Question 83323
A(x) = f'(x) = x^2 where as A(x) is the area under the curve
{{{int(x^2,dx,0,4) = x^3/3}}} | 4/0
4^3/3 - 0^3/3
64/3
{{{graph(300,300,0,4,0,16,x^2)}}}