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  1. Jhannybean
    • one year ago
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    \[\sf \frac{d}{dx}\int_1^{4x} \sqrt{t^2+1}dt = \frac{d}{dx}\left[F(4x)-F(1)\right]\]

  2. Jhannybean
    • one year ago
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    Remember that \(\sf (F(t))' = f(t)\) and \(\sf f(t) =\sqrt{t^2+1}\) So what would \(\sf (F(x))'\) = ?

  3. misty1212
    • one year ago
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    HI!!

  4. misty1212
    • one year ago
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    replace the \(t\) in the integrand by \(4x\)

  5. misty1212
    • one year ago
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    then, via the chain rule, multiply the whole thing by 4

  6. misty1212
    • one year ago
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    \[\sqrt{\left(\text{ put 4\(x\) here}\right)^2+1}\times 4\]

  7. SolomonZelman
    • one year ago
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    Anytime, in general: \(\large\color{black}{ \displaystyle \frac{d}{dx}\left[\int_{\rm C}^{g(x)} F'(t)dt\right] }\) \(\large\color{black}{ \displaystyle \frac{d}{dx}\left[F(t)~{\Huge |}^{g(x)}_{\rm C}~\right] }\) \(\large\color{black}{ \displaystyle \frac{d}{dx}\left[~F(g(x))-F({\rm C})~\right] }\) \(\large\color{black}{ \displaystyle F'(g(x))\cdot g'(x)-0=F'(g(x))\cdot g'(x) }\) ------------------------------------------------- Conclusion: \(\large\color{blue}{ \displaystyle \frac{d}{dx}\left[\int_{\rm C}^{g(x)} F'(t)dt\right]=F'(g(x))\cdot g'(x) }\)

  8. anonymous
    • one year ago
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    Sorry for the late reply. So the answer would be C?

  9. SolomonZelman
    • one year ago
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    yes, correct

  10. anonymous
    • one year ago
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    @SolomonZelman How would I solve an integral for an absolute value?

  11. SolomonZelman
    • one year ago
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    You would split the function for x>0 and x<0

  12. anonymous
    • one year ago
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    https://gyazo.com/a6de7e8243555fc7379828fcf09d1fd9

  13. SolomonZelman
    • one year ago
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    Oh, then you don't need splitting

  14. SolomonZelman
    • one year ago
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    The function from 0 to -3 is never positive. So you can just write \(\displaystyle \int_{-3}^{0}-(x+2)dx \)

  15. SolomonZelman
    • one year ago
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    because absolute value function on the interval (0,3) is negative (at least not positive). So, it is just a line y=-(x+2) at this interval.

  16. SolomonZelman
    • one year ago
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    Did I make sense just now?

  17. anonymous
    • one year ago
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    Yeah, I pretty much got it. Solving atm

  18. SolomonZelman
    • one year ago
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    atm ?

  19. SolomonZelman
    • one year ago
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    atmosphere?

  20. anonymous
    • one year ago
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    At the moment

  21. anonymous
    • one year ago
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    lol

  22. SolomonZelman
    • one year ago
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    No seriously, I am terrible at text language. I have hopefully managed to learn proper English, but text language is not for me. I guess it is too informal.

  23. anonymous
    • one year ago
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    Oh well, lol meant laugh out loud, just in case you didn't know.

  24. SolomonZelman
    • one year ago
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    I know "lol" "jk" and some others perhaps, but I think this is pretty much it.

  25. SolomonZelman
    • one year ago
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    Anyway, what did you get for your integral ?

  26. anonymous
    • one year ago
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    I got 3 for the final answer.

  27. SolomonZelman
    • one year ago
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    yes, that is right

  28. anonymous
    • one year ago
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    Thanks. Wish I could give you another medal.

  29. SolomonZelman
    • one year ago
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    I don't really care about medals, I got over 12K:)

  30. anonymous
    • one year ago
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    LOL, then you wouldn't mind helping me with another? :p

  31. SolomonZelman
    • one year ago
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    I guess not, if I will be able to:)

  32. anonymous
    • one year ago
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    This one should be easy, just can't remember what to do. https://gyazo.com/eb75236776b0079f4699d2e338fe6ec5

  33. anonymous
    • one year ago
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    Just plug in 0-2?

  34. SolomonZelman
    • one year ago
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    I refreshed the page, I have to do it again

  35. anonymous
    • one year ago
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    Ok, well anyways, if I recall properly the answer should be D.

  36. SolomonZelman
    • one year ago
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    Actually not D.

  37. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle e^0=1 }\)

  38. anonymous
    • one year ago
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    Ohhhhh

  39. SolomonZelman
    • one year ago
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    Proof: \(\large\color{black}{ \displaystyle a^0=a^{b-b}=\frac{a^b}{a^b}=\frac{\cancel{a^b}}{\cancel{a^b}}=1 }\)

  40. anonymous
    • one year ago
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    By the way would I plug in 0,1, and 2 or just 0 and 2?

  41. SolomonZelman
    • one year ago
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    Just plug in 2, and then 0 F(2)-F(0)

  42. anonymous
    • one year ago
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    This is what I did. \[\frac{ 1 }{ 4}e ^{4x}|_{0}^{2}\]

  43. anonymous
    • one year ago
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    Should be C?

  44. SolomonZelman
    • one year ago
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    yes, correct, that is what you had to do:)

  45. SolomonZelman
    • one year ago
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    yes, it should be C.

  46. SolomonZelman
    • one year ago
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    Good job!

  47. anonymous
    • one year ago
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    One last one please. Actually two but one I already solved. https://gyazo.com/f8561038cb4b1113268276fbaf824c2e https://gyazo.com/e7aea1e2a2b3ca9e81319fcbe8b70cb1

  48. anonymous
    • one year ago
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    Should be C for first and A for second?

  49. SolomonZelman
    • one year ago
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    yes, correct for both.

  50. SolomonZelman
    • one year ago
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    But, you have to still know that as \(x\rightarrow0\), then \(\ln(x)\rightarrow \infty\).

  51. anonymous
    • one year ago
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    Appreciate the help. Just so you know I got the absolute value one wrong, but I see where my mistake is. I added instead of subtracted.

  52. SolomonZelman
    • one year ago
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    just in case... and this you can easily demonstrate by defining zero as: \(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\left[10^{-n} \right] }\) And then, \(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\left[~~\log_{10}\left(10^{-n}\right) \right] = \lim_{n\rightarrow \infty} \left[-n\log_{10}(10)\right]=\lim_{n\rightarrow \infty} \left[-n\right]=-\infty}\)

  53. SolomonZelman
    • one year ago
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    You added what for absolute value/

  54. SolomonZelman
    • one year ago
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    ?

  55. anonymous
    • one year ago
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    instead of F[0]-F[-3] I put F[0]+F[-3] which...........now that I look at it I did do right no? That would give me 3. :/

  56. SolomonZelman
    • one year ago
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    yes, because F[0]+F[-3]

  57. SolomonZelman
    • one year ago
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    I mean because - - = +

  58. SolomonZelman
    • one year ago
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    I have to go right now. Be well:)

  59. anonymous
    • one year ago
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    Bye. Thanks for the help.

  60. SolomonZelman
    • one year ago
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    You are welcome!

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