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