$$\int\limits_{ - 1}^2 {|x\sin \pi x|dx = \int\limits_{ - 1}^1 {|x\sin \pi x|dx + \int\limits_1^2 {|x\sin \pi x|dx = 2\int\limits_0^1 {|x\sin \pi x|dx + \int\limits_1^2 {( - x\sin \pi x)dx} } } } } $$

[$$\because$$ $$|x\sin \pi x|$$ is an even function; For $$1 < x < 2,\,x\sin \pi x < 0$$]

$$ = 2\int\limits_0^1 {x\sin \pi x\,.\,dx - \int\limits_1^2 {x\sin \pi x\,dx} } $$ [For $$0 < x < 1,\,x\sin \pi x > 0$$]

$$ = 2\left[ {x\left( {{{ - \cos \pi x} \over \pi }} \right)} \right]_0^2 = 2\int\limits_0^1 {1\left( {{{ - \cos \pi x} \over \pi }} \right)dx - \left[ {x\left( {{{ - \cos \pi x} \over \pi }} \right)} \right]_1^2 + \int\limits_1^2 {1\left( {{{ - \cos \pi x} \over \pi }} \right)dx} } $$

$$ = - {2 \over \pi }[\cos \pi - 0] + {2 \over \pi }\left[ {{{\sin \pi x} \over \pi }} \right]_0^1 + {1 \over \pi }[2\cos 2\pi - \cos \pi ] - {1 \over \pi }\left[ {{{\sin \pi x} \over \pi }} \right]_1^2$$

$$ = {2 \over \pi } + {2 \over {{\pi ^2}}}(0) + {1 \over \pi }(2 + 1) - {1 \over {{\pi ^2}}}(0) = {5 \over \pi }$$