Question

If the area bounded by the graph of $y=x{e}^{-ax}(a>0)$ and the abscissa is $\frac{1}{9}$ sq. unit, then the value of $a$ is equal to

• A. 3
• B. 03
• C. 3.0
• D. 3.00

Answer 1

Answer: (A), (B), (C), (D)

$y=x{e}^{-ax}$
$\Rightarrow y^{\prime }=e^{-ax}-xa{e}^{-ax}$
$=e^{-ax}(1-ax)=0$
$\therefore x=\frac{1}{a}$
As $x\to \infty ,y\to 0$
$x\to -\infty ,y\to -\infty$

Required area is
$A={\int }_0^{\infty }\begin{array}{c}\underset{}{x}\\ I\end{array}\begin{array}{c}\underset{}{e^{-ax}}\\ II\end{array}dx={\left[\frac{x{e}^{-ax}}{-a}\right]}_0^{\infty }+\frac{1}{a}{\int }_0^{\infty }{e}^{-ax}dx={\left[\frac{-1}{a^2}{e}^{-ax}\right]}_0^{\infty }b=-\frac{1}{a^2}[0-1]=\frac{1}{a^2}$
$\therefore \frac{1}{a^2}=\frac{1}{9}\Rightarrow a=3.$