Answer
$\dfrac{1-e^{-25}}{2} \approx 0.5$
Work Step by Step
We are given that $y=xe^{-x^2} $ and $x=0$ to $x=5$
Here, the area $(A)$ can be expressed as: $A=\int_{0}^{5} xe^{-x^2} \ dx$
Suppose that $u=-x^2 \implies dx=\dfrac{-du}{2x}$
Now, we have $A=\int_{0}^{5} xe^{u} (\dfrac{-du}{2x})$
or, $=-\dfrac{1}{2} [e^u]_0^5$
or, $=-\dfrac{1}{2} [e^{-x^2}]_0^5$
or, $=-\dfrac{1}{2} [e^{-(5)^2}] +\dfrac{1}{2} [e^{-(0)^2}]$
Therefore, the required area is: $Area=\dfrac{1-e^{-25}}{2} \approx 0.5$
