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ChillOut

  • one year ago

Fun question

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  1. Flvs.net
    • one year ago
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    Hit me.

  2. Flvs.net
    • one year ago
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    Oh lord

  3. ChillOut
    • one year ago
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    Oops! Wrong one!

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  4. Flvs.net
    • one year ago
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    Oh lol

  5. Flvs.net
    • one year ago
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    God dang, what grade is this?

  6. ChillOut
    • one year ago
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    College Calculus

  7. Flvs.net
    • one year ago
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    I am only in 7th grade. sorry

  8. idku
    • one year ago
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    sen t ?

  9. ChillOut
    • one year ago
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    It's in portuguese. It's the same as "sine".

  10. idku
    • one year ago
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    \(\large\color{slate}{\displaystyle \frac{d^2}{dx^2}\int\limits_{0}^{x}\left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt}\)

  11. ChillOut
    • one year ago
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    Yep, That's it.

  12. idku
    • one year ago
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    lets work this part I suppose: \(\large\color{slate}{\displaystyle \left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt}\) u will get \(\large\color{slate}{ \int\limits_{1}^{\sin(t)}\sqrt{1+\sin^4t}dt}\) is that right, or am I off?

  13. idku
    • one year ago
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    too many variables, lol...

  14. ChillOut
    • one year ago
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    The inner integral gives me a strange answer in wolfram. There should be some change of coordinates or trig substitution that I'm missing.

  15. ChillOut
    • one year ago
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    http://www.wolframalpha.com/input/?i=int+sqrt%281%2Bu^4%29dudt

  16. idku
    • one year ago
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    Jesus was crucified on this answer, it is not the Romans or Jews that killed him, but for real, this is where my good feelings towards math stops.

  17. idku
    • one year ago
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    sorry, not a fan of world's most complex and longest expressions.

  18. ChillOut
    • one year ago
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    Haha, as I said, I'm missing something on this question, it can't be that hard.

  19. anonymous
    • one year ago
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    @Chillout remember me its Peaches I see dat u blocked y

  20. anonymous
    • one year ago
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    i only skipped 2 grades srry i can't help tip I'm on the next chapter xD

  21. ChillOut
    • one year ago
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    Some mates of mine were saying that all that is needed is FTC.

  22. ganeshie8
    • one year ago
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    \[\cos x\sqrt{1+\sin^4 x}\]

  23. ChillOut
    • one year ago
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    Is really that simple? Just plug in sin and integrate over dt?

  24. ganeshie8
    • one year ago
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    thats because of fundamental theorem of calculus : integral is the inverse of derivative

  25. ChillOut
    • one year ago
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    Yeah that's what I said earlier.

  26. ganeshie8
    • one year ago
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    \[\dfrac{d}{dx} \int\limits_a^{g(x)}f(t)~dt~~=~f(g(x))*g'(x)\]

  27. ChillOut
    • one year ago
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    I can solve calculus problems just fine, but yeah, forgetting that is like... unforgivable XD

  28. ganeshie8
    • one year ago
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    \(\large\color{slate}{\displaystyle \frac{d^2}{dx^2}\int\limits_{0}^{x}\left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt\\~\\~\\ =\dfrac{d}{dx} \int\limits_1^{\sin(x)} \sqrt{1+u^4}du\\~\\~\\ =\sqrt{1+\sin^4 x} * (\sin x)'\\~\\~\\ } \)

  29. ChillOut
    • one year ago
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    Yeah, that was just naive of me, but thanks for your help!

  30. anonymous
    • one year ago
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    sill need help

  31. IrishBoy123
    • one year ago
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    Holy Cow

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