Question

f(x) is a differentiable function satisfying the relation $f\left(x\right)=x^2+{\int }_0^x{e}^{-t}f(x-t)dt$, then

• A. 1060
• B. 1260
• C. 960
• D. 1224

Answer 1

The correct option is C

960

$f\left(x\right)=x^2+{\int }_0^x{e}^{-t}f(x-t)dt=x^2+{\int }_0^x{e}^{-(x-t)}f\left(t\right)dt=x^2+e^{-x}{\int }_0^x{e}^{t}f\left(t\right)dt\therefore e^{x}f\left(x\right)=x^2{e}^x+{\int }_0^x{e}^{t}f\left(t\right)dt$
Differentiating both sides w.r.tx
$e^{x}f\left(x\right)+e^x{f}^{\prime }\left(x\right)=2x{e}^x+x^2{e}^x+e^{x}f\left(x\right)\therefore f^{\prime }\left(x\right)=2x+x^2\therefore f\left(x\right)=x^2+\frac{x^3}{3}+cf\left(0\right)=0\Rightarrow c=0\therefore f\left(x\right)=x^2+\frac{x^3}{3}\therefore {\sum }_{k=1}^{9}f\left(k\right)=\frac{\mathrm{9.10.19}}{6}+\frac{91.100}{12}=960$