Question

JEE Mains Maths question is in the image

Answer 1

we have to evaluate $\int\limits^{\pi}_{-\pi}{|\pi-|x||}\,dx$

solution : concept : $\int\limits^a_{-a}f(x)\,dx=2\int\limits^a_0f(x)\,dx$ where f(x) is an even function i.e., f(x) = f(-x)

here |π - |x|| is an even function.

so, $\int\limits^{\pi}_{-\pi}{|\pi-|x||}\,dx=2\int\limits^{\pi}_0{|\pi-|x||}\,dx$

as 0 < x < π ⇒|x| = x

and |π - x| = (π - x)

so, $2\int\limits^{\pi}_0{|\pi-|x||}\,dx=2\int\limits^{\pi}_0{(\pi-x)}\,dx$

= $2\left[\pi x-\frac{x}{2}\right]^{\pi}_0$

= 2[π² - π²/2]

= π²

Therefore the value of $\int\limits^{\pi}_{-\pi}{|\pi-|x||}\,dx$ is π²