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Integrals help

  1. Apr 5, 2014 #1
    I have these integrals to find:

    ∫ (5x^2 + sqrt(x) - 4/x^2) dx

    ∫ [cos(x/2) - sin(3x/2)] dx

    ∫ s/sqrt(s^2 + 4) ds (upper coordinate is 5 lower coordinate is 1)

    I have worked it out as:

    ∫〖(5x^2+√x〗-4/x^2) dx=5x^3/(2+1)+x^(1/2+1)/(1+1/2)-4x^(-2+1)/(-2+1)+C=5/3 x^3+2/3x^(3/2)+4/x+C


    ∫〖(cos⁡(x/2)-sin⁡(3x/2) )dx=2 sin⁡(x/2)+2/3 cos⁡(3x/2)+C〗


    ∫_1^5〖s/√(s^2+4)〗ds=1/2 √(s^2+4) (1≤s≤5)=1/2 (5^2+1)^(1/2)-1/2(1^2+1)^(1/2)=√26/2-√2/2

    Do these look right?
     
  2. jcsd
  3. Apr 5, 2014 #2
    The first one is kind of messy but you know you can split integrals like that up?
    ∫ (5x^2 + sqrt(x) - 4/x^2) dx = ∫5x^2 dx + ∫sqrt(x) dx - ∫4/x^2 dx
    That should make it easier. The second integral is right.
    By coordinate, do you mean bounds in a definite integral? so ∫ s/sqrt(s^2 + 4) ds = sqrt(s^2 + 4) + C
    sqrt(5^2+4)-sqrt(1^2+4), so sqrt(29) - sqrt(5) should be the last one.
     
    Last edited: Apr 6, 2014
  4. Apr 6, 2014 #3

    Mark44

    Staff: Mentor

    No, this is incorrect. The OP had it right the first time. Your version is missing the factor of 1/2 in the antiderivative.
     
  5. Apr 6, 2014 #4
    Are you talking about this? ∫ s/sqrt(s^2 + 4) ds = sqrt(s^2 + 4) + C
    Look
    u = s^2 + 4
    du = 2s ds


    (1/2)∫du/sqrt(u)

    (2/2) sqrt(u)

    sqrt(s^2 + 4) + C
    Where am I missing the factor of 1/2?
     
  6. Apr 6, 2014 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You missed a factor of 2.

    ehild
     
  7. Apr 6, 2014 #6

    Curious3141

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    Homework Helper

    Panphobia is correct.
     
  8. Apr 6, 2014 #7

    Mark44

    Staff: Mentor

    I shouldn't do integration in my head.:cry:
     
  9. Apr 6, 2014 #8

    Curious3141

    User Avatar
    Homework Helper

    Why not? I did, and it was right. :tongue:

    Just messin' with you. :biggrin:
     
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