Evaluate the following integrals: ^ 1 _0 |x |dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits^{1}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}$

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Answer

For $0<\text{x}<1,\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$
$\therefore\ \int\limits^{1}_0|\text{x}\sin\pi\text{x}|\text{ dx}=\int\limits^{1}_0\text{x}\sin\pi\text{x}\text{ dx}$
Let $\text{I}=\int\text{x}\sin\pi\text{x}\text{ dx}$
$=\text{x}\int\sin\pi\text{x}\text{ dx}-\int\Big(\frac{\text{d}}{\text{dx}}\text{x}\int\sin\pi\text{x dx}\Big)\text{dx}$
$=\text{x}\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)-\int\Big(\frac{-\text{cos}\pi\text{x}}{\pi}\Big)\text{dx}$
$=\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}$
Applying the limits, we get
$\int\limits^{1}_0|\text{x}\sin\pi\text{x}|\text{ dx}=\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^1_0$
$=\Big(\frac{-\cos\pi}{\pi}+\frac{\sin\pi}{\pi^2}\big)-(0+0)$
$=\frac{1}{\pi}+0-0$
$=\frac{1}{\pi}$

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