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

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inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Given that \(\Huge F(x)=\int_0^{x^2}\frac{1}{t^2}dt\)

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1I 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.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1if \[\Large F(x) = \int_{a}^{x} f(t) dt \] then \[\Large F ^{\prime}(x) = f(x) \]  More generally, if \[\Large F(x) = \int_{a}^{u(x)} f(t) dt \] then \[\Large F ^{\prime}(x) = f(x)*u^{\prime}(x) \]

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1This 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?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1yes 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 \[\Large F(x) = \int_{a}^{x} f(t) dt\] \[\Large F(x) = G(x)  G(a)\] \[\Large F^{\prime}(x) = \frac{d}{dx}\left[G(x)  G(a)\right]\] \[\Large F^{\prime}(x) = G ^{\prime} (x)  G ^{\prime} (a)\] \[\Large F^{\prime}(x) = G ^{\prime} (x)  0\] \[\Large F^{\prime}(x) = G ^{\prime} (x)\] \[\Large F^{\prime}(x) = f(x)\]

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1and 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

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1oops 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

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1but yes, you use the chain rule

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Ahh. 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.
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