Question

If 'a' is constant, then the value of the integral $a^2\underset{0}{\overset{\infty }{\int }}x{e}^{-ax}dx$ is _______

• A. 1

Answer 1

The correct option is A 1
From integration by parts rule,
$\int udv=uv-\int vdu$
Let u = x and $dv=e^{-ax}dx$. Then, du = dx and $v=\frac{e^{-ax}}{-a}$. Therefore

$\int x{e}^{-ax}dx=x\left(\frac{e^{-ax}}{-a}\right)-\int \left(\frac{e^{-ax}}{-a}\right)dx$

$=\frac{-x{e}^{-ax}}{a}-\frac{1}{a^2}{e}^{-ax}$

$Hence,{\int }_0^{\infty }x{e}^{-ax}dx={\left[\frac{-x{e}^{-ax}}{a}\right]}_0^{\infty }-{\left[\frac{1}{a^2}{e}^{-ax}\right]}_0^{\infty }$

$=(0-0)-\frac{1}{a^2}(0-1)=\frac{1}{a^2}$

$a^2\underset{0}{\overset{\infty }{\int }}x{e}^{-ax}dx=a^2\cdot \frac{1}{a^2}=1$