<s>[INST] Calculate the area between the function $\sqrt{x}$ and $x^2$ in the interval $[0 , 2]$. Calculate the area between the function 
$\sqrt{x}
$and 
$x^2$
in the interval [0 , 2].
Ok,  $0<x<1$, $\sqrt{x}$$>x^2$
and  $1<x<2$, $\sqrt{x}$$<x^2$
$$ Area = \int_{0}^1(\sqrt{x}-x^2)dx + \int_{1}^2(x^2-\sqrt{x})dx$$
I know that $$ \int(\sqrt{x})dx = \frac 23x^\frac 32$$
and $$ \int(x^2)dx = \frac 13x^3 $$
So the Area is $$ \frac {10}{3} - \frac {{4}\sqrt{2}}{3} $$
Is this correct? I'm not sure if I'm doing the right thing
 [/INST] To begin to answer your question one can first check have a graph of the equations handy, or understand that $y=\sqrt{x}$ raises faster than $y=x^2$ in the interval from $[0,1]$, after that $y=x^2$ is above the $y=\sqrt{x}$ from $[1,2]$. So know to setup the integrals:
\begin{align*}A_{\mathrm{tot}}&=A_1+A_2\\A_1&=\int_{0}^1(\sqrt{x}-x^2)dx\\A_2&=\int_1^2(x^2-\sqrt{x})dx\\ A_1&=[\frac{2}{3}x^{\frac{3}2}-\frac{x^3}3]\Bigg\vert_0^1=\frac{1}3 \\ A_2&=[\frac{x^3}{3}-\frac{2}{3}x^\frac{3}2]\Bigg\vert_1^2=\frac{9-4\sqrt2}{3}\\A_\mathrm{tot}&=\frac{10-4\sqrt{2}}{3}\end{align*}
</s>