Question

If ${\int }_0^{x}f\left(t\right)dt=x^2+{\int }_x^{1}tf\left(t\right)dt$, then the value of $f\left(1\right)$ is:

• A. $0$
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $1$

Answer 1

The correct option is D $1$

Differentiating both sides-

$\frac{d}{dx}{\int }_0^{x}f\left(t\right)dt=\frac{d}{dx}{x}^2+\frac{d}{dx}{\int }_x^{1}tf\left(t\right)dt$

Now using Leibnitz integral rule-

$\frac{d}{dx}{\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x\right)dx$

$=f\left(b\right(x\left)\right)\frac{d}{dx}b\left(x\right)-f\left(a\right(x\left)\right)\frac{d}{dx}a\left(x\right)$

Hence, we get-

$f\left(x\right)=2x+(0-xf(x\left)\right)=2x-xf\left(x\right)$

putting x=1:-

$f\left(1\right)=2-f\left(1\right)⟹2f\left(1\right)=2$

$f\left(1\right)=1$

Hence, answer is option-(D).