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

Let

$I = {∫}_{0}^{\pi}f\left(x\right)sinxdx . . . \left(1\right)$

Now using the property ${∫}_{a}^{b}f\left(x\right)dx = {∫}_{a}^{b}f\left(a + b - x\right)dx$ we get,

$⇒ I = {∫}_{0}^{\pi}f\left(\pi - x\right)\sin \left(\pi - x\right)dx$

$⇒ I = {∫}_{0}^{\pi}f\left(\pi - x\right)sinxdx . . . \left(2\right)$

Adding \left(1\right) \text{ and } \left(2\right)$, we get</p><p>$2I = {∫}_{0}^{\pi}\left[f\left(x\right) + f\left(\pi - x\right)\right]sinxdx$</p><p>$⇒ 2I = {\pi}^{2}{∫}_{0}^{\pi}sinxdx${as given f\left(x\right) + f\left(\pi - x\right) = {\pi}^{2}}$

$⇒ 2I = {\pi}^{2}{\left[- \cos x\right]}_{0}^{\pi}$

$⇒ 2I = 2{\pi}^{2}$

⇒ I = {\pi}^{2}

Hence this is the correct option.