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$\int x\sqrt{4{x}^{2}+9}dx=\frac{1}{12}{\left(4{x}^{2}+9\right)}^{3\text{/}2}+C};u=4{x}^{2}+9$
$\int \frac{x}{\sqrt{4{x}^{2}+9}}dx=\frac{1}{4}\sqrt{4{x}^{2}+9}+C};u=4{x}^{2}+9$
$du=8xdx;f\left(u\right)=\frac{1}{8\sqrt{u}}$
$\int \frac{x}{{(4{x}^{2}+9)}^{2}}dx=-\frac{1}{8(4{x}^{2}+9)}};u=4{x}^{2}+9$
In the following exercises, find the antiderivative using the indicated substitution.
$\int {\left(x+1\right)}^{4}dx};u=x+1$
$\frac{1}{5}{\left(x+1\right)}^{5}+C$
$\int {\left(x-1\right)}^{5}dx};u=x-1$
$\int {\left(2x-3\right)}^{\mathrm{-7}}dx};u=2x-3$
$-\frac{1}{12{\left(3-2x\right)}^{6}}+C$
$\int {\left(3x-2\right)}^{\mathrm{-11}}dx};u=3x-2$
$\int \frac{x}{\sqrt{{x}^{2}+1}}dx};u={x}^{2}+1$
$\sqrt{{x}^{2}+1}+C$
$\int \frac{x}{\sqrt{1-{x}^{2}}}dx};u=1-{x}^{2$
$\int \left(x-1\right){\left({x}^{2}-2x\right)}^{3}dx};u={x}^{2}-2x$
$\frac{1}{8}{\left({x}^{2}-2x\right)}^{4}+C$
$\int \left({x}^{2}-2x\right){\left({x}^{3}-3{x}^{2}\right)}^{2}dx};u={x}^{3}=3{x}^{2$
$\int {\text{cos}}^{3}\theta d\theta};u=\text{sin}\phantom{\rule{0.1em}{0ex}}\theta $ $\text{(}Hint\text{:}\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{2}\theta =1-{\text{sin}}^{2}\theta \text{)}$
$\text{sin}\phantom{\rule{0.1em}{0ex}}\theta -\frac{{\text{sin}}^{3}\theta}{3}+C$
$\int {\text{sin}}^{3}\theta d\theta};u=\text{cos}\phantom{\rule{0.1em}{0ex}}\theta $ $\text{(}Hint\text{:}\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\theta =1-{\text{cos}}^{2}\theta \text{)}$
In the following exercises, use a suitable change of variables to determine the indefinite integral.
$\int x{\left(1-x\right)}^{99}dx$
$\frac{{\left(1-x\right)}^{101}}{101}-\frac{{\left(1-x\right)}^{100}}{100}+C$
$\int t{\left(1-{t}^{2}\right)}^{10}dt$
$\int {\left(11x-7\right)}^{\mathrm{-3}}dx$
$-\frac{1}{22\left(7-11{x}^{2}\right)}+C$
$\int {\left(7x-11\right)}^{4}dx$
$\int {\text{cos}}^{3}\theta \phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta d\theta$
$-\frac{{\text{cos}}^{4}\theta}{4}+C$
$\int {\text{sin}}^{7}\theta \phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\theta d\theta$
$\int {\text{cos}}^{2}\left(\pi t\right)\text{sin}\left(\pi t\right)dt$
$-\frac{{\text{cos}}^{3}\left(\pi t\right)}{3\pi}+C$
$\int {\text{sin}}^{2}x{\text{cos}}^{3}xdx$ $\text{(}Hint\text{:}\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}x+{\text{cos}}^{2}x=1\text{)}$
$\int t\phantom{\rule{0.1em}{0ex}}\text{sin}\left({t}^{2}\right)\text{cos}\left({t}^{2}\right)dt$
$-\frac{1}{4}\phantom{\rule{0.05em}{0ex}}{\text{cos}}^{2}\left({t}^{2}\right)+C$
$\int {t}^{2}}{\text{cos}}^{2}\left({t}^{3}\right)\text{sin}\left({t}^{3}\right)dt$
$\int \frac{{x}^{2}}{{\left({x}^{3}-3\right)}^{2}}dx$
$-\frac{1}{3({x}^{3}-3)}+C$
$\int \frac{{x}^{3}}{\sqrt{1-{x}^{2}}}dx$
$\int \frac{{y}^{5}}{{\left(1-{y}^{3}\right)}^{3\text{/}2}}dy$
