• # question_answer If the area bounded by the graph of $y=x\,{{e}^{-ax}}(a>0)$ and the abscissa axis is $\frac{1}{9}$ then the value of 'a' is equal to A)  1                     B)  4 C)  6                                 D)  9

$y=x{{e}^{-ax}}$ $\Rightarrow \,{{y}^{t}}={{e}^{-ax}}-xa{{e}^{-ax}}$ $={{e}^{-ax}}(1-ax)=0$ $\therefore \,x=\frac{1}{a}$ as $x\to \infty ,\,y\to 0$ $x\to -\infty ,\,y\to \infty$ $A=\left. \int\limits_{a}^{\infty }{\underbrace{x}_{{}}\underbrace{{{e}^{-ax}}}_{{}}dx\,=\frac{x.{{e}^{-ax}}}{-a}} \right]_{a}^{\infty }$ $\left. +\frac{1}{a}\,\int_{0}^{\infty }{{{e}^{-ax}}dx=(0)\,-\frac{1}{{{a}^{2}}}{{e}^{-ax}}\,} \right]_{0}^{\infty }\,=\frac{-1}{{{a}^{2}}}[0-1]\,=\frac{1}{{{a}^{2}}}$ $\therefore \,\,\frac{1}{{{a}^{2}}}\,=\frac{1}{9}\Rightarrow \,\,a=3$