Question

Let $f:R\to R$ be a differentiable function such that $f\left(x\right)=x^2+\underset{0}{\overset{x}{\int }}{e}^{-t}f(x-t)dt.$ Then, the value of $\underset{0}{\overset{1}{\int }}f\left(x\right)dx$ is

• A. $\frac{1}{4}$
• B. $-\frac{1}{12}$
• C. $\frac{5}{12}$
• D. $\frac{12}{7}$

Answer 1

The correct option is C $\frac{5}{12}$

$f\left(x\right)=x^2+\underset{0}{\overset{x}{\int }}{e}^{-t}f(x-t)dt\cdots \left(i\right)$
Using the property:
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\Rightarrow f\left(x\right)=x^2+\underset{0}{\overset{x}{\int }}{e}^{-(x-t)}f(x-(x-t))dt$
$=x^2+e^{-x}\underset{0}{\overset{x}{\int }}{e}^{t}f\left(t\right)dt\cdots \left(ii\right)$
Differentiating w.r.t. $x,$ we get
$\Rightarrow f^{\prime }\left(x\right)=2x-e^{-x}\underset{0}{\overset{x}{\int }}{e}^{t}f\left(t\right)dt+e^{-x}{e}^{x}f\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=2x-e^{-x}\underset{0}{\overset{x}{\int }}{e}^{t}f\left(t\right)dt+f\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=2x+x^2\text{[Using equation}\left(ii\right)]$
$\Rightarrow f\left(x\right)=\frac{x^3}{3}+x^2+c$
Also, $f\left(0\right)=0\text{[From equation}\left(i\right)]$
$\Rightarrow f\left(x\right)=\frac{x^3}{3}+x^2$
$\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\underset{0}{\overset{1}{\int }}(\frac{x^3}{3}+x^2)dx$
$={[\frac{x^4}{12}+\frac{x^3}{3}]}_0^1=\frac{1}{12}+\frac{1}{3}=\frac{5}{12}$