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 $\displaystyle\int_{\pi/2}^\pi x\cos x\,dx=\left(x\sin x+\cos
x\right)\left|\begin{array}{c} ^{\pi} \\ _{\pi/2} \end{array}\right
.=\pi\sin\pi+\cos\pi-\left(\tfrac \pi 2 \sin\tfrac\pi 2+\cos\tfrac
\pi 2\right)=0-1-\tfrac\pi 2 \cdot 1-0=-1-\tfrac \pi 2$.
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$}\\}{=}\displaystyle
uv\left|\begin{array}{c} ^\pi \\ _{\pi/2} \end{array}\right
.-\int_{\pi/2}^\pi v\,du\displaystyle =\left(x\sin
x\left|\begin{array}{c} ^\pi \\ _{\pi/2}
\end{array}\right.\right)-\int_{\pi/2}^\pi\sin x\,dx$
$\displaystyle = \left(\pi\sin\pi-\tfrac \pi 2 \sin\tfrac\pi
2\right)-\left(-\cos x\right)\left|\begin{array}{c} ^\pi \\ _{\pi/2}
\end{array}\right .=\pi\sin\pi-\tfrac \pi 2 \sin\tfrac\pi 2+\cos
\pi-\cos\tfrac \pi 2=0-\tfrac \pi 2-1-0=-1-\tfrac \pi 2$