Question

Evaluate : 

\[\int\limits_0^{3/2} \left| x \sin \pi x \right|dx\]

Answer 1

\[\text{For }0 < x < 1, x > 0\text{ and }\sin\pi x > 0 \Rightarrow x\sin\pi x > 0\]
\[\text{For }1 < x < \frac{3}{2}, x > 0\text{ and }\sin\pi x < 0 \Rightarrow x\sin\pi x < 0\]

\[\therefore \int_0^\frac{3}{2} \left| x\sin\pi x \right|dx = \int_0^1 x\sin\pi x dx - \int_1^\frac{3}{2} x\sin\pi x dx\]
\[Let I = \int x\sin\pi x dx\]
\[ = x\int \sin\pi x dx - \int\left( \frac{d}{dx}x\int \sin\pi x dx \right)dx\]
\[ = x\left( \frac{- \cos\pi x}{\pi} \right) - \int\left( \frac{- \cos\pi x}{\pi} \right)dx\]

\[= \frac{- x\cos\pi x}{\pi} + \frac{\sin\pi x}{\pi^2}\]

Applying the limits, we get

\[\int_0^\frac{3}{2} \left| x\sin\pi x \right|dx = \left[ \frac{- x\cos\pi x}{\pi} + \frac{\sin\pi x}{\pi^2} \right]_0^1 - \left[ \frac{- x\cos\pi x}{\pi} + \frac{\sin\pi x}{\pi^2} \right]_1^\frac{3}{2} \]
\[ = \left[ \left( \frac{- \cos\pi}{\pi} + \frac{\sin\pi}{\pi^2} \right) - \left( 0 + 0 \right) \right] - \left[ \left( \frac{- \frac{3}{2}\cos\frac{3\pi}{2}}{\pi} + \frac{\sin\frac{3\pi}{2}}{\pi^2} \right) - \left( \frac{- \cos\pi}{\pi} + \frac{\sin\pi}{\pi^2} \right) \right]\]

\[= \left[ \left( \frac{1}{\pi} + 0 \right) \right] - \left[ \left( 0 - \frac{1}{\pi^2} \right) - \left( \frac{1}{\pi} + 0 \right) \right]\]
\[ = \frac{1}{\pi} + \frac{1}{\pi^2} + \frac{1}{\pi}\]
\[ = \frac{2}{\pi} + \frac{1}{\pi^2}\]
\[ = \frac{2\pi + 1}{\pi^2}\]