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March 27, 2017

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Find the area of the region between y = x sin x and y = x for 0 ≤ x ≤ pi/2

  • Calculus II - ,

    well, the first question is do they cross?

    xsinx=x
    sinx=1
    x=PI/2
    so no crossing.

    Then area is

    int (x-xsinx)dx over limits.

    x^2/2 -sinx+xcosx

    PI/2)^2/2-1 check that.

  • Calculus II - ,

    I so happens that the two functions intersect at
    x = 0 and x = π/2, the domain of our area

    area
    = [integral] (xsinx - x)dx from 0 to π/2
    = sinx - xcosx - (1/2)x^2 | from 0 to π/2
    = sinπ/2 - π/2(cosπ/2) - π^2/8 - (sin0 - 0 - 0)
    = 1 - 0 - π^2/8
    = 1 - π^2/8
    = - .2337

    OOPS, just realized that my assumption that the trig curve was above the straight line was false, so we have to reverse the integrand

    area = integral (x - xsinx)dx

    make the necessary changes, only the signs will be affected, or
    we could just take the absolute value of each of my lines.

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