calculus

Find the area bounded by the given curves y= 7cos(x) and y=7cos^2(x) between x=0 and x= π/2.

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asked by love
  1. first we need their intersection:
    7cos^2(x) = 7cos(x)
    cos^2 x - cosx = 0
    cosx(cosx- 1)
    cosx = 0 or cosx = 1
    x = π/2, or x = 0
    ahhh, our boundaries happen to be their intersection.
    let's look at the graph:
    www.wolframalpha.com/input/?i=plot+y%3D+7cos(x)+,+y%3D7cos%5E2(x)+from+-.2+to+(%CF%80%2F2+%2B+1)

    looks like 7cos^2 x is above 7cosx
    so area =∫ (7cos^2x -7cosx) dx from 0 to π/2
    = (1/2)(x + sinx coxx) + 7sinx ] from 0 to π/2
    = ..... (you so the substitution)

    (the integrals of sin^2 x and cos^2 x should be in your repertoire of standard integrals)

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    posted by Reiny
  2. I think you make a mistake with which one above and below

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    posted by love
  3. You are right, so just turn the terms around to
    area =∫ (7cosx - 7cos^2x) dx from 0 to π/2

    I should have looked at my own graph a bit closer.

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    posted by Reiny
  4. This are how I solve it. Do you think my answer correct?
    [0,π/2]∫7cos(x)-7cos^2(x)dx
    [0,π/2]∫7cos(x)-7(1-sin^2(x))dx
    [0,π/2]∫7cos(x)-7(1-(1/2(1-cos(2x)))dx
    [0,π/2]7sin(x)-7x/2 -7sin(2x)/2]
    [0,π/2]7sin(x)-7x/2 -7*2sin(x)cos(x)/2]
    7*π/2(-7*1-7*1*0)]
    =7π/2

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    posted by love
  5. In your step:
    [0,π/2]∫7cos(x)-7cos^2(x)dx
    [0,π/2]∫7cos(x)-7(1-sin^2(x))dx

    why did you change cos^2 x to (1 - sin^2 x)
    to integrate sin^2 x is just as difficult as cos^2 x

    did you mean to use: cos(2x) = 2cos^2 x - 1 ??
    then 2cos^2 x = cos(2x) + 1
    cos^2 x = (1/2)cos(2x) + 1/2
    ∫7cos^2 x = ∫7(1/2)cos(2x) + 7/2
    = (7/4)sin(2x) + 7x/2

    so [0,π/2]∫7cos(x)-7cos^2(x)dx
    = [7sinx - (7/4)sin(2x) - 7x/2 ] from 0 to π/2
    = (7(1) - (0 - 7π/4)) - (0 - ((7/4)(0) - 0))
    = 7 + 7π/4

    Wolfram shows 7 - 7π/4, I can't seem to find my arithmetic error

    www.wolframalpha.com/input/?i=%E2%88%AB7cos(x)-7cos%5E2(x)dx+from+0+to+%CF%80%2F2

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    posted by Reiny

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