{k=1}^{\pi} \frac{n}{n^{2}+k^{*} x^{3}}=\frac{\tan ^{-1} x}{x} \text { if } x \neq 0
$$

Wendi Zhao
Wendi Zhao
Numerade Educator
03:41

Problem 44

$$
\text { Prove }(a) \text { Property 2, (b) Property } 3 \text { on Page 81. }
$$

Babita Kumari
Babita Kumari
Numerade Educator
03:01

Problem 45

$$
\text { If } f(x) \text { is integrable in }(a, c) \text { and }(c, b), \text { prove that } \int_{a}^{b} f(x) d x=\int_{a}^{e} f(x) d x+\int_{c}^{b} f(x) d x \text {. }
$$

Uma Kumari
Uma Kumari
Numerade Educator
02:08

Problem 46

$$
\text { If } f(x) \text { and } g(x) \text { are integrable in }[a, b] \text { and } f(x) \leqq g(x), \text { prove that } \int_{0}^{0} f(x) d x \leqq \int_{a}^{0} g(x) d x \text {. }
$$

Regina Hays
Regina Hays
Numerade Educator
01:11

Problem 47

$$
\text { Prove that } 1-\cos x \geqq x^{2} / \pi \text { for } 0 \leqq x \leqq \pi / 2
$$

Hoan Nguyen
Hoan Nguyen
Numerade Educator
01:08

Problem 48

$$
\text { Prove that }\left|\int_{0}^{1} \frac{\cos n x}{x+1} d x\right| \leqq \ln 2 \text { for all } n \text {, }
$$

Carson Merrill
Carson Merrill
Numerade Educator
07:15

Problem 49

$$
\text { Prove that }\left|\int_{1}^{\sqrt{3}} \frac{e^{-\pi} \sin x}{x^{2}+1} d x\right| \leqq \frac{\pi}{12 e} \text {. }
$$

Dorcas Attuabea Addo
Dorcas Attuabea Addo
Numerade Educator
04:28

Problem 50

Prove the result (5), Page 82. [Hint: If $m \leqq f(x) \leqq M$, then $m g(x) \leqq f(x) g(x) \leqq M g(x)$. Now integrate and apply Property 7 , Page 82.]

Foster Wisusik
Foster Wisusik
Numerade Educator
01:36

Problem 51

Prove that there exist values $\xi_{1}$ and $\xi_{8}$ in $0 \leqq x \leqq 1$ such that
$$
\int_{0}^{1} \frac{\sin \pi x}{x^{2}+1} d x=\frac{2}{\pi\left(\xi_{1}^{2}+1\right)}=\frac{\pi}{4} \sin \pi \xi_{\mathbf{z}}
$$

Natalie Anderson
Natalie Anderson
Numerade Educator
02:11

Problem 52

$$
\text { Prove that there is a value } \xi \text { in } 0 \leqq x \leqq \pi \text { such that } \int_{0}^{\pi} e^{-x} \cos x d x=\text { sin } \xi \text {. }
$$

Natalie Anderson
Natalie Anderson
Numerade Educator
02:15

Problem 53

Evaluate:
(a) $\int x^{2} e^{\sin x^{3}} \cos x^{3} d x$
(b) $\int_{0}^{1} \frac{\tan ^{-1} t}{1+t^{2}} d t$
(c) $\int_{1}^{3} \frac{d x}{\sqrt{4 x-x^{3}}}$
(d) $\int \frac{\operatorname{csch}^{2} \sqrt{u}}{\sqrt{u}} d u$,
(e) $\int_{-2}^{2} \frac{d x}{16-x^{2}}$
Ans. (a) $\frac{1}{8} e^{\sin x^{3}}+c$
(b) $\pi^{3} / 32$,
(c) $\pi / 3$,
(d) $-2$ coth $\sqrt{u}+c$
(e) $t \ln 3$

Naman Kumar
Naman Kumar
Numerade Educator
09:42

Problem 54

Show that
(a) $\int_{0}^{1} \frac{d x}{\left(3+2 x-x^{3}\right)^{3 / 2}}=\frac{\sqrt{3}}{12}$
(b) $\int \frac{d x}{x^{2} \sqrt{x^{2}-1}}=\frac{\sqrt{x^{3}-1}}{x}+c$.

