<s>[INST] Evaluate $\int_0^1 \sqrt{2x-1} - \sqrt{x}$ $dx$ I'm trying to calculate the area between the curves $y = \sqrt{x}$ and $y= \sqrt{2x-1}$
Here's the graph: 
I've already tried calculating the area with respect to $y$, i.e. 
$\int_0^1 (\frac{y^2+1}{2} - y^2)$ $ dx$  
[since $y^2=x$ for the first curve and $\frac{y^2+1}{2}=x$ for the second curve]
And the result for that integral is $1/3$ which should be the same for the result of the integral in question
$\int_0^1 (\sqrt{2x-1} - \sqrt{x})$ $dx$
But, $\int_0^1 (\sqrt{2x-1} - \sqrt{x})$ $dx$ = $\int_0^1 (2x-1)^{1/2}$ $dx$ - $\int_0^1 {x}^{1/2} $ $dx$ 
[let $2x-1 = u$ and $2 dx = du$]
$\Rightarrow$ $\frac 12 \int_0^1 (u)^{1/2}$ $du$ - $\frac{2x^{3/2}}{3}$
$=\frac{u^{3/2}}{3}$ - $\frac{2x^{3/2}}{3}$
$=\frac{(2x-1)^{3/2}}{3}$ - $\frac{2x^{3/2}}{3}$
Evaluated at $x=1$ and $x=0$ and subtracting:
$(\frac{-1}{3})-(\frac{-1}{3})$
Why?
 [/INST] I would say that you can integrate as you have suggested, however, despite the variable y:
$\int_0^1 (\frac{y^2+1}{2} - y^2)\color{red}{dy}=\frac{1}{2}[y-\frac{y^3}{3}]_0^1=\frac{1}{3}$
There is no reason to go back to the variable x.
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