Mathematics - Differential Equation Question with Solution | TestHub

Let ${\int }_{-\pi }^t(f(x)+x)dx={\pi }^2-t^2$ $$\begin{array}{rl}& \Rightarrow {\int }_{-\pi }^{t}f(x)dx+{\int }_{-\pi }^{t}xdx={\pi }^2-t^2\\ & \Rightarrow {\int }_{-\pi }^{t}f(x)dx+\left(\frac{t^2}{2}-\frac{{\pi }^2}{2}\right)={\pi }^2-t^2\\ & \Rightarrow {\int }_{-\pi }^{t}f(x)dx=\frac{3}{2}\left({\pi }^2-t^2\right)\end{array}$$ differentiating with respect to $t$ $\frac{d}{dt}\left[{\int }_{-\pi }^{t}f(x)dx\right]=\frac{3}{2}\frac{d}{dt}\left({\pi }^2-t^2\right)$ $f(t)\cdot \frac{dt}{dt}-f(-\pi )\frac{d}{dt}(-\pi )=-3t$ $f(t)=-3t$ $f\left(-\frac{\pi }{3}\right)=-3\left(-\frac{\pi }{3}\right)=\pi$