Perry Roeder
Perry Roeder
Numerade Educator
View

Problem 55

Prove that (a) $\int \sqrt{u^{2} \pm a^{2}} d u=\frac{1}{2} u \sqrt{u^{2} \pm a^{2}} \pm \frac{1}{2} a^{2} \ln \left|u+\sqrt{u^{2} \pm a^{2}}\right|$
(b) $\int \sqrt{a^{2}-u^{2}} d u=\frac{1}{2} u \sqrt{a^{2}-u^{2}}+\frac{1}{2} a^{x} \sin ^{-1} u / a+c, a>0$

Alex Mangiapane
Alex Mangiapane
Numerade Educator
02:03

Problem 56

Find $\int \frac{x d x}{\sqrt{x^{2}+2 x+5}}$

Nafis Fuad
Nafis Fuad
Numerade Educator
01:29

Problem 57

$$
\text { Establish the validity of the method of integration by parts. }
$$

Robert Daugherty
Robert Daugherty
Numerade Educator
01:45

Problem 58

Evaluate
(a) $\int_{1}^{\pi} x \cos 3 x d x$
(b) $\int x^{3} e^{-y x} d x$

Gio Maya
Gio Maya
Numerade Educator
00:34

Problem 59

Show that (a) $\int_{0}^{1} x^{2} \tan ^{-1} x d x=\frac{1}{12} \pi-\frac{1}{6}+\frac{1}{6} \ln 2$

Fuzail Shakir
Fuzail Shakir
Numerade Educator
01:23

Problem 60

(a) If $u=f(x)$ and $v=g(x)$ have continuous $n$th derivatives, prove that
$$
\int u v^{(n)} d x=u v^{(n-1)}-u^{\prime} v^{(n-s)}+u^{\prime \prime} v^{(n-8)}-\cdots(-1)^{n} \int u^{(n)} v d x
$$
called generalized integration by parts. (b) What simplifications occur if $u^{(n)}=0 ?$ Discusg.
(a) to evaluate $\int_{0}^{\pi} x^{4} \sin x d x . \quad$

Manik Pulyani
Manik Pulyani
Numerade Educator
01:48

Problem 61

Show that $\int_{0}^{1} \frac{x d x}{(x+1)^{2}\left(x^{2}+1\right)}=\frac{\pi-2}{8}$

Patrick Vaughn
Patrick Vaughn
Numerade Educator
02:32

Problem 62

$$
\text { Prove that } \int_{0}^{\pi} \frac{d x}{\alpha-\cos x}=\frac{\pi}{\sqrt{\alpha^{2}-1}}, \alpha>1 \text {. }
$$

Kian Manafi
Kian Manafi
Numerade Educator
01:03

Problem 63

Evaluate $\int_{0}^{1} \frac{d x}{1+x}$ approximately, using $(a)$ the trapezoidal rule, (b) Simpson's rule, taking $n=4$. Compare with the exact value, $\ln 2=0.6931$

Tyler Moulton
Tyler Moulton
Numerade Educator
02:07

Problem 64

Using ( $a$ ) the trapezoidal rule, (b) Simpson's rule evaluate $\int_{0}^{\pi / 2} \sin ^{2} x d x$ by obtaining the values of $\sin ^{2} x$ at $x=0^{\circ}, 10^{\circ}, \ldots, 90^{\circ}$ and compare with the exact value $\pi / 4 .$

Mohamed Raafat Mohamed
Mohamed Raafat Mohamed
Numerade Educator
04:33

Problem 65

Prove the (a) rectangular rule, (b) trapezoidal rule, i.e. (16) and (17) of Page 85.

Stephen Hobbs
Stephen Hobbs
Numerade Educator
04:30

Problem 66

Prove Simpson's rule.

Bailey Meade
Bailey Meade
Numerade Educator
02:12

Problem 67

Evaluate to 3 decimal places using numerical integration:
(a) $\int \frac{d x}{1+x^{2}}$
(b) $\int^{+} \cosh x^{2} d x$

Steven Clarke
Steven Clarke
Numerade Educator
05:31

Problem 68

Find the $(a)$ area and $(b)$ moment of inertia about the $y$ axis of the region in the $x y$ plane bounded by $y=\sin x, 0 \leqq x \leqq \pi$ and the $x$ axis, assuming unit density. An8. $(a) 2,(b) \pi^{2}-4$

Wendi Zhao
Wendi Zhao
Numerade Educator
05:31

Problem 68

Find the $(a)$ area and $(b)$ moment of inertia about the $y$ axis of the region in the $x y$ plane bounded by $y=\sin x, 0 \leqq x \leqq \pi$ and the $x$ axis, assuming unit density.
An8. (a) $2,(b) \pi^{2}-4$

Wendi Zhao
Wendi Zhao
Numerade Educator
01:34

Problem 69

Find the moment of inertia about the $x$ axis of the region bounded by $y=x^{2}$ and $y=x$, if the density is proportional to the distance from the $x$ axis. Ans. $1 M$, where $M=$ mass of the region.

