Question

If $f\left(x\right)={\int }_0^{x}t\sin tdt$, the write the value of $f\text{'}\left(x\right)$. [CBSE 2014]

Answer 1

$$\begin{gathered} f\left(x\right)={\int }_0^{x}t\sin tdt \\ \Rightarrow f\left(x\right)={\overline{)t\left(-\cos t\right)}}_0^x-{\int }_0^x\frac{d}{dt}\left(t\right)\times \left(-\cos t\right)dt \\ \Rightarrow f\left(x\right)=-\left(x\cos x-0\right)+{\int }_0^x\cos tdt \\ \Rightarrow f\left(x\right)=-x\cos x+{\overline{)\sin t}}_0^x \end{gathered}$$
$$\begin{gathered} \Rightarrow f\left(x\right)=-x\cos x+\left(\sin x-0\right) \\ \Rightarrow f\left(x\right)=-x\cos x+\sin x \end{gathered}$$

Differentiating both sides with respect to x, we get

$$\begin{gathered} f\text{'}\left(x\right)=-\left[x\times \left(-\sin x\right)+\cos x\times 1\right]+\cos x \\ \Rightarrow f\text{'}\left(x\right)=-\left(-x\sin x\right)-\cos x+\cos x \\ \Rightarrow f\text{'}\left(x\right)=x\sin x \end{gathered}$$

Thus, the value of $f\text{'}\left(x\right)$ is x sinx.