FTC Part II: Solving for t*cos t|42x

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Bazzinga
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(Not quite sure how to use Latex so I print screened it :-p)
[PLAIN]http://img87.imageshack.us/img87/5991/calc1.png

So I've been staring at this question, and I think I might have it but I'm not 100% sure, is the answer just

t*cos t |42x = (4)*cos(4) - (2x)*cos(2x) ?

Or am I looking at the question wrong?

EDIT:
Oh I think I've got it, would the answer just be 4*cos 4 by FTC part 2?
 
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Oh, I guess I was reading FTC part II wrong =P so I would have to integrate t*cos(t), making the answer...
dy/dx[4*sin(4) + cos(4)] - [2x*sin(2x) + cos(2x)] ?
 
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Bazzinga said:
Oh, I guess I was reading FTC part II wrong =P so I would have to integrate t*cos(t), making the answer...
[4*sin(4) + cos(4)] - [2x*sin(2x) + cos(2x)] ?

No. You DON'T have to integrate (and you haven't done it right anyway). You are finding the DERIVATIVE of the integral. Do you agree with my previous post? Is F(4)-F(2x) the integral? Can you differentiate F(4)-F(2x)?
 
Dick said:
No. You DON'T have to integrate (and you haven't done it right anyway). You are finding the DERIVATIVE of the integral. Do you agree with my previous post? Is F(4)-F(2x) the integral? Can you differentiate F(4)-F(2x)?
I'm still not seeing what you're saying... You can differentiate F(4)-F(2x), it would be F'(4)-F'(2x), meaning the answer is what I said first? 4*cos(4) - 2x*cos(2x) ?
Ahh all this theory stuff makes my head spin :cry:
 
Bazzinga said:
I'm still not seeing what you're saying... You can differentiate F(4)-F(2x), it would be F'(4)-F'(2x), meaning the answer is what I said first? 4*cos(4) - 2x*cos(2x) ?
Ahh all this theory stuff makes my head spin :cry:

Don't cry. You're close with that answer. But F(4) is a constant, right? d/dx of a constant is zero isn't it? And for the second term you need to differentiate F(2x). You need to use the chain rule. It's a function of a function.
 
All you have to do is put the integral in the form of the theorem:
[tex] \int_{2x}^{4}t\cos tdt=\int_{4}^{2x}-t\cos tdt[/tex]
Now you see that a=4 and X=2x, so from here the derivative is...
The derivative of F(4) is zero as F(4) is a constant.
 
I think I get it! F'(4) is 0, so the answer is -F'(x) = -2x*cos(2x) ? Where am I doing something with the chain rule though?
 
Bazzinga said:
I think I get it! F'(4) is 0, so the answer is -F'(x) = -2x*cos(2x) ? Where am I doing something with the chain rule though?

Not correct. The chain rule says the derivative of f(g(x)) is f'(g(x))*g'(x). Here f is F and g(x) is 2x. You are missing the g' part.
 
Ohh, I get it, so my equation will be F'(2x)*'(2x), meaning the final answer would be
-4x*cos(2x) !
 
Bazzinga said:
Ohh, I get it, so my equation will be F'(2x)*'(2x), meaning the final answer would be
-4x*cos(2x) !

Right. F'(2x)*(2x)'. If you want to check that, the antiderivative of t*cos(t) is F(t)=t*sin(t)+cos(t). You can check that you are right by actually finding F(4)-F(2x) and differentiating it.
 
Dick said:
Right. F'(2x)*(2x)'. If you want to check that, the antiderivative of t*cos(t) is F(t)=t*sin(t)+cos(t). You can check that you are right by actually finding F(4)-F(2x) and differentiating it.

Awesome! Thanks a lot for your help, you really helped me understand it instead of just throwing an answer at me