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mlddmlnogBest ResponseYou've already chosen the best response.0
1. \[\frac{ d }{ dx }[\int\limits_{1}^{x}t ^{3}dt] = ?\] 2. \[\frac{ d }{ dx }[\int\limits_{3}^{sinx}\frac{ 1 }{ 1+t ^{2} }dt] = ?\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
For the first one, we can apply the FTC Part 1.\[\large \color{cornflowerblue}{\frac{d}{dx}\int\limits_a^x f(t)dt=f(x)}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\huge \frac{d}{dx}\int\limits_1^x t^3 dt=x^3\]Understand how we applied that? :D We can go into a little bit further detail if it's confusing.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Ok ok lemme explain it a tiny bit further, because it will help us on the next problem anyway.
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
hm.. can we go into further detail please? haha i always need to see the process in oro understand something :p
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
We're basically, Integrating, and then undoing the integration by differentiating. But the variable of our function changes due to the LIMITS on our integral.
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
hm.. give me a few minutes to think please.. haha you can go on to the next problem while i try to understand this one :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
BIG F represents the antiderivative of f. Then it changed back into little f when we took the derivative after integrating.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
In our problem, here is what would actually be happening.\[\large \frac{d}{dx}\int\limits_1^x t^3 dt= \frac{d}{dx}\left[\frac{1}{4}t^4_1^x\right]\] \[\large \frac{d}{dx}\left[\frac{1}{4}x^41^4\right]\] Taking the derivative gives us,\[\large x^30\]
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
OH. i'm slow. :/ haha i getnow :)
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
dw:1358301319768:dwactuall,y no wait.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
The derivative of 1/4*1^4 will still give you 0 though right? :D So no big deal.
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
ohh yes. thank you :) now let's go on to the next one :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
The same rule will apply, but we'll have one extra step due to the chain rule. See how the in the upper LIMIT, it's no longer X, it's a lil more than that? That will force us to apply the chain rule when we DIFFERENTIATE.
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
oh... ok. i will try to do this one by myself. :)
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
yea.. uh nevermind. haha i'm stuck at the firt step. :(
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Lemme show you the answer real quick.. this one is a little hard to swallow.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large \frac{d}{dx}\int\limits_3^{\sin x} \frac{1}{1+t^2}dt \qquad = \qquad \frac{1}{1+\sin^2x}\left(\frac{d}{dx}\sin x\right)\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
See how, similar to the last problem, the UPPER LIMIT replaces our t? But the extra step is  since we have more than just X in the upper limit, we have to apply the chain rule, taking the derivative of sin x.
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
ok. but i think i need to see the process :/ this one is hard to understand ...
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large \frac{d}{dx}\int\limits\limits_3^{\sin x} \frac{1}{1+t^2}dt \qquad = \qquad \frac{d}{dx}\left[\arctan t_3^{\sin x}\right]\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large =\frac{d}{dx}\left[\arctan(\sin x)\arctan 3\right]\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Taking the derivative gives us,\[\large \frac{1}{1+(\sin x)^2}(\sin x)'+0\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
You'll want to try and get used to understanding these WITHOUT seeing the process because more than likely on a test, you'll be given something that can't actually be integrated very easily, so you'll want to remember the FTC.
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
yea i know :( but i just need to memorize the differentiation and integration formulas, because i got stuck at where we had to differentiate arctan. :p
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
i have a question. what is the integration of cost/t?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
It's called the "Cosine Integral". There are many integrals such as that one which can not be solved using elementary methods. Integrals are not nice and neat like derivatives. Many many integrals just can't be solved :D
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
oh ..really? because i was going to do the next problem by myself but i guess i need help with it .. it says : given \[F(x)=\int\limits_{1}^{x ^{2}}\frac{ cost }{ t }dt \] find F'(x) and F'(1).
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Ok this is a perfect example of what I was talking about. This is an integral that can't be solved using normal methods. But if we integrate, and then differentiate again, it will give us back what we started with, just with a different variable due to the limits of integration.
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
so, ,, for F'(x), we could just substitute x for t?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large F'(x)=\frac{d}{dx}\int\limits\limits_1^{x^2} \frac{\cos t}{t}dt\]No you would substitute t with whatever is in your UPPER LIMIT.
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
ok, so i understand that part now :) but how would you find F'(1)? do you just replace x with 1? :)
 one year ago
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