# How do you use the Fundamental Theorem of Calculus to find the derivative of int (u^3) / (1+u^2) du from 2-3x to 5?

Feb 17, 2017

#### Answer:

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

#### Explanation:

The following assumes you mean the derivative wrt $x$.

We start with the formal statement of the FTC, Part I:

$\frac{d}{\mathrm{du}} \left({\int}_{a}^{u} f \left(t\right) \setminus \mathrm{dt}\right) = f \left(u\right)$ where $a =$ some constant.

If in fact $u = u \left(x\right)$, we can use the Chain Rule so that:

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

We can also play with the intervals ..... in summary ${\int}_{a}^{b} = {\int}_{a}^{0} + {\int}_{0}^{b} = {\int}_{0}^{b} - {\int}_{0}^{a}$.... in order to reach this point:

$\frac{d}{\mathrm{dx}} \left({\int}_{{u}_{1} \left(x\right)}^{{u}_{2} \left(x\right)} f \left(t\right) \setminus \mathrm{dt}\right) = f \left({u}_{2} \left(x\right)\right) \frac{{\mathrm{du}}_{2}}{\mathrm{dx}} - f \left({u}_{1} \left(x\right)\right) \frac{{\mathrm{du}}_{1}}{\mathrm{dx}}$

Gasp! With all that, we just need to process this:

$\frac{d}{\mathrm{dx}} \left({\int}_{2 - 3 x}^{5} \frac{{u}^{3}}{1 + {u}^{2}} \mathrm{du}\right)$

$= \frac{{5}^{3}}{1 + {5}^{2}} \frac{d \left(5\right)}{\mathrm{dx}} - \frac{{\left(2 - 3 x\right)}^{3}}{1 + {\left(2 - 3 x\right)}^{2}} \frac{d \left(2 - 3 x\right)}{\mathrm{dx}}$

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

Maybe you can further simplify that.