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Integration by Parts
Integration by Parts
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Integration by Parts with a definite integral
Going in Circles
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Integration by Parts
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What is a Double Integral?
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Examples 14
Examples 57
Swapping the Order of Integration
Area and Volume Revisited
Double integrals in polar coordinates
dA = r dr (d theta)
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Multiple integrals in physics
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Integrals in Probability and Statistics
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Change of Variables
Review: Change of variables in 1 dimension
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Bonus: Cylindrical and spherical coordinates


Integration by Parts with a definite integral
Previously, we found $\displaystyle \int x \ln(x)\,dx=x\ln x 
\tfrac 1 4 x^2+c$.
In order to compute the definite integral $\displaystyle \int_1^e x
\ln(x)\,dx$, it is probably easiest to
compute the antiderivative $\displaystyle \int x \ln(x)\,dx$
without the limits of itegration (as we computed
previously), and then use FTC II
to evalute the definite integral.
$$\int_1^e x \ln(x)\,dx=\left(x\ln x  \tfrac 1 4 x^2\right)
\left\begin{array}{c} ^e \\ _1 \end{array}\right. =e\ln
e\tfrac 1 4 e^2(1\cdot 0\tfrac 1 4)=e\tfrac 1 4 e^2+\tfrac 1 4$$
DO: Compute
$\displaystyle\int_{\pi/2}^\pi x\cos x\,dx$ by first computing the
antiderivative, then evaluating the definite integral. Work
on this before looking ahead!
$\int x\cos\,dx\overset{\fbox{$\,\,u\,=\,x\quad\, v\,=\,\sin
x\\du\,=\,dx\,\,\, dv\,=\,\cos x\,dx$}\\}{=} uv\int
v\,du=x\sin x\int\sin x\,dx=x\sin x+\cos x+c$
Then $\displaystyle\int_{\pi/2}^\pi x\cos x\,dx=\left(x\sin x+\cos
x\right)\left\begin{array}{c} ^{\pi} \\ _{\pi/2} \end{array}\right
.=\pi\sin\pi+\cos\pi\left(\tfrac \pi 2 \sin\tfrac\pi 2+\cos\tfrac
\pi 2\right)=01\tfrac\pi 2 \cdot 10=1\tfrac \pi 2$.
Alternatively, we can evaluate at the
limits of integration as we go.
$\displaystyle\int_ {\pi/2}^\pi
x\cos\,dx\overset{\fbox{$\,\,u\,=\,x\quad\, v\,=\,\sin
x\\du\,=\,dx\,\,\, dv\,=\,\cos x\,dx$}\\}{=}\displaystyle
uv\left\begin{array}{c} ^\pi \\ _{\pi/2} \end{array}\right
.\int_{\pi/2}^\pi v\,du\displaystyle =\left(x\sin
x\left\begin{array}{c} ^\pi \\ _{\pi/2}
\end{array}\right.\right)\int_{\pi/2}^\pi\sin x\,dx$
$\displaystyle = \left(\pi\sin\pi\tfrac \pi 2 \sin\tfrac\pi
2\right)\left(\cos x\right)\left\begin{array}{c} ^\pi \\ _{\pi/2}
\end{array}\right .=\pi\sin\pi\tfrac \pi 2 \sin\tfrac\pi 2+\cos
\pi\cos\tfrac \pi 2=0\tfrac \pi 210=1\tfrac \pi 2$
