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mlddmlnog
 2 years ago
Best ResponseYou've already chosen the best response.01. \[\frac{ d }{ dx }[\int\limits_{1}^{x}t ^{3}dt] = ?\] 2. \[\frac{ d }{ dx }[\int\limits_{3}^{sinx}\frac{ 1 }{ 1+t ^{2} }dt] = ?\]

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

zepdrix
 2 years ago
Best 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.

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

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

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

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

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0In 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,\[\large x^30\]

mlddmlnog
 2 years ago
Best ResponseYou've already chosen the best response.0OH. i'm slow. :/ haha i getnow :)

mlddmlnog
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1358301319768:dwactuall,y no wait.

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

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0The 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.

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

mlddmlnog
 2 years ago
Best ResponseYou've already chosen the best response.0yea.. uh nevermind. haha i'm stuck at the firt step. :(

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

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0See 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.

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

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

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

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0You'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.

mlddmlnog
 2 years ago
Best ResponseYou've already chosen the best response.0yea 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

mlddmlnog
 2 years ago
Best ResponseYou've already chosen the best response.0i have a question. what is the integration of cost/t?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0It'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

mlddmlnog
 2 years ago
Best ResponseYou've already chosen the best response.0oh ..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).

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Ok 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.

mlddmlnog
 2 years ago
Best ResponseYou've already chosen the best response.0so, ,, for F'(x), we could just substitute x for t?

zepdrix
 2 years ago
Best 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.

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