Evaluate the following integrals: ^1_ -1 |x |dx

Question

Evaluate the following integrals:
$\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$

✓

Answer

Let $\text{I}=\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$
Consider $\text{f(x)}=|\text{x}\cos\pi\text{x}|$
$\text{f}(-\text{x})=\big|(\text{x})\cos\pi(-\text{x})\big|=|-\text{x}\cos\pi\text{x}|=|\text{x}\cos\pi\text{x}|\text{f(x)}$
$\therefore\ \text{I}=\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$
$=2\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
Now,
$|\text{x}\cos\pi\text{x}|=\begin{cases}\text{x}\cos\pi\text{x},&\text{if }0\leq\text{x}\leq\frac{1}{2}\\-\text{x}\cos\pi\text{x},&\text{if }\frac{1}{2}\text{x}\leq1\end{cases}$
$\therefore\ \text{I}=2\Bigg[\int\limits^{{1}/{2}}_0\text{x}\cos\pi\text{x dx}+\int\limits^1_{1/2}(-\text{x}\cos\text{dx})\text{dx}\Bigg]$
$=2\Big(\frac{1}{2\pi}\sin\frac{\pi}{2}-0\Big)-\frac{2}{\pi}\times\Big[\Big(-\frac{\cos\pi\text{x}}{\pi}\Big)\Big]^{\frac{1}{2}}_0\\-2\Big(\frac{1}{\pi}\sin\pi-\frac{1}{2\pi}\sin\frac{\pi}{2}\Big)+\frac{2}{\pi}\times\Big(-\frac{\cos\pi\text{x}}{\pi}\Big)\Big]^1_{\frac{1}{2}}$
$=\frac{1}{\pi}+\frac{2}{\pi^2}\Big(\cos\frac{\pi}{2}-\cos0\Big)+\frac{1}{\pi}-\frac{2}{\pi}\Big(\cos\pi-\cos\frac{\pi}{2}\Big)$
$=\frac{1}{\pi}-\frac{2}{\pi^2}+\frac{1}{\pi}+\frac{2}{\pi^2}$
$=\frac{2}{\pi}$

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