Here's the question you clicked on:
mlddmlnog
evaluate the following:
1. \[\frac{ d }{ dx }[\int\limits_{1}^{x}t ^{3}dt] = ?\] 2. \[\frac{ d }{ dx }[\int\limits_{3}^{sinx}\frac{ 1 }{ 1+t ^{2} }dt] = ?\]
For the first one, we can apply the FTC Part 1.\[\large \color{cornflowerblue}{\frac{d}{dx}\int\limits_a^x f(t)dt=f(x)}\]
\[\huge \frac{d}{dx}\int\limits_1^x t^3 dt=x^3\]Understand how we applied that? :D We can go into a little bit further detail if it's confusing.
Ok ok lemme explain it a tiny bit further, because it will help us on the next problem anyway.
hm.. can we go into further detail please? haha i always need to see the process in oro understand something :p
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.
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 :)
BIG F represents the anti-derivative of f. Then it changed back into little f when we took the derivative after integrating.
In our problem, here is what would actually be happening.\[\large \frac{d}{dx}\int\limits_1^x t^3 dt= \frac{d}{dx}\left[\frac{1}{4}t^4|_1^x\right]\] \[\large \frac{d}{dx}\left[\frac{1}{4}x^4-1^4\right]\] Taking the derivative gives us,\[\large x^3-0\]
OH. i'm slow. :/ haha i getnow :)
|dw:1358301319768:dw|actuall,y no wait.
The derivative of 1/4*1^4 will still give you 0 though right? :D So no big deal.
ohh yes. thank you :) now let's go on to the next one :)
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.
oh... ok. i will try to do this one by myself. :)
yea.. uh nevermind. haha i'm stuck at the firt step. :(
Lemme show you the answer real quick.. this one is a little hard to swallow.
\[\large \frac{d}{dx}\int\limits_3^{\sin x} \frac{1}{1+t^2}dt \qquad = \qquad \frac{1}{1+\sin^2x}\left(\frac{d}{dx}\sin x\right)\]
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.
ok. but i think i need to see the process :/ this one is hard to understand ...
\[\large \frac{d}{dx}\int\limits\limits_3^{\sin x} \frac{1}{1+t^2}dt \qquad = \qquad \frac{d}{dx}\left[\arctan t|_3^{\sin x}\right]\]
\[\large =\frac{d}{dx}\left[\arctan(\sin x)-\arctan 3\right]\]
Taking the derivative gives us,\[\large \frac{1}{1+(\sin x)^2}(\sin x)'+0\]
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.
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
i have a question. what is the integration of cost/t?
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
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).
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.
so, ,, for F'(x), we could just substitute x for t?
\[\large F'(x)=\frac{d}{dx}\int\limits\limits_1^{x^2} \frac{\cos t}{t}dt\]No you would substitute t with whatever is in your UPPER LIMIT.
ok, so i understand that part now :) but how would you find F'(1)? do you just replace x with 1? :)