If for a continuous function $\mathrm{f}(\mathrm{x})$, $\int_{-\pi}^t(f(x)+x d x)=\pi^2-t^2$, for all $t…
If for a continuous function $\mathrm{f}(\mathrm{x})$, $\int_{-\pi}^t(f(x)+x d x)=\pi^2-t^2$, for all $t \geq-\pi$, then $\mathrm{f}\left(-\frac{\pi}{3}\right)$ is equal to:
$\pi$
$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{6}$
Solution
Let $\int_{-\pi}^t(f(x)+x) d x=\pi^2-t^2$
$
\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}
$
differentiating with respect to $t$
$\frac{d}{d t}\left[\int_{-\pi}^t f(x) d x\right]=\frac{3}{2} \frac{d}{d t}\left(\pi^2-t^2\right)$
$f(t) \cdot \frac{d t}{d t}-f(-\pi) \frac{d}{d t}(-\pi)=-3 t$
$f(t)=-3 t$
$f\left(-\frac{\pi}{3}\right)=-3\left(-\frac{\pi}{3}\right)=\pi$