Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Evaluate $\int_{-1}^{\frac{3}{2}}|x \sin \ (\pi x)| d x$
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Answer
Here $f (x) = | x \sin \pi x| = \left\{\begin{array}{l} {x \sin \pi x \text { for }-1 \leq x \leq 1} \\ {-x \sin \pi x \text { for } 1 \leq x \leq \frac{3}{2}} \end{array}\right.$
Therefore $\int_{-1}^{\frac{3}{2}}|x \sin \pi x| \ d x = \int_{-1}^{1} x \sin \pi x\ d x+\int_{1}^{\frac{3}{2}}-x \sin \pi x\ d x$
$= \int_{-1}^{1} x \sin \pi x\ d x-\int_{1}^{\frac{3}{2}} x \sin \pi x\ d x$
Integrating both integrals on righthand side, we get
$\int_{-1}^{\frac{3}{2}}|x \sin \pi x| d x = \left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{-1}^{1}-\left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{1}^{\frac{3}{2}}$
$= \frac{2}{\pi}-\left[-\frac{1}{\pi^{2}}-\frac{1}{\pi}\right]$
$=\frac{3}{\pi}+\frac{1}{\pi^{2}}$
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