Question
The function $f$ is defined on $[0, 1]$ by $$f(x) = e^{2\sqrt x}\ .$$
Use the substitution $x = u^2$ to find the exact area of the region enclosed by the graph of $f$, the axes and the line $x = 1$, then use your answer to deduce the exact value of $$\int_{1}^{e^2} (\ln y)^2\ \mathrm{d}y\ .$$
My working
Let the required area be $A$.
\begin{align} A & = \int_{0}^{1} e^{2\sqrt x}\ \mathrm{d}x \\[5 mm] & = \int_{0}^{1} 2ue^{2u}\ \mathrm{d}u \end{align}
Let $s = 2u$ and $t' = e^{2u}\ .$
$\implies s' = 2$ and $t = \frac 1 2 e^{2u}\ .$
\begin{align} A & = [ue^{2u} - \frac 1 2 e^{2u}]^{u = 1}_{u = 0} \\[5 mm] & = \frac 1 2 (e^2 + 1) \end{align}
I am able to complete the first part (finding the area), but I do not know how to proceed with the second part (evaluating the integral). I have tried to use another substitution, where $x = (\ln y)^2$, but I am not getting anywhere. Any help/intuition will be greatly appreciated :)
The required integral is connected with the inverse function $$y=e^{2 \sqrt{x}}\to 2\sqrt{x}=\ln y\to x=\frac{\ln^2 y}{4}$$ The area of the inverse function (red in the picture) is the area of the rectangle less the blue area calculated by OP. $$\frac{1}{4}\int_1^{e^2}\ln^2 y\,dy=e^2-\frac{1+e^2}{2}=\frac{1}{2} \left(e^2-1\right)$$ Therefore requested integral is $$\int_1^{e^2}\ln^2 y\,dy=2\left(e^2-1\right)$$