If $\int_e^x t f(t) d t=\sin x-x \cos x-\frac{x^2}{2}$, for all $x \in R-\{0\}$, then the value of…
If $\int_e^x t f(t) d t=\sin x-x \cos x-\frac{x^2}{2}$, for all $x \in R-\{0\}$, then the value of $f\left(\frac{\pi}{6}\right)$ is
1/2
1
0
$-1 / 2$
Solution
Let $\int_e^x t f(t) d t=\sin x-x \cos x-\frac{x^2}{2}$
By using Leibnitz rule, we get
$
\begin{aligned}
& \frac{d}{d x}\left[\int_e^x t f(t) d t\right]=\frac{d}{d x}\left[\sin x-x \cos x-\frac{x^2}{2}\right] \\
& \Rightarrow x f(x)-e f(e) \cdot 0=x \sin x-x
\end{aligned}
$
Now, put $x=\frac{\pi}{6}$, we get
$
\begin{aligned}
& \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{aligned}
$