Mathematics - Definite Integration Question with Solution | TestHub

Let ${\int }_e^{x}tf(t)dt=\sin x-x\cos x-\frac{x^2}{2}$ By using Leibnitz rule, we get $$\begin{array}{rl}& \frac{d}{dx}\left[{\int }_e^{x}tf(t)dt\right]=\frac{d}{dx}\left[\sin x-x\cos x-\frac{x^2}{2}\right]\\ & \Rightarrow xf(x)-ef(e)\cdot 0=x\sin x-x\end{array}$$ Now, put $x=\frac{\pi }{6}$, we get $$\begin{array}{rl}& \frac{\pi }{6}\cdot f\left(\frac{\pi }{6}\right)=\frac{\pi }{6}\cdot \sin \frac{\pi }{6}-\frac{\pi }{6}\\ & \Rightarrow f\left(\frac{\pi }{6}\right)=\sin \frac{\pi }{6}-1=\frac{1}{2}-1=-\frac{1}{2}\end{array}$$