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Fundamental theorem of calculus:
Use the second FTC to find F'(x)
 one year ago
 one year ago
Fundamental theorem of calculus: Use the second FTC to find F'(x)
 one year ago
 one year ago

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

inkyvoydBest ResponseYou've already chosen the best response.1
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.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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 \[\Large F ^{\prime}(x) = f(x)*u^{\prime}(x) \]
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
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?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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 \[\Large F(x) = \int_{a}^{x} f(t) dt\] \[\Large F(x) = G(x)  G(a)\] \[\Large F^{\prime}(x) = \frac{d}{dx}\left[G(x)  G(a)\right]\] \[\Large F^{\prime}(x) = G ^{\prime} (x)  G ^{\prime} (a)\] \[\Large F^{\prime}(x) = G ^{\prime} (x)  0\] \[\Large F^{\prime}(x) = G ^{\prime} (x)\] \[\Large F^{\prime}(x) = f(x)\]
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
but yes, you use the chain rule
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
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.
 one year ago
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