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Jgeurts

  • 2 years ago

Evaluate the following integrals. Show your work.

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  1. zepp
    • 2 years ago
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    1/cos^2(x) is sec^2(x) and the integral of that is tan(x)

  2. Jgeurts
    • 2 years ago
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    \[\int\limits_{0}^{2}x \sqrt{9-2x^2}dx\]

  3. Jgeurts
    • 2 years ago
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    \[\frac{ d }{ dx }\int\limits_{x^2}^{sinx}\sqrt{1+t^2}dt\]

  4. primeralph
    • 2 years ago
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    no, just substitute, then everything falls out

  5. Best_Mathematician
    • 2 years ago
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    use calculator so easy

  6. Jgeurts
    • 2 years ago
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    I need to do it without @best_mathematician

  7. Best_Mathematician
    • 2 years ago
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    ok here...

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  8. modphysnoob
    • 2 years ago
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    integral and derivative cancel each other out

  9. Best_Mathematician
    • 2 years ago
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    now u know wht to do right @Jgeurts

  10. Jgeurts
    • 2 years ago
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    yes thanks BM

  11. Jgeurts
    • 2 years ago
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    @modphysnoob show me please

  12. modphysnoob
    • 2 years ago
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    \[\frac{ d }{ dx }\int\limits_{x^2}^{sinx}\sqrt{1+t^2}dt\] after our supposed integration F(Sin(x)) - F ( x^2) take dervaitve d/dx of F( Sin(x)) = cos(x) f(sin(x) d/dx of F(x^2) = f (x^2) 2x go back and plug into your funciton

  13. modphysnoob
    • 2 years ago
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    f(t)= Sqrt[1+t^2] f(sin(x)= sqrt[1+ sin^2(x)] f(sin(x)*cos(x)= sqrt[1+ sin^2(x)]* cos(x)

  14. modphysnoob
    • 2 years ago
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    now the bottom part f(x^2)2x=sqrt[1+x^2]2x

  15. modphysnoob
    • 2 years ago
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    sqrt[1+ sin^2(x)]* cos(x)-sqrt[1+x^2]2x

  16. modphysnoob
    • 2 years ago
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    question?

  17. Jgeurts
    • 2 years ago
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    Im formulating my question, haha

  18. modphysnoob
    • 2 years ago
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    ok

  19. Jgeurts
    • 2 years ago
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    so the d/dx means derive the interval?

  20. modphysnoob
    • 2 years ago
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    yes, since we are taking a function integrate it and derive , it undoes the integration

  21. Jgeurts
    • 2 years ago
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    haha great! i get it, thank you, that one was super hard!

  22. modphysnoob
    • 2 years ago
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    I can show you an easy example if you like?

  23. Jgeurts
    • 2 years ago
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    @modphysnoob sure that would be great, im learning for my final :)

  24. modphysnoob
    • 2 years ago
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    let's to a problem like yours but with easier integrand |dw:1367723123193:dw|

  25. Jgeurts
    • 2 years ago
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    so we would derive sinx to cosx and x2 to 2x

  26. Jgeurts
    • 2 years ago
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    then plug in for t and subtract them from each other right?

  27. Jgeurts
    • 2 years ago
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    so cosx-2x?

  28. modphysnoob
    • 2 years ago
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    so in this case f(t)= t when we integrate we will get a function which we will call F(t) F(t) is evaluated at F(sin(x)) - F( x^2) then taken d/dx of d/dx( F(sin(x)) - F( x^2) ) using chain rule d/dx ( F(sin(x)) = cos(x) f( sin(x)

  29. modphysnoob
    • 2 years ago
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    which is cos(x) sin(x)

  30. Jgeurts
    • 2 years ago
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    i see

  31. modphysnoob
    • 2 years ago
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    let's do botttom part d/dx( F (x^2) = 2x f( x^2) 2x x^2= 2x^3

  32. modphysnoob
    • 2 years ago
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    so cos(x) sin(x)-2x^3

  33. modphysnoob
    • 2 years ago
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    this is easy integrand so we could actually integrate it and then differntiate it to confirm out result. Integrate t t^2/2 plug in limits (sin^2 x)/2 - x^4/2 take derivative d/dx cos(x) sin(x)- 2x^3

  34. modphysnoob
    • 2 years ago
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    got it?

  35. Jgeurts
    • 2 years ago
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    Yes thats great! thank you so much! @modphysnoob

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