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:05

Problem 70

$$
\text { Show that the arc length of the catenary } y=\cosh x \text { from } x=0 \text { to } x=\ln 2 \text { is 8 }
$$

R M
R M
Numerade Educator
06:40

Problem 71

$$
\text { Show that the length of one arch of the cycloid } x=a(\theta-\sin \theta), y=a(1-\cos \theta),(0 \leqq \theta \leqq 2 \pi) \text { is } 8 a \text {. }
$$

Regina Hays
Regina Hays
Numerade Educator
02:54

Problem 72

$$
\text { Prove that the area bounded by the ellipse } x^{2} / a^{2}+y^{2} / b^{2}=1 \text { is a ab. }
$$

Sriram Soundarrajan
Sriram Soundarrajan
Numerade Educator
04:26

Problem 73

Find the volume of the region obtained by revolving the curve $y=\sin x, 0 \leqq x \leqq \pi$, about the $x$ axis. Ans. $\pi^{2} / 2$

Prakash Hampole
Prakash Hampole
Numerade Educator
03:41

Problem 74

Prove that the centroid of the region bounded by $y=\sqrt{a^{2}-x^{3}},-a \leqq x \leqq a$ and the $x$ axis is located at $(0,4 a / 3 \pi)$.

James Kiss
James Kiss
Numerade Educator
01:17

Problem 75

$$
\text { (a) If } \rho=f(\phi) \text { is the equation of a curve in polar coordinates, show that the area bounded by this }
$$
$$
\text { curve and the lines } \phi=\phi_{1} \text { and } \phi=\phi_{2} \text { is } \frac{1}{0} \rho^{\Delta 2} d \phi . \quad \text { (b) Find the area bounded by one loop of the }
$$
$$
\text { lemniscate } \rho^{2}=a^{2} \cos 2 \phi
$$

Carson Merrill
Carson Merrill
Numerade Educator
01:17

Problem 76

(a) Prove that the arc length of the curve in Problem $75(a)$ is
$\int^{\phi_{2}} \sqrt{\rho^{2}+\left(d_{\rho} / d \phi\right)^{2}} d \phi . \quad$ (b) Find the
$$
\text { length of arc of the cardioid } \rho=a(1-\cos \phi)
$$

Joseph Liao
Joseph Liao
Numerade Educator
01:10

Problem 77

$$
\text { Establish the theorem of the mean for derivatives from the first mean value theorem for integrals. }
$$

Amrita Bhasin
Amrita Bhasin
Numerade Educator
03:22

Problem 78

Prove that
(a) $\lim _{\epsilon \rightarrow 0+} \int_{0}^{1-\epsilon} \frac{d x}{\sqrt{4-x}}=4$
(b) $\lim _{\epsilon \rightarrow 0+} \int_{\epsilon}^{0} \frac{d x}{\sqrt[3]{x}}=6$
(c) $\lim _{\epsilon \rightarrow 0+} \int_{0}^{1-\epsilon} \frac{d x}{\sqrt{1-x^{2}}}=\frac{\pi}{2}$ and give a geometric interpretation of the results.
[These limits, denoted usually by $\int_{0}^{4} \frac{d x}{\sqrt{4-x}}, \int_{0}^{8} \frac{d x}{\sqrt[3]{x}}$ and $\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$ respectively, are called improper integrals of the second kind (see Problem 33) since the integrands are not bounded in the range of integration. For further discussion of improper integrals, see Chapter 12.]

Naresh Bagrecha
Naresh Bagrecha
Numerade Educator
03:08

Problem 79

Prove that
(a) $\lim _{x \rightarrow \infty} \int_{0}^{m} x^{s} e^{-x} d x=4 !=24$
(b) $\lim _{\epsilon \rightarrow 0+} \int_{1}^{2-\epsilon} \frac{d x}{\sqrt{x(2-x)}}=\frac{\pi}{2}$

Matthew Allcock
Matthew Allcock
Numerade Educator
01:49

Problem 80

Evaluate
(a) $\int_{0}^{\infty} \frac{d x}{1+x^{2}}$,
(b) $\int_{0}^{\pi / 2} \frac{\sin 2 x}{(\sin x)^{4 / 8}} d x$,
(c) $\int_{0}^{\infty} \frac{d x}{x+\sqrt{x^{2}+1}}$.
(b) 3
(c) does not exist

Naman Kumar
Naman Kumar
Numerade Educator
02:03

Problem 81

Evaluate $\lim _{x \rightarrow \pi / 2} \frac{e x^{2} / \pi-e \pi / 4+\int_{z}^{n / 2} e^{x i n t} d t}{1+\cos 2 x}$

Nick Johnson
Nick Johnson
Numerade Educator
01:13

Problem 82

Prove: $(a) \frac{d}{d x} \int_{z^{2}}^{x^{3}}\left(t^{3}+t+1\right) d t=3 x^{8}+x^{6}-2 x^{3}+3 x^{2}-2 x$
(b) $\frac{d}{d x} \int_{x}^{x^{4}} \cos t^{2} d t=2 x \cos x^{4}-\cos x^{2}$

