# 7.1 2nd Fundamental Theorem Of Calculusap Calculus

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! AP Calculus Unit 7 – Advanced Integration & Applications Day 1 Notes: Second Fundamental Theorem of Calculus Given the functions, f(t), below, use F x ³ x f t dt 1 ( ) to find F(x) and F’(x) in terms of x. F(t) = 4t – t2 2. F(t) = cos t Given the functions, f(t), below, use ³ 2 1 ( ) x F x f t dt to find F(x) and F’(x) in terms of x. ### 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 derivative is ${\left(2 {x}^{7} - 1\right)}^{3} \left(7 {x}^{6}\right) - {\left(2 {x}^{2} - 1\right)}^{3} \left(2 x\right)$

• 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.
• Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena.
• First Fundamental Theorem of Calculus If ƒ(x) is continuous on a,b, then there is a c such that ∫(from a to b) ƒ(x) dx = ƒ(c)(b - a) Mean Value Theorem for Integrals.

#### 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

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

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

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

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

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