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(Second fundamental therom)

check FTC 2

:)

Suppose you have the following: \[
F(x) = \int_a^{g(x)}f(t)dt
\]

Let \(y = g(x)\)

You would have to split the integral at some constant (say a)

Then suppose : \[
H(y) = \int_a^yf(t)dt
\]Notice that \(H(y) = H(g(x)) = F(x)\)

By the fundamental theorem: \[
\frac{d}{dy}H(y) = f(y) = f(g(x))
\]

See where I'm going with this?

and \[\int\limits\limits_{\cos 3x}^{a} \sqrt{1+t^2}=-\int\limits\limits_{a}^{\cos3x} \sqrt{1+t^2}\]

Basically we have: \[
\frac{d}{dy}F(x) = f(g(x))
\]

@kirbykirby Ok, thanks!

When we say \(y=g(x)\) we're just saying they are interchangeable.

The reason I did that was to make it so you can see that a function is sort of just like a variable.

also there is a difference between \(F(x)\) and \(f(x)\)

Sorry, my reply was written wrong. Let me restate it.

I can see that H(y) = H (g(x)) (y=g(x)), but I /don't/ see how H(y)=F(x) (or H(g(x)=F(x))

By the chain rule: \[
\frac{d}{dy}F(x) \frac{dy}{dx} = \frac{d}{dx}F(x) = f(y)\frac{dy}{dx}
\]

Yes I got the chain rule step (esp after you rewrote it in a different way).

I've done this thing like 2 or 3 times and every time I it gets a bit easier to teach.