Gregory Higby
Gregory Higby
Numerade Educator
02:07

Problem 83

$$
\text { Prove the result (18) on Page } 83 \text {. }
$$

Ankur S
Ankur S
Numerade Educator
04:03

Problem 84

Prove that
(a) $\int_{0}^{\pi} \sqrt{1+\sin x} d x=4$
(b) $\int_{0}^{\pi / 2} \frac{d x}{\sin x+\cos x}=\sqrt{2} \ln (\sqrt{2}+1)$

Steven Clarke
Steven Clarke
Numerade Educator
01:56

Problem 85

Explain the fallacy: $\quad I=\int_{-1}^{2} \frac{d x}{1+x^{3}}=-\int_{-1}^{1} \frac{d y}{1+y^{2}}=-1$, using the transformation $x=1 / y$. Hence $I=0$. But $1=\tan ^{-1}(1)-\tan ^{-1}(-1)=\pi / 4-(-\pi / 4)=\pi / 2$. Thus $\pi / 2=0$.

Jack Chen
Jack Chen
Numerade Educator
02:06

Problem 86

$$
\text { Prove that } \int_{0}^{1 / 2} \frac{\cos \pi x}{\sqrt{1+x^{2}}} d x \leqq \frac{1}{4} \tan ^{-1} \frac{1}{2} \text {. }
$$

Vipender Yadav
Vipender Yadav
Numerade Educator
01:52

Problem 87

$$
\text { Evaluate } \lim _{n \rightarrow \infty}\left\{\frac{\sqrt{n+1}+\sqrt{n+2}+\cdots+\sqrt{2 n-1}}{n^{3 / 2}}\right\} \text {. }
$$

Dwijendra Rao
Dwijendra Rao
Numerade Educator
01:09

Problem 88

$$
\text { Prove that } f(x)=\left\{\begin{array}{l}
1 \text { if } x \text { is irrational } \\
0 \text { if } x \text { is rational }
\end{array} \text { is not Riemann integrable in }[0,1] .\right.
$$

Carson Merrill
Carson Merrill
Numerade Educator
06:08

Problem 89

Prove the result ( $(3)$ of Problem 35. [Hint: First consider the effect of only one additional point of subdivision.]

Anas Venkitta
Anas Venkitta
Numerade Educator
00:21

Problem 90

In Problem 35, prove that $8 \leqq \underline{S}$

AG
Ankit Gupta
Numerade Educator
04:10

Problem 91

If $f(x)$ is sectionally continuous in $[a, b]$, prove that $\int_{a} f(x) d x$ exists. [Hint: Enclose each point of discontinuity in an interval, noting that the sum of the lengths of such intervals can be made arbitrarily small. Then consider the difference between the upper and lower sums.]

Harmender Singh Yadav
Harmender Singh Yadav
Numerade Educator
01:07

Problem 92

$$
\text { If } f(x)=\left\{\begin{array}{ll}
2 x & 0<x<1 \\
3 & x=1 \\
6 x-1 & 1<x<2
\end{array}, \text { find } \int_{0}^{2} f(x) d x .\right. \text { Interpret the result graphically. }
$$

Manik Pulyani
Manik Pulyani
Numerade Educator
01:01

Problem 93

$$
\text { Evaluate } \int_{0}^{3}\left\{x-[x]+\frac{1}{2}\right\} d x \text { where }[x] \text { denotes the greatest integer less than or equal to } x . \text { Inter- }
$$
$$
\text { pret the result graphically. }
$$

Tyler Moulton
Tyler Moulton
Numerade Educator
11:11

Problem 94

(a) Prove that $\int_{0}^{\pi / 2} \frac{\sin ^{m} x}{\sin ^{m} x+\cos ^{m} x} d x=\frac{\pi}{4}$ for all real values of $m$.
(b) Prove that $\int_{0}^{2 \pi} \frac{d x}{1+\tan ^{4} x}=\pi$

Anthony Ramos
Anthony Ramos
Numerade Educator
11:11

Problem 95

(a) Prove that $\int_{0}^{\pi / 2} \frac{\sin ^{m} x}{\sin ^{m} x+\cos ^{m} x} d x=\frac{\pi}{4}$ for all real values of $m$.
(b) Prove that $\int_{0}^{2 \pi} \frac{d x}{1+\tan ^{4} x}=\pi$.

Anthony Ramos
Anthony Ramos
Numerade Educator
03:56

Problem 96

$$
\text { Show that } \int_{0}^{0.8} \frac{\tan ^{-1} x}{x} d x=0.4872 \text { approximately. }
$$

M Hassan Anwar
M Hassan Anwar
Numerade Educator
02:06

Problem 97

$$
\text { Show that } \int_{0}^{\pi} \frac{x d x}{1+\cos ^{2} x}=\frac{\pi^{2}}{2 \sqrt{2}} \text {. }
$$

Vipender Yadav
Vipender Yadav
Numerade Educator