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inkyvoyd

  • one year ago

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

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  1. inkyvoyd
    • one year ago
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    Given that \(\Huge F(x)=\int_0^{x^2}\frac{1}{t^2}dt\)

  2. inkyvoyd
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    but yes, you use the chain rule

  9. inkyvoyd
    • one year ago
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    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
    • one year ago
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    np

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