Problem source

Show that 
\[
\int_{-1}^1 \vert \, x\e^x \,\vert  \d x =- \int_{-1}^0  x\e^x  \d x +
 \int_0^1  x\e^x   \d x
\]
and hence evaluate the integral.
Evaluate the following integrals:
\begin{questionparts}
\item $\displaystyle \int_0^4 \vert\, x^3-2x^2-x+2 \,\vert \, \d x\,;$
\item $\displaystyle  \int_{-\pi}^\pi \vert\, \sin x  +\cos x \,\vert \; \d x\,.$
\end{questionparts}

Solution source

\begin{align*}
&& \int_{-1}^1 |x e^x |\d x &= \int_{-1}^0 |xe^x| \d x + \int_0^1 |xe^x| \d x \\
&&&= \int_{-1}^0 -xe^x \d x + \int_0^1 x \e^x \d x \\
&&&= -\int_{-1}^0 xe^x \d x + \int_0^1 x \e^x \d x \\
\\
&& \int xe^x \d x &= xe^x - \int e^x \d x \\
&&&= xe^x - e^x \\
\\
\Rightarrow &&  \int_{-1}^1 |x e^x |\d x  &= \left [ xe^x - e^x \right]_0^{-1}+ \left [ xe^x - e^x \right]_0^{1} \\
&&&= -e^{-1}-e^{-1} +e^{0} + e^1 - e^1 +e^0 \\
&&&= 2-2e^{-1}
\end{align*}

\begin{questionparts}
\item $\,$ \begin{align*}
&& I &= \int_0^4 | x^3-2x^2-x+2| \d x \\
&&&= \int_0^4 |(x-2)(x-1)(x+1)| \d x\\
&&&= \int_0^1( x^3-2x^2-x+2) \d x- \int_1^2 ( x^3-2x^2-x+2) \d x + \int_2^4 ( x^3-2x^2-x+2) \d x \\
&&&= \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_0^1 -  \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_1^2 +  \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_2^4 \\
&&&= 2 \left  ( \frac14 - \frac23 -\frac12 + 2\right) - 2 \left ( \frac14 2^4 - \frac23 2^3 -\frac12 2^2 + 2 \cdot 2\right)+ \left ( \frac14 4^4 - \frac23 4^3 -\frac12 4^2 + 2 \cdot 4\right) \\
&&&= \frac{133}{6}
\end{align*}

\item $\,$ \begin{align*}
&& J &= \int_{-\pi}^\pi | \sin x + \cos x | \d x \\
&&&= \int_{-\pi}^{\pi} | \sqrt{2} \sin(x + \tfrac{\pi}{4})| \d x \\
&&&= 2\sqrt{2}\int_0^\pi \sin x \d x \\
&&&= 4\sqrt{2}
\end{align*}

\end{questionparts}