# 7.1 2nd Fundamental Theorem Of Calculusap Calculus

1. In this wiki, we will see how the two main branches of calculus, differential and integral calculus, are related to each other. While the two might seem to be unrelated to each other, as one arose from the tangent problem and the other arose from the area problem, we will see that the fundamental theorem of calculus does indeed create a link between the two.
2. Now, what I want to do in this video is connect the first fundamental theorem of calculus to the second part, or the second fundamental theorem of calculus, which we tend to use to actually evaluate definite integrals. So let's think about what F of b minus F of a is, what this is, where both b and a are also in this interval.

### 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]#?

The first part of the fundamental theorem stets that when solving indefinite integrals between two points a and b, just subtract the value of the integral at a from the value of the integral at b. The second part of the theorem gives an indefinite integral of a function.

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

$F \left(x\right) = {\int}_{p \left(x\right)}^{q \left(x\right)} f \left(t\right) ' d ' t$

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

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

## 7.1 2nd Fundamental Theorem Of Calculus Ap Calculus Free

Plug it in to get

$\frac{' d '}{' d ' x} \left({\int}_{{x}^{2}}^{{x}^{7}} {\left(2 t - 1\right)}^{3} ' d ' t\right)$

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

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

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