The area between two curves can be more easily states as the area between two graphs.

In order to do this you must take the antiderivative of the two functions.

This is denoted by have a capital letter such as $F(x)\quad and\quad G(x)$

Let's take the graphs $f(x)=x^{2}$ and $g(x)=x^{3}$

Let us also take the area between 2 and 4 as denoted by $2\leq A\leq 4$

$\int _{2}^{4}x^{2}\,dx$

and

$F(x)=\left({\frac {1}{3}}\right)*x^{3}$

$G(x)=\left({\frac {1}{4}}\right)*x^{4}$

Now evaluate the antiderivatives from 2 to 4

$F(x)=[\left({\frac {1}{3}}\right)*4^{3}]-[\left({\frac {1}{3}}\right)*2^{3}]$

$F(x)=[\left({\frac {1}{3}}\right)*64]-[\left({\frac {1}{3}}\right)*8]$

$F(x)=[\left({\frac {64}{3}}\right)]-[\left({\frac {8}{3}}\right)]$

$F(x)=\left({\frac {56}{3}}\right)$

Now we will evaluate the antiderivative G(x)

$G(x)=[\left({\frac {1}{4}}\right)*4^{4}]-[\left({\frac {1}{4}}\right)*2^{4}]$

$G(x)=[\left({\frac {1}{4}}\right)*256]-[\left({\frac {1}{4}}\right)*16]$

$G(x)=[64]-[4]$

$G(x)=60$

Now we take **F(x)-G(x)** to find the area between the two curves

$\left({\frac {56}{3}}\right)-\left({\frac {180}{3}}\right)$

$\left(-{\frac {124}{3}}\right)$

Therefore, $2\leq A\leq 4$ between $f(x)=x^{2}$ and $g(x)=x^{3}$ is $\left(-{\frac {124}{3}}\right)$