$-\frac{2\left({y}^{3}-2\right)}{3\sqrt{1-{y}^{3}}}$
${{\displaystyle \int \text{cos}\phantom{\rule{0.1em}{0ex}}\theta \left(1-\text{cos}\phantom{\rule{0.1em}{0ex}}\theta \right)}}^{99}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta d\theta $
${{\displaystyle \int \left(1-{\text{cos}}^{3}\theta \right)}}^{10}{\text{cos}}^{2}\theta \phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta d\theta $
$\frac{1}{33}{\left(1-{\text{cos}}^{3}\theta \right)}^{11}+C$
$\int \left(\text{cos}\phantom{\rule{0.1em}{0ex}}\theta -1\right){\left({\text{cos}}^{2}\theta -2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\theta \right)}^{3}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta d\theta$
$\int \left({\text{sin}}^{2}\theta -2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta \right){\left({\text{sin}}^{3}\theta -3{\text{sin}}^{2}\theta \right)}^{3}\text{cos}\phantom{\rule{0.1em}{0ex}}\theta d\theta$
$\frac{1}{12}{\left({\text{sin}}^{3}\theta -3{\text{sin}}^{2}\theta \right)}^{4}+C$
In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.
[T] $y=3{\left(1-x\right)}^{2}$ over $\left[0,2\right]$
[T] $y=x{\left(1-{x}^{2}\right)}^{3}$ over $\left[\mathrm{-1},2\right]$
${L}_{50}=\mathrm{-8.5779}.$ The exact area is $\frac{\mathrm{-81}}{8}$
[T] $y=\text{sin}\phantom{\rule{0.1em}{0ex}}x{\left(1-\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)}^{2}$ over $\left[0,\pi \right]$
[T] $y=\frac{x}{{\left({x}^{2}+1\right)}^{2}}$ over $\left[\mathrm{-1},1\right]$
${L}_{50}=\mathrm{-0.006399}$ … The exact area is 0.
In the following exercises, use a change of variables to evaluate the definite integral.
${\int}_{0}^{1}x\sqrt{1-{x}^{2}}}dx$
${\int}_{0}^{1}\frac{x}{\sqrt{1+{x}^{2}}}dx$
$u=1+{x}^{2},du=2xdx,\frac{1}{2}{\displaystyle {\int}_{1}^{2}{u}^{\mathrm{-1}\text{/}2}du=\sqrt{2}-1}$
${\int}_{0}^{2}\frac{t}{\sqrt{5+{t}^{2}}}dt$
${\int}_{0}^{1}\frac{t}{\sqrt{1+{t}^{3}}}dt$
$u=1+{t}^{3},du=3{t}^{2},\frac{1}{3}{\displaystyle {\int}_{1}^{2}{u}^{\mathrm{-1}\text{/}2}du=\frac{2}{3}\left(\sqrt{2}-1\right)}$
${\int}_{0}^{\pi \text{/}4}{\text{sec}}^{2}\theta \phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta d\theta$
${\int}_{0}^{\pi \text{/}4}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}\theta}{{\text{cos}}^{4}\theta}d\theta$
$u=\text{cos}\phantom{\rule{0.1em}{0ex}}\theta ,du=\text{\u2212}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta d\theta ,{\displaystyle {\int}_{1\text{/}\sqrt{2}}^{1}{u}^{\mathrm{-4}}du=\frac{1}{3}\left(2\sqrt{2}-1\right)}$
In the following exercises, evaluate the indefinite integral $\int f\left(x\right)dx$ with constant $C=0$ using u -substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of C that would need to be added to the antiderivative to make it equal to the definite integral $F\left(x\right)={\displaystyle {\int}_{a}^{x}f\left(t\right)dt},$ with a the left endpoint of the given interval.
