# How do you use part I of the Fundamental Theorem of Calculus to find the derivative of: f(x)= int(((1/3)t^2)-1)^7dt from x to 3?

Apr 6, 2018

$f ' \left(x\right) = - {\left(\frac{1}{3} {x}^{2} - 1\right)}^{7}$

#### Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

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

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

$f ' \left(x\right)$ where $f \left(x\right) = {\int}_{x}^{3} {\left(\frac{1}{3} {t}^{2} - 1\right)}^{7} \setminus \mathrm{dt}$

ie, we want:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} {\int}_{x}^{3} {\left(\frac{1}{3} {t}^{2} - 1\right)}^{7} \setminus \mathrm{dt}$

We must change the limits of integration to get the function into the correct form:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(- {\int}_{3}^{x} {\left(\frac{1}{3} {t}^{2} - 1\right)}^{7} \setminus \mathrm{dt}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \frac{d}{\mathrm{dx}} {\int}_{3}^{x} {\left(\frac{1}{3} {t}^{2} - 1\right)}^{7} \setminus \mathrm{dt}$

And so we can directly apply the FTOC, giving:

$f ' \left(x\right) = - {\left(\frac{1}{3} {x}^{2} - 1\right)}^{7}$