Question

Let $f\left(x\right)$ be a function satisfying $f^{\prime }\left(x\right)=\frac{x^3-f\left(x\right)}{x}withf\left(2\right)=2$ then the value of $({\int }_1^2{e}^{x^2}.f\left(x\right)dx)$ is -

• A. $\frac{{5e}^2}{8}$
• B. $\frac{e^2}{8}$
• C. $\frac{{3e}^4}{8}$
• D. $\frac{5}{8}{3e}^4$

Answer 1

The correct option is C $\frac{{3e}^4}{8}$

given $x{f}^{\prime }\left(x\right)+f\left(x\right)=x^3$
$\Rightarrow f\left(x\right).e^{\int \frac{1}{x}dx}=\int x^2.e^{\int \frac{1}{x}dx}dx$
$\Rightarrow xf\left(x\right)=\int x^{3}dx$
$\Rightarrow xf\left(x\right)=\frac{x^4}{4}+C$
given $f\left(2\right)=2\Rightarrow c=0$
$\therefore f\left(x\right)=\frac{x^3}{4}$
$I={\int }_1^2{e}^{x^2}f\left(x\right)dx={\int }_1^2{e}^{x^2}.\frac{x^2.x}{4}dx=\frac{{3e}^4}{8}$