[T] $\int \left(2x+1\right){e}^{{x}^{2}+x-6}dx$ over $\left[\mathrm{-3},2\right]$
[T] $\int \frac{\text{cos}\left(\text{ln}\left(2x\right)\right)}{x}dx$ on $\left[0,2\right]$
The antiderivative is
$y=\text{sin}\left(\text{ln}\left(2x\right)\right).$ Since the antiderivative is not continuous at
$x=0,$ one cannot find a value of
C that would make
$y=\text{sin}\left(\text{ln}\left(2x\right)\right)-C$ work as a definite integral.
[T] $\int \frac{3{x}^{2}+2x+1}{\sqrt{{x}^{3}+{x}^{2}+x+4}}dx$ over $\left[\mathrm{-1},2\right]$
[T] $\int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{{\text{cos}}^{3}x}dx$ over $\left[-\frac{\pi}{3},\frac{\pi}{3}\right]$
The antiderivative is
$y=\frac{1}{2}\phantom{\rule{0.05em}{0ex}}{\text{sec}}^{2}x.$ You should take
$C=\mathrm{-2}$ so that
$F\left(-\frac{\pi}{3}\right)=0.$
[T] $\int \left(x+2\right){e}^{\text{\u2212}{x}^{2}-4x+3}dx$ over $\left[\mathrm{-5},1\right]$
[T] $\int 3{x}^{2}\sqrt{2{x}^{3}+1}}dx$ over $\left[0,1\right]$
The antiderivative is
$y=\frac{1}{3}{\left(2{x}^{3}+1\right)}^{3\text{/}2}.$ One should take
$C=-\frac{1}{3}.$
If $h\left(a\right)=h\left(b\right)$ in ${\int}_{a}^{b}g\text{'}\left(h\left(x\right)\right)h\left(x\right)dx},$ what can you say about the value of the integral?
Is the substitution $u=1-{x}^{2}$ in the definite integral ${\int}_{0}^{2}\frac{x}{1-{x}^{2}}dx$ okay? If not, why not?
No, because the integrand is discontinuous at $x=1.$
In the following exercises, use a change of variables to show that each definite integral is equal to zero.
${\int}_{0}^{\pi}{\text{cos}}^{2}\left(2\theta \right)\text{sin}\left(2\theta \right)d\theta$
${\int}_{0}^{\sqrt{\pi}}t\phantom{\rule{0.1em}{0ex}}\text{cos}\left({t}^{2}\right)\text{sin}\left({t}^{2}\right)dt$
$u=\text{sin}\left({t}^{2}\right);$ the integral becomes $\frac{1}{2}{\displaystyle {\int}_{0}^{0}udu}.$
${\int}_{0}^{1}\left(1-2t\right)dt$
${\int}_{0}^{1}\frac{1-2t}{\left(1+{\left(t-\frac{1}{2}\right)}^{2}\right)}dt$
$u=\left(1+{\left(t-\frac{1}{2}\right)}^{2}\right);$ the integral becomes $\text{\u2212}{\displaystyle {\int}_{5\text{/}4}^{5\text{/}4}\frac{1}{u}du}.$
${\int}_{0}^{\pi}\text{sin}\left({\left(t-\frac{\pi}{2}\right)}^{3}\right)\text{cos}\left(t-\frac{\pi}{2}\right)dt$
${\int}_{0}^{2}\left(1-t\right)\text{cos}\left(\pi t\right)dt$
$u=1-t;$ the integral becomes
$\begin{array}{l}{\displaystyle {\int}_{1}^{\mathrm{-1}}u\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\pi \left(1-u\right)\right)du}\hfill \\ ={\displaystyle {\int}_{1}^{\mathrm{-1}}u\left[\text{cos}\phantom{\rule{0.1em}{0ex}}\pi \phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}u-\text{sin}\phantom{\rule{0.1em}{0ex}}\pi \phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}u\right]du}\hfill \\ =\text{\u2212}{\displaystyle {\int}_{1}^{\mathrm{-1}}u\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}udu}\hfill \\ ={\displaystyle {\int}_{\mathrm{-1}}^{1}u\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}udu=0}\hfill \end{array}$
since the integrand is odd.
