Question

Find the area bounded by the curve $y=x{e}^{-x^2}$, the x-axis, and the line $x=c$ where $y\left(c\right)$ is maximum

• A. $\frac{1}{2}(1-e^{-1/2})$
• B. $\frac{1}{4}(1-e^{-1/2})$
• C. $\frac{1}{2}(1-e^{1/2})$
• D. $\frac{1}{4}(1-e^{1/2})$

Answer 1

The correct option is A $\frac{1}{2}(1-e^{-1/2})$

$y=x{e}^{-x^2}$
$\frac{dy}{dx}=e^{-x^2}-2x^2{e}^{-x^2}$
$\frac{dy}{dx}=0$
$\Rightarrow e^{-x^2}(1-2x^2)=0$
$\Rightarrow x=\frac{\pm 1}{\sqrt{2}}$
$f\left(\frac{-1}{\sqrt{2}}\right)=-\frac{1}{\sqrt{2}}\left(e^{\frac{-1}{2}}\right)$.
$f\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}\left(e^{\frac{-1}{2}}\right)$.
So, we get a maximum at $x=\frac{1}{\sqrt{2}}$
Thus, the required area
$A={\int }_0^{\frac{1}{\sqrt{2}}}ydx$
$={\int }_0^{\frac{1}{\sqrt{2}}}x{e}^{-x^2}dx$
$=\frac{1}{2}{\int }_0^{\frac{1}{\sqrt{2}}}2x{e}^{-x^2}dx$
Let $x^2=t$ $\Rightarrow 2xdx=dt$
At $x=0,t=0$ and $x=\frac{1}{\sqrt{2}},t=\frac{1}{2}$.
Substituting in the above expression, we get
$A=\frac{1}{2}{\int }_0^{\frac{1}{2}}{e}^{-t}.dt$
$=\frac{1}{2}[e^{-t}{]}_{\frac{1}{2}}^0$
$=\frac{1}{2}(1-e^{\frac{-1}{2}})$ sq units.