# How do you use the Fundamental Theorem of Calculus to find the derivative of int (2t - 1)^3dt from x^5 to x^6?

Jul 17, 2016

$= 6 {x}^{5} {\left(2 {x}^{6} - 1\right)}^{3} - 5 {x}^{4} {\left(2 {x}^{5} - 1\right)}^{3}$

#### Explanation:

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

by chain rule where $u = u \left(x\right)$

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

in this case we have $\frac{d}{\mathrm{dx}} {\int}_{{x}^{5}}^{{x}^{6}} {\left(2 t - 1\right)}^{3} \mathrm{dt}$

.........and we need to rewrite this is the forms given in $\triangle$ and $\star$

$\frac{d}{\mathrm{dx}} {\int}_{{x}^{5}}^{{x}^{6}} {\left(2 t - 1\right)}^{3} \mathrm{dt}$

$= \frac{d}{\mathrm{dx}} {\int}_{{x}^{5}}^{a} {\left(2 t - 1\right)}^{3} \mathrm{dt} + {\int}_{a}^{{x}^{6}} {\left(2 t - 1\right)}^{3} \mathrm{dt}$

$= - \frac{d}{\mathrm{dx}} {\int}_{a}^{{x}^{5}} {\left(2 t - 1\right)}^{3} \mathrm{dt} + {\int}_{a}^{{x}^{6}} {\left(2 t - 1\right)}^{3} \mathrm{dt}$

and applying FTC and the chain rule

$= - {\left(2 \left({x}^{5}\right) - 1\right)}^{3} \frac{d}{\mathrm{dx}} {x}^{5} + {\left(2 \left({x}^{6}\right) - 1\right)}^{3} \frac{d}{\mathrm{dx}} {x}^{6}$

$= - {\left(2 {x}^{5} - 1\right)}^{3} \cdot 5 {x}^{4} + {\left(2 {x}^{6} - 1\right)}^{3} \cdot 6 {x}^{5}$

$= 6 {x}^{5} {\left(2 {x}^{6} - 1\right)}^{3} - 5 {x}^{4} {\left(2 {x}^{5} - 1\right)}^{3}$