Question

If ${\int }_0^{x}f\left(t\right)dt=x^2+{\int }_x^1{t}^{2}f\left(t\right)dt;$ then the value of $I={\int }_0^{e}f\left(x\right)dx$ is

• A. $lne$
• B. $1$
• C. $0$
• D. $ln(1+e^2)$

Answer 1

The correct option is D $ln(1+e^2)$

${\int }_0^{x}f\left(t\right)dt=x^2+{\int }_x^1{t}^{2}f\left(t\right)dt$
Differentiating both sides w.r.t x
$f\left(x\right)=2x-x^{2}f\left(x\right)\Rightarrow f\left(x\right)(1+x^2)=2x$
$\Rightarrow f\left(x\right)=\frac{2x}{1+x^2}$
Therefore
$I={\int }_0^{e}f\left(x\right)dx={\int }_0^e\frac{2x}{1+x^2}dx$
$={\left[\log (1+x^2)\right]}_0^e=\log (1+e^2)$