Gujarat 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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