${\int}_{\pi \text{/}4}^{3\pi \text{/}4}{\text{sin}}^{2}t\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}tdt$
Show that the average value of $f\left(x\right)$ over an interval $\left[a,b\right]$ is the same as the average value of $f\left(cx\right)$ over the interval $\left[\frac{a}{c},\frac{b}{c}\right]$ for $c>0.$
Setting $u=cx$ and $du=cdx$ gets you $\frac{1}{\frac{b}{c}-\frac{a}{c}}{\displaystyle {\int}_{a\text{/}c}^{b\text{/}c}f\left(cx\right)dx=\frac{c}{b-a}}\phantom{\rule{0.2em}{0ex}}{\displaystyle {\int}_{u=a}^{u=b}f\left(u\right)\frac{du}{c}=\frac{1}{b-a}{\displaystyle {\int}_{a}^{b}f\left(u\right)du}}.$
Find the area under the graph of $f\left(t\right)=\frac{t}{{\left(1+{t}^{2}\right)}^{a}}$ between $t=0$ and $t=x$ where $a>0$ and $a\ne 1$ is fixed, and evaluate the limit as $x\to \infty .$
Find the area under the graph of $g\left(t\right)=\frac{t}{{\left(1-{t}^{2}\right)}^{a}}$ between $t=0$ and $t=x,$ where $0<x<1$ and $a>0$ is fixed. Evaluate the limit as $x\to 1.$
${\int}_{0}^{x}g\left(t\right)dt=\frac{1}{2}{\displaystyle {\int}_{u=1-{x}^{2}}^{1}\frac{du}{{u}^{a}}}}=\frac{1}{2\left(1-a\right)}{u}^{1-a}{|}_{u=1-{x}^{2}}^{1}=\frac{1}{2\left(1-a\right)}\left(1-{\left(1-{x}^{2}\right)}^{1-a}\right).$ As $x\to 1$ the limit is $\frac{1}{2\left(1-a\right)}$ if $a<1,$ and the limit diverges to +∞ if $a>1.$
The area of a semicircle of radius 1 can be expressed as ${\int}_{\mathrm{-1}}^{1}\sqrt{1-{x}^{2}}dx}.$ Use the substitution $x=\text{cos}\phantom{\rule{0.1em}{0ex}}t$ to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.
The area of the top half of an ellipse with a major axis that is the x -axis from $x=\mathrm{-1}$ to a and with a minor axis that is the y -axis from $y=\text{\u2212}b$ to b can be written as ${\int}_{\text{\u2212}a}^{a}b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}dx}.$ Use the substitution $x=a\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t$ to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.
${\int}_{t=\pi}^{0}b\sqrt{1-{\text{cos}}^{2}t}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}(\text{\u2212}a\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t)dt={\displaystyle {\int}_{t=0}^{\pi}ab{\text{sin}}^{2}tdt$
[T] The following graph is of a function of the form $f(t)=a\phantom{\rule{0.1em}{0ex}}\text{sin}(nt)+b\phantom{\rule{0.1em}{0ex}}\text{sin}(mt).$ Estimate the coefficients a and b , and the frequency parameters n and m . Use these estimates to approximate ${\int}_{0}^{\pi}f\left(t\right)dt}.$
[T] The following graph is of a function of the form $f\left(x\right)=a\phantom{\rule{0.1em}{0ex}}\text{cos}\left(nt\right)+b\phantom{\rule{0.1em}{0ex}}\text{cos}\left(mt\right).$ Estimate the coefficients a and b and the frequency parameters n and m . Use these estimates to approximate ${\int}_{0}^{\pi}f\left(t\right)dt}.$
$f\left(t\right)=2\phantom{\rule{0.1em}{0ex}}\text{cos}\left(3t\right)-\text{cos}\left(2t\right);{\displaystyle {\int}_{0}^{\pi \text{/}2}\left(2\phantom{\rule{0.1em}{0ex}}\text{cos}\left(3t\right)-\text{cos}\left(2t\right)\right)=-\frac{2}{3}}$
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