Question

\[\int_0^1 | x\sin \pi x | dx\]

Answer 1

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

\[\Rightarrow\ x\sin\pi x > 0\]

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

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

Applying the limits, we get

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

\[= \frac{1}{\pi} + 0 - 0\]
\[ = \frac{1}{\pi}\]