## inkyvoyd 2 years ago Fundamental theorem of calculus: Use the second FTC to find F'(x)

1. inkyvoyd

Given that $$\Huge F(x)=\int_0^{x^2}\frac{1}{t^2}dt$$

2. inkyvoyd

I tried to find the integral of t^2, then used basic substitution, only to realize that f(x) is not defined for x=0. Can someone give me a brief refresher on what the FTC is? I realize they search for F'(x), but I guess I just don't know what I am doing.

3. jim_thompson5910

if $\Large F(x) = \int_{a}^{x} f(t) dt$ then $\Large F ^{\prime}(x) = f(x)$ -------------------------------------- More generally, if $\Large F(x) = \int_{a}^{u(x)} f(t) dt$ then $\Large F ^{\prime}(x) = f(x)*u^{\prime}(x)$

4. inkyvoyd

This which resembles the chain rule is derived from there I'm guessing? Is all of this under the fundamental theorem of calculus? Or is there another name I'm missing?

5. jim_thompson5910

yes this is all under the FTC because $\Large F(x) = \int_{a}^{x} f(t) dt = G(x) - G(a)$ where G ' (x) = f(x) so $\Large F(x) = \int_{a}^{x} f(t) dt$ $\Large F(x) = G(x) - G(a)$ $\Large F^{\prime}(x) = \frac{d}{dx}\left[G(x) - G(a)\right]$ $\Large F^{\prime}(x) = G ^{\prime} (x) - G ^{\prime} (a)$ $\Large F^{\prime}(x) = G ^{\prime} (x) - 0$ $\Large F^{\prime}(x) = G ^{\prime} (x)$ $\Large F^{\prime}(x) = f(x)$

6. jim_thompson5910

and if $\Large F(x) = \int_{a}^{u(x)} f(t) dt = G(x) - G(a)$ where G ' (x) = f(x), we can say $\Large F(x) = \int_{a}^{u(x)} f(t) dt$ $\Large F(x) = G(u(x)) - G(a)$ $\Large F^{\prime}(x) = \frac{d}{dx}\left[G(u(x)) - G(a)\right]$ $\Large F^{\prime}(x) = G ^{\prime} (u(x)) * u^{\prime} (x) - G ^{\prime} (a)$ $\Large F^{\prime}(x) = G ^{\prime} (u(x)) * u^{\prime} (x) - 0$ $\Large F^{\prime}(x) = G ^{\prime} (u(x)) * u^{\prime} (x)$ $\Large F^{\prime}(x) = f(u(x)) * u^{\prime} (x)$ Sorry I made a typo before, but I think I fixed it

7. jim_thompson5910

oops meant to say $\Large F(x) = \int_{a}^{u(x)} f(t) dt = G(u(x)) - G(a)$ for the second case, but you get the idea

8. jim_thompson5910

but yes, you use the chain rule

9. inkyvoyd

Ahh. Thank you very much for the clarification - I should've seen this on my part (tbh I'm reaping the rewards of procrastination). Hehe, I stil have to finish a mean value theorem for integrals and some improper integral problems :S. I guess at least I know how to sort of do improper integrals.

10. jim_thompson5910

np