Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

inkyvoyd

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

  • one year ago
  • one year ago

  • This Question is Closed
  1. inkyvoyd
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  2. inkyvoyd
    Best Response
    You've already chosen the best response.
    Medals 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
  3. jim_thompson5910
    Best Response
    You've already chosen the best response.
    Medals 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
  4. inkyvoyd
    Best Response
    You've already chosen the best response.
    Medals 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
  5. jim_thompson5910
    Best Response
    You've already chosen the best response.
    Medals 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
  6. jim_thompson5910
    Best Response
    You've already chosen the best response.
    Medals 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
  7. jim_thompson5910
    Best Response
    You've already chosen the best response.
    Medals 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
  8. jim_thompson5910
    Best Response
    You've already chosen the best response.
    Medals 1

    but yes, you use the chain rule

    • one year ago
  9. inkyvoyd
    Best Response
    You've already chosen the best response.
    Medals 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
  10. jim_thompson5910
    Best Response
    You've already chosen the best response.
    Medals 1

    np

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.