If for a continuous function f( x), ∫_-π^t(f(x)+x d x)=π^2-t^2, for all t ≥-π, then f (- π 3 ) is equal to:

Let $\displaystyle \int_{-\pi}^t(f(x)+x) d x=\pi^2-t^2$∫−πt​(f(x)+x)dx=π2−t2 $$\begin{aligned} &\Rightarrow \int_{-\pi}^t f(x) d x+\int_{-\pi}^t x d x=\pi^2-t^2 \\ &\Rightarrow \int_{-\pi}^t f(x) d x+\left(\frac{t^2}{2}-\frac{\pi^2}{2}\right)=\pi^2-t^2 \\ &\Rightarrow \int_{-\pi}^t f(x) d x=\frac{3}{2}\left(\pi^2-t^2\right) \end{aligned}$$​⇒∫−πt​f(x)dx+∫−πt​xdx=π2−t2⇒∫−πt​f(x)dx+(2t2​−2π2​)=π2−t2⇒∫−πt​f(x)dx=23​(π2−t2)​ differentiating with respect to $t$t $\displaystyle \frac{d}{d t}\left[\int_{-\pi}^t f(x) d x\right]=\frac{3}{2} \frac{d}{d t}\left(\pi^2-t^2\right)$dtd​[∫−πt​f(x)dx]=23​dtd​(π2−t2) $f(t) \cdot \frac{d t}{d t}-f(-\pi) \frac{d}{d t}(-\pi)=-3 t$f(t)⋅dtdt​−f(−π)dtd​(−π)=−3t $f(t)=-3 t$f(t)=−3t $f\left(-\frac{\pi}{3}\right)=-3\left(-\frac{\pi}{3}\right)=\pi$f(−3π​)=−3(−3π​)=π