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Argonx16 Group Title

Calculus! (See comments)

  • one year ago
  • one year ago

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  1. Argonx16 Group Title
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    \[\frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2}\] Preferably with steps+rationale shown. Thanks!

    • one year ago
  2. Argonx16 Group Title
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    If the equation is too small to see: Find the derivative of the integral of squrt(1+t^2)dt from cos3x to tan 3x

    • one year ago
  3. Argonx16 Group Title
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    (Second fundamental therom)

    • one year ago
  4. mathg8 Group Title
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    check FTC 2

    • one year ago
  5. mathg8 Group Title
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    :)

    • one year ago
  6. wio Group Title
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    Suppose you have the following: \[ F(x) = \int_a^{g(x)}f(t)dt \]

    • one year ago
  7. Argonx16 Group Title
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    So @mathg8 , so far in my class we have only solved problems with one of the limits of integration being an expression, and one would be a constant (ie 3), how would this differ.

    • one year ago
  8. wio Group Title
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    Let \(y = g(x)\)

    • one year ago
  9. kirbykirby Group Title
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    You would have to split the integral at some constant (say a)

    • one year ago
  10. wio Group Title
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    Then suppose : \[ H(y) = \int_a^yf(t)dt \]Notice that \(H(y) = H(g(x)) = F(x)\)

    • one year ago
  11. wio Group Title
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    By the fundamental theorem: \[ \frac{d}{dy}H(y) = f(y) = f(g(x)) \]

    • one year ago
  12. kirbykirby Group Title
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    \[\frac {d}{dx} \int\limits\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2}dt = \frac {d}{dx} \int\limits\limits_{\cos 3x}^{a} \sqrt{1+t^2}dt+\frac {d}{dx} \int\limits\limits_{a}^{\tan 3x} \sqrt{1+t^2}dt\]

    • one year ago
  13. wio Group Title
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    See where I'm going with this?

    • one year ago
  14. wio Group Title
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    @Argonx16

    • one year ago
  15. kirbykirby Group Title
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    and \[\int\limits\limits_{\cos 3x}^{a} \sqrt{1+t^2}=-\int\limits\limits_{a}^{\cos3x} \sqrt{1+t^2}\]

    • one year ago
  16. Argonx16 Group Title
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    @wio I'm reading through it (:

    • one year ago
  17. kirbykirby Group Title
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    If you know how to work with an expression other than just "x" on the upper integration bound, then you should have no problem computing it the way I presented it by splitting your integral into 2

    • one year ago
  18. wio Group Title
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    Basically we have: \[ \frac{d}{dy}F(x) = f(g(x)) \]

    • one year ago
  19. Argonx16 Group Title
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    @kirbykirby Ok, thanks!

    • one year ago
  20. Argonx16 Group Title
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    @wio here, let me put what you said in words to see if I get it

    • one year ago
  21. wio Group Title
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    @Argonx16 I'm not finished yet but sure.

    • one year ago
  22. Argonx16 Group Title
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    If you have f(x) = integral of f(t) from a to g(x), and y=g(x), and \[H(y)=\int\limits_{a}^{y}f(t)dt\] Then we can see that H(y)=H(g(x)) because of the fact that y=g(x) (Not quite sure how H(y)=H(g(x)) though) @wio

    • one year ago
  23. wio Group Title
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    When we say \(y=g(x)\) we're just saying they are interchangeable.

    • one year ago
  24. wio Group Title
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    The reason I did that was to make it so you can see that a function is sort of just like a variable.

    • one year ago
  25. wio Group Title
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    also there is a difference between \(F(x)\) and \(f(x)\)

    • one year ago
  26. Argonx16 Group Title
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    Sorry, my reply was written wrong. Let me restate it.

    • one year ago
  27. Argonx16 Group Title
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    I can see that H(y) = H (g(x)) (y=g(x)), but I /don't/ see how H(y)=F(x) (or H(g(x)=F(x))

    • one year ago
  28. wio Group Title
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    It's the transitive property of equality. \[ H(y)=\int\limits_{a}^{y}f(t)dt = \int\limits_{a}^{g(x)}f(t)dt =F(x) \]I'm not saying they are the same function though. I'm just saying they are equal.

    • one year ago
  29. wio Group Title
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    Look if you want, just ignore \(H(y)\). If I said: \[ F(x)=\int\limits_{a}^{y}f(t)dt \]Do you understand that by the fundamental theorem: \[ \frac{d}{dy}F(x) = f(y) \]

    • one year ago
  30. Argonx16 Group Title
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    Ok, I see what you did to make F(x)=H(y)=H(g(x)) And, yes I understand the second fundamental theorem.

    • one year ago
  31. wio Group Title
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    Now the trick here is to multiply both sides by \(dy/dx\) \[ \frac{d}{dy}F(x) \frac{dy}{dx} = f(y)\frac{dy}{dx} \]

    • one year ago
  32. wio Group Title
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    By the chain rule: \[ \frac{d}{dy}F(x) \frac{dy}{dx} = \frac{d}{dx}F(x) = f(y)\frac{dy}{dx} \]

    • one year ago
  33. wio Group Title
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    @Argonx16 this is sort of the tricky part, but do you get it?

    • one year ago
  34. wio Group Title
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    Another way to write this is: \[ \frac{dF(x)}{dy}\frac{dy}{dx} = \frac{dF(x)}{dx} \]Or \[ \frac{dF}{dy}\frac{dy}{dx} = \frac{dF}{dx} \]

    • one year ago
  35. Argonx16 Group Title
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    Yes I got the chain rule step (esp after you rewrote it in a different way).

    • one year ago
  36. wio Group Title
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    So we got:\[ \frac{d}{dx}F(x) = \frac{d}{dx}\int\limits_{a}^{y}f(t)dt = \frac{d}{dx}\int\limits_{a}^{g(x)}f(t)dt \]And: \[ f(y)\frac{dy}{dx} = f(g(x))\frac{dg(x)}{dx} = f(g(x))g'(x) \]

    • one year ago
  37. wio Group Title
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    Putting it all together: \[ \frac{d}{dx} \left[ \int\limits_{a}^{g(x)}f(t)dt \right] = f(g(x))g′(x) \]

    • one year ago
  38. wio Group Title
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    \[ \frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2} = \frac {d}{dx} \int\limits_{z}^{\tan 3x} \sqrt{1+t^2} -\frac {d}{dx} \int\limits^{\cos 3x}_{a} \sqrt{1+t^2} \]

    • one year ago
  39. wio Group Title
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    To expand on the earlier realization: \[ \frac{d}{dx} \left[ \ \int\limits_{h(x)}^{g(x)}f(t)dt \right] = f(g(x))g′(x) -f(h(x))h'(x) \]

    • one year ago
  40. wio Group Title
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    @Argonx16 Does it all make sense now?

    • one year ago
  41. Argonx16 Group Title
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    Yes! @wio Thanks for spending so much time helping me on this! I know how much patience/time/effort it takes to help so in depth, so I am very appreciative of your help :D

    • one year ago
  42. wio Group Title
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    I've done this thing like 2 or 3 times and every time I it gets a bit easier to teach.

    • one year ago
  43. Argonx16 Group Title
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    Yeah. It definitely helps so much more. OS is also a good way to retain math concepts taught earlier :D

    • one year ago
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