# How do you use the second fundamental theorem of Calculus to find the derivative of given int (2t-1)^3 dt from [x^2, x^7]?

Feb 23, 2016

The derivative is ${\left(2 {x}^{7} - 1\right)}^{3} \left(7 {x}^{6}\right) - {\left(2 {x}^{2} - 1\right)}^{3} \left(2 x\right)$

#### Explanation:

Suppose $f \left(x\right)$ is continuous on $\left[a , b\right]$ and both $p \left(x\right)$ and $q \left(x\right)$ are differentiable on $\left(a , b\right)$

Define the function

$F \left(x\right) = {\int}_{p \left(x\right)}^{q \left(x\right)} f \left(t\right) \text{d} t$

Combining the Second Fundamental Theorem of Calculus and the Chain Rule implies that $F \left(x\right)$ is differentiable and

$F ' \left(x\right) = f \left(q \left(x\right)\right) \cdot q ' \left(x\right) - f \left(p \left(x\right)\right) \cdot p ' \left(x\right)$

So, in this case we have

$f \left(t\right) = {\left(2 t - 1\right)}^{3}$
$p \left(x\right) = {x}^{2}$
$q \left(x\right) = {x}^{7}$

Plug it in to get

frac{"d"}{"d"x}(int_(x^2)^(x^7) (2t-1)^3 "d"t)

$= {\left(2 \left({x}^{7}\right) - 1\right)}^{3} \frac{\text{d"}{"d"x}(x^7) - (2(x^2)-1)^3 frac{"d"}{"d} x}{{x}^{2}}$

$= {\left(2 {x}^{7} - 1\right)}^{3} \left(7 {x}^{6}\right) - {\left(2 {x}^{2} - 1\right)}^{3} \left(2 x\right)$