Calculus! (See comments)

- anonymous

Calculus! (See comments)

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- anonymous

\[\frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2}\]
Preferably with steps+rationale shown.
Thanks!

- anonymous

If the equation is too small to see: Find the derivative of the integral of squrt(1+t^2)dt from cos3x to tan 3x

- anonymous

(Second fundamental therom)

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## More answers

- anonymous

check FTC 2

- anonymous

:)

- anonymous

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

- anonymous

So @mathg8 , so far in my class we have only solved problems with one of the limits of integration being an expression, and one would be a constant (ie 3), how would this differ.

- anonymous

Let \(y = g(x)\)

- kirbykirby

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

- anonymous

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

- anonymous

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

- kirbykirby

\[\frac {d}{dx} \int\limits\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2}dt = \frac {d}{dx} \int\limits\limits_{\cos 3x}^{a} \sqrt{1+t^2}dt+\frac {d}{dx} \int\limits\limits_{a}^{\tan 3x} \sqrt{1+t^2}dt\]

- anonymous

See where I'm going with this?

- anonymous

@Argonx16

- kirbykirby

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

- anonymous

@wio I'm reading through it (:

- kirbykirby

If you know how to work with an expression other than just "x" on the upper integration bound, then you should have no problem computing it the way I presented it by splitting your integral into 2

- anonymous

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

- anonymous

@kirbykirby Ok, thanks!

- anonymous

@wio here, let me put what you said in words to see if I get it

- anonymous

@Argonx16 I'm not finished yet but sure.

- anonymous

If you have f(x) = integral of f(t) from a to g(x), and y=g(x), and \[H(y)=\int\limits_{a}^{y}f(t)dt\]
Then we can see that H(y)=H(g(x)) because of the fact that y=g(x)
(Not quite sure how H(y)=H(g(x)) though)
@wio

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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))

- anonymous

It's the transitive property of equality. \[
H(y)=\int\limits_{a}^{y}f(t)dt = \int\limits_{a}^{g(x)}f(t)dt =F(x)
\]I'm not saying they are the same function though. I'm just saying they are equal.

- anonymous

Look if you want, just ignore \(H(y)\). If I said: \[
F(x)=\int\limits_{a}^{y}f(t)dt
\]Do you understand that by the fundamental theorem: \[
\frac{d}{dy}F(x) = f(y)
\]

- anonymous

Ok, I see what you did to make F(x)=H(y)=H(g(x))
And, yes I understand the second fundamental theorem.

- anonymous

Now the trick here is to multiply both sides by \(dy/dx\) \[
\frac{d}{dy}F(x) \frac{dy}{dx} = f(y)\frac{dy}{dx}
\]

- anonymous

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

- anonymous

@Argonx16 this is sort of the tricky part, but do you get it?

- anonymous

Another way to write this is: \[
\frac{dF(x)}{dy}\frac{dy}{dx} = \frac{dF(x)}{dx}
\]Or \[
\frac{dF}{dy}\frac{dy}{dx} = \frac{dF}{dx}
\]

- anonymous

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

- anonymous

So we got:\[
\frac{d}{dx}F(x) = \frac{d}{dx}\int\limits_{a}^{y}f(t)dt = \frac{d}{dx}\int\limits_{a}^{g(x)}f(t)dt
\]And: \[
f(y)\frac{dy}{dx} = f(g(x))\frac{dg(x)}{dx} = f(g(x))g'(x)
\]

- anonymous

Putting it all together: \[
\frac{d}{dx} \left[ \int\limits_{a}^{g(x)}f(t)dt \right] = f(g(x))g′(x)
\]

- anonymous

\[
\frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2} = \frac {d}{dx} \int\limits_{z}^{\tan 3x} \sqrt{1+t^2} -\frac {d}{dx} \int\limits^{\cos 3x}_{a} \sqrt{1+t^2}
\]

- anonymous

To expand on the earlier realization: \[
\frac{d}{dx} \left[ \ \int\limits_{h(x)}^{g(x)}f(t)dt \right] = f(g(x))g′(x) -f(h(x))h'(x)
\]

- anonymous

@Argonx16 Does it all make sense now?

- anonymous

Yes! @wio Thanks for spending so much time helping me on this! I know how much patience/time/effort it takes to help so in depth, so I am very appreciative of your help :D

- anonymous

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

- anonymous

Yeah. It definitely helps so much more. OS is also a good way to retain math concepts taught earlier :D

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