Answer
$\dfrac{1}{2}-\dfrac{1}{2e} \approx 0.3160$
Work Step by Step
We are given that $y=xe^{x^2-1} $ and $x=0$ to $x=1$
Here, the area $(A)$ can be expressed as: $A=\int_0^{1} xe^{x^2-1} \ dx$
Let us consider that $u=x^2-1 \implies dx=\dfrac{du}{2x}$
Now, we have $A=\dfrac{1}{2} \int_0^{1} e^u \ du $
or, $=[\dfrac{1}{2} e^u ]_0^{1}+C$
or, $=[\dfrac{1}{2} e^{x^2-1}]_0^{1} +C$
or, $=\dfrac{1}{2} [e^{1-1}-e^{0-1}]$
or, $=\dfrac{1}{2} [1-e^{-1}]$
Therefore, the required area is: $Area=\dfrac{1}{2}-\dfrac{1}{2e} \approx 0.3160$
