inkyvoyd
  • inkyvoyd
Fundamental theorem of calculus: Use the second FTC to find F'(x)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
inkyvoyd
  • inkyvoyd
Given that \(\Huge F(x)=\int_0^{x^2}\frac{1}{t^2}dt\)
inkyvoyd
  • 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.
jim_thompson5910
  • 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 \[\Large F ^{\prime}(x) = f(x)*u^{\prime}(x) \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

inkyvoyd
  • 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?
jim_thompson5910
  • 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 \[\Large F(x) = \int_{a}^{x} f(t) dt\] \[\Large F(x) = G(x) - G(a)\] \[\Large F^{\prime}(x) = \frac{d}{dx}\left[G(x) - G(a)\right]\] \[\Large F^{\prime}(x) = G ^{\prime} (x) - G ^{\prime} (a)\] \[\Large F^{\prime}(x) = G ^{\prime} (x) - 0\] \[\Large F^{\prime}(x) = G ^{\prime} (x)\] \[\Large F^{\prime}(x) = f(x)\]
jim_thompson5910
  • 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
jim_thompson5910
  • 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
jim_thompson5910
  • jim_thompson5910
but yes, you use the chain rule
inkyvoyd
  • 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.
jim_thompson5910
  • jim_thompson5910
np

Looking for something else?

Not the answer you are looking for? Search for more explanations.