# How do you use the fundamental theorem of calculus to find F'(x) given F(x)=int t^2sqrt(1-t^3)dt from [0,x]?

Mar 3, 2017

$F ' \left(x\right) = {x}^{2} \sqrt{1 - {x}^{3}}$

#### Explanation:

If asked to find the derivative of an integral using the fundamental theorem of Calculus, you should not evaluate the integral

The Fundamental Theorem of Calculus tells us that:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{a}^{x} \setminus f \left(t\right) \setminus \mathrm{dt} = f \left(x\right)$

(ie the derivative of an integral gives us the original function back).

$F ' \left(x\right) = \frac{d}{\mathrm{dx}} F \left(x\right)$
$\text{ } = \frac{d}{\mathrm{dx}} \setminus {\int}_{0}^{x} \setminus {t}^{2} \sqrt{1 - {t}^{3}} \setminus \mathrm{dt}$
$\text{ } = f \left(x\right)$, where $f \left(t\right) = {t}^{2} \sqrt{1 - {t}^{3}} \setminus \setminus \setminus$ (using FTOC)
$\text{ } = {x}^{2} \sqrt{1 - {x}^{3}}$