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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following integrals:
$\int\limits^{\frac{3}{2}}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}$
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Answer

For $0<\text{x}<1,\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$
For $1<\text{x}<\frac{3}{2},\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$
$\therefore\ \int\limits^{\frac{3}{2}}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}=\int\limits^1_0\text{x}\sin\pi\text{x dx}-\int\limits^{\frac{3}{2}}_1\text{x}\sin\pi\text{x dx}$
Let $\text{I}=\int\text{x}\sin\pi\text{ dx}$
$=\text{x}\int\sin\pi\text{x}-\int\Big(\frac{\text{d}}{\text{dx}}\text{x}\int\sin\pi\text{x dx}\Big)\text{dx}$
$=\text{x}\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)-\int\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)\text{dx}$
$=\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}$
Applying the limits, we get
$\int\limits^{\frac{3}{2}}_0|\text{x}\sin\pi\text{x}|\text{dx}=\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^1_0-\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^{\frac{3}{2}}_1$
$=\Big[\Big(\frac{\cos\text{x}}{\pi}+\frac{\sin\pi}{\pi^2}\Big)-(0+0)\Big]-\Bigg[\bigg(\frac{-\frac{3}{2}\cos\frac{3\pi}{2}}{\pi}+\frac{\sin\frac{3\pi}{2}}{\pi^2}\bigg)-\Big(\frac{-\cos\pi}{\pi}+\frac{\sin\pi}{\pi^2}\Big)\Bigg]$
$=\bigg[\Big(\frac{1}{\pi}+0\Big)\bigg]-\bigg[\Big(0-\frac{1}{\pi^2}\Big)-\Big(\frac{1}{\pi}+0\Big)\bigg]$
$=\frac{1}{\pi}+\frac{1}{\pi^2}+\frac{1}{\pi}$
$=\frac{2\pi+1}{\pi^2}$

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