Let f(x) = x + a π ^ 2 - 4 x + b π ^ 2 - 4 x , x ∈ ℝ be a function...

Given,

$f\left(x\right) = x + {∫}_{0}^{\pi / 2}\sin \left(x + y\right)f\left(y\right)dy$

$⇒ f\left(x\right) = x + {∫}_{0}^{\pi / 2}\left(\sin x \cos y + \cos x \sin y\right)f\left(y\right)dy$

$⇒ f\left(x\right) = x + {∫}_{0}^{\pi / 2}\left(\left(cosyf\left(y\right)dy\right)\sin x + \left(sinyf\left(y\right)dy\right)\cos x\right)$

And,$f\left(x\right) = x + \dfrac{a}{{\pi}^{2} - 4}\sin x + \dfrac{b}{{\pi}^{2} - 4}\cos x, x \in \mathbb{R}$

On comparing both the equations of $f ( x )$ we get,

$⇒ \dfrac{a}{{\pi}^{2} - 4} = {∫}_{0}^{\pi / 2}cosyf\left(y\right)dy \ldots \left(1\right)$

$⇒ \dfrac{b}{{\pi}^{2} - 4} = {∫}_{0}^{\pi / 2}sinyf\left(y\right)dy \ldots \left(2\right)$

Adding equation \left(1\right)$and equation \left(2\right)$ we get,

$\dfrac{a + b}{{\pi}^{2} - 4} = {∫}_{0}^{\pi / 2}\left(\sin y + \cos y\right)f\left(y\right)dy \ldots \left(3\right)$

$∵ {∫}_{0}^{a}f ( x ) d x = {∫}_{0}^{a}f ( a - x ) d x$

$∴ \dfrac{a + b}{{\pi}^{2} - 4} = {∫}_{0}^{\pi / 2}\left(\sin y + \cos y\right)f\left(\dfrac{\pi}{2} - y\right)dy \ldots \left(4\right)$

Add equation \left(3\right)$and equation \left(4\right)$ we get,

$\dfrac{2\left(a + b\right)}{{\pi}^{2} - 4} = {∫}_{0}^{\pi / 2}\left(\sin y + \cos y\right)\left(\dfrac{\pi}{2} + \dfrac{\left(a + b\right)}{{\pi}^{2} - 4}\left(\sin y + \cos y\right)\right)dy$

$⇒ \dfrac{2\left(a + b\right)}{{\pi}^{2} - 4} = \pi + \dfrac{a + b}{{\pi}^{2} - 4}\left(\dfrac{\pi}{2} + 1\right)$

$⇒ \left(a + b\right) = - 2\pi\left(\pi + 2\right)$