# 7.1 2nd Fundamental Theorem Of Calculusap Calculus

### 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 Fundamental Theorems of Calculus I. If f is continuous on a, b, then the function x a. The Integral Evaluation Theorem. Don’t overlook the obvious! In contrast with the above theorem, which every calculus student knows, the Second Fundamental Theorem is more obscure and seems less useful. The purpose of this chapter is to explain it, show its use and importance, and to show how the two theorems are related. To start things oﬀ, here it is. Second Fundamental Theorem. Let f(x) be continuous, and ﬁx a. (2) Set F(x) = Z x a f(t)dt; then F′(x) = f(x). Understand the Fundamental Theorem of Calculus. The first theorem of calculus, also referred to as the first fundamental theorem of calculus, is an essential part of this subject that you need to work on seriously in order to meet great success in your math-learning journey. Practice, Practice, and Practice! Practice makes perfect.

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

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

Plug it in to get

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

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

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