Question

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

Answer 1

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

Answer 2

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.

Answer 3

if

\[\Large F(x) = \int_{a}^{x} f(t) dt \]

then

\[\Large F ^{\prime}(x) = f(x) \]


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More generally,

if

\[\Large F(x) = \int_{a}^{u(x)} f(t) dt \]

then

\[\Large F ^{\prime}(x) = f(x)*u^{\prime}(x) \]

Answer 4

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?

Answer 5

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

\[\Large F(x) = \int_{a}^{x} f(t) dt\]

\[\Large F(x) = G(x) - G(a)\]

\[\Large F^{\prime}(x) = \frac{d}{dx}\left[G(x) - G(a)\right]\]

\[\Large F^{\prime}(x) = G ^{\prime} (x) - G ^{\prime} (a)\]

\[\Large F^{\prime}(x) = G ^{\prime} (x) - 0\]

\[\Large F^{\prime}(x) = G ^{\prime} (x)\]

\[\Large F^{\prime}(x) = f(x)\]

Answer 6

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

Answer 7

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

Answer 8

but yes, you use the chain rule

Answer 9

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.

Answer 10

np