Question

Evaluate : ${\int }_0^{\frac{\pi }{2}}|1-x|dx$

Answer 1

${\int }_0^{\frac{\pi }{2}}|1-x|dx$

$={\int }_0^1(1-x)dx+{\int }_1^{\frac{\pi }{2}}(x-1)dx$

$={[x-\frac{x^2}{2}]}_0^1+{[\frac{x^2}{2}-x]}_1^{\frac{\pi }{2}}$

$=1-\frac{1}{2}-0+\frac{{\pi }^2}{4}-\frac{\pi }{2}-\frac{1}{2}+1$

$=1+\frac{{\pi }^2}{4}-\frac{\pi }{2}$