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mlddmlnog Group Title

evaluate the following:

  • one year ago
  • one year ago

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  1. mlddmlnog Group Title
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    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
  2. mlddmlnog Group Title
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    @zepdrix

    • one year ago
  3. zepdrix Group Title
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    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
  4. zepdrix Group Title
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    \[\huge \frac{d}{dx}\int\limits_1^x t^3 dt=x^3\]Understand how we applied that? :D We can go into a little bit further detail if it's confusing.

    • one year ago
  5. zepdrix Group Title
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    Ok ok lemme explain it a tiny bit further, because it will help us on the next problem anyway.

    • one year ago
  6. mlddmlnog Group Title
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    hm.. can we go into further detail please? haha i always need to see the process in oro understand something :p

    • one year ago
  7. mlddmlnog Group Title
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    thanks!

    • one year ago
  8. zepdrix Group Title
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    |dw:1358300665316:dw|

    • one year ago
  9. zepdrix Group Title
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    |dw:1358300764593:dw|

    • one year ago
  10. zepdrix Group Title
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    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
  11. mlddmlnog Group Title
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    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
  12. zepdrix Group Title
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    BIG F represents the anti-derivative of f. Then it changed back into little f when we took the derivative after integrating.

    • one year ago
  13. zepdrix Group Title
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    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,\[\large x^3-0\]

    • one year ago
  14. mlddmlnog Group Title
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    OH. i'm slow. :/ haha i getnow :)

    • one year ago
  15. mlddmlnog Group Title
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    |dw:1358301319768:dw|actuall,y no wait.

    • one year ago
  16. zepdrix Group Title
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    Yes you do, my bad :)

    • one year ago
  17. zepdrix Group Title
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    The derivative of 1/4*1^4 will still give you 0 though right? :D So no big deal.

    • one year ago
  18. mlddmlnog Group Title
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    ohh yes. thank you :) now let's go on to the next one :)

    • one year ago
  19. zepdrix Group Title
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    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
  20. mlddmlnog Group Title
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    oh... ok. i will try to do this one by myself. :)

    • one year ago
  21. zepdrix Group Title
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    kk

    • one year ago
  22. mlddmlnog Group Title
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    yea.. uh nevermind. haha i'm stuck at the firt step. :(

    • one year ago
  23. zepdrix Group Title
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    Lemme show you the answer real quick.. this one is a little hard to swallow.

    • one year ago
  24. zepdrix Group Title
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    \[\large \frac{d}{dx}\int\limits_3^{\sin x} \frac{1}{1+t^2}dt \qquad = \qquad \frac{1}{1+\sin^2x}\left(\frac{d}{dx}\sin x\right)\]

    • one year ago
  25. zepdrix Group Title
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    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
  26. zepdrix Group Title
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    brbs :O

    • one year ago
  27. mlddmlnog Group Title
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    ok. but i think i need to see the process :/ this one is hard to understand ...

    • one year ago
  28. zepdrix Group Title
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    \[\large \frac{d}{dx}\int\limits\limits_3^{\sin x} \frac{1}{1+t^2}dt \qquad = \qquad \frac{d}{dx}\left[\arctan t|_3^{\sin x}\right]\]

    • one year ago
  29. zepdrix Group Title
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    \[\large =\frac{d}{dx}\left[\arctan(\sin x)-\arctan 3\right]\]

    • one year ago
  30. zepdrix Group Title
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    Taking the derivative gives us,\[\large \frac{1}{1+(\sin x)^2}(\sin x)'+0\]

    • one year ago
  31. zepdrix Group Title
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    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
  32. mlddmlnog Group Title
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    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
  33. mlddmlnog Group Title
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    i have a question. what is the integration of cost/t?

    • one year ago
  34. mlddmlnog Group Title
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    @zepdrix

    • one year ago
  35. zepdrix Group Title
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    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
  36. mlddmlnog Group Title
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    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
  37. zepdrix Group Title
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    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
  38. mlddmlnog Group Title
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    so, ,, for F'(x), we could just substitute x for t?

    • one year ago
  39. zepdrix Group Title
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    \[\large F'(x)=\frac{d}{dx}\int\limits\limits_1^{x^2} \frac{\cos t}{t}dt\]No you would substitute t with whatever is in your UPPER LIMIT.

    • one year ago
  40. mlddmlnog Group Title
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    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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