Answer
$All$ $three$ $areas$ $are$ $equal.$
Work Step by Step
$Area$ $(1)$ $=$ $\int_{0}^{1}e^\sqrt{x}dx$
Let $u$ $=$ $\sqrt{x}$
Then $du$ $=$ $\frac{1}{2}$$x^\frac{-1}{2}$$dx$ $=$ $\frac{1}{2\sqrt x}$ $dx$ $=$ $\frac{1}{2u}$ $dx$
When $x$ $=$ $0$, then $u$ $=$ $0$
When $x$ $=$ $1$, then $u$ $=$ $1$
So, we have $Area$ $(1)$ $=$ $\int_{0}^{1}2{u}e^{u}du$
$Area$ $(2)$ $=$ $\int_{0}^{1}2{x}e^{x}dx$ [same as $Area$ $(1)$]
$Area$ $(3)$ $=$ $\int_{0}^{1}e^\sqrt{sin(x)}sin(2x)dx$
$=$ $\int_{0}^{1}e^\sqrt{sin(x)}2sin(x)cos(x)dx$
Let $u$ $=$ $\sin(x)$
Then $du$ $=$ $\cos(x)dx$
When $x$ $=$ ${\pi/2}$, then $u$ $=$ $\sin(\pi/2)$ $=$ $1$
When $x$ $=$ $0$, then $u$ $=$ $\sin(0)$ $=$ $0$
So, we have $Area$ $(3)$ $=$ $\int_{0}^{1}2{u}e^{u}du$ [same as $Area$ $(1)$ and $Area$ $(2)$]
$\Rightarrow$$All$ $three$ $areas$ $are$ $equal$
