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inkyvoyd

  • 3 years ago

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

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

  2. inkyvoyd
    • 3 years 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
    • 3 years 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
    • 3 years 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
    • 3 years 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
    • 3 years 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
    • 3 years 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
    • 3 years ago
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    but yes, you use the chain rule

  9. inkyvoyd
    • 3 years 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
    • 3 years ago
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    np

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