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Math Help - Help with this Calculus Problem

  1. #1
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    Help with this Calculus Problem

    This is a piecewise function
    Let f(x) = 2/pi*x when [0,pi/2]
    2sinx when (pi/2, 3pi/2]
    1 when (3pi/2, 2pi]


    a. find a continuous function, F(x), satisfying F(x) = the integral f(t) dt from 0 to x, on [0,2pi]

    b. Give the values of x on [0,2pi] for which F(x) is not differentiable.

    c. compute the integral f(x) dx from 0 to 2pi.
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  2. #2
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    Hello, Susie38!

    Here are parts (b) and (c).

    f(x) \:=  \:\begin{Bmatrix}\frac{2}{\pi}x & 0 \leq x \leq \frac{\pi}{2} \\ 2\sin x & \frac{\pi}{2} < x \leq \frac{3\pi}{2} \\ 1 & \frac{3\pi}{2} < x \leq 2\pi\end{Bmatrix}


    a. Find a continuous function, F(x), satisfying F(x) = \int^x_0 f(t)\,dt

    b. Give the values of x on [0,2\pi] for which F(x) is not differentiable.

    c. Compute: \int^{2\pi}_0 f(x)\,dx
    On \left[0,\,\frac{\pi}{2}\right]\], the function is a straight line.

    On \left(\frac{\pi}{2},\,\frac{3\pi}{2}\right], the function is a sine curve.

    On \left(\frac{3\pi}{2},\,2\pi\right], the function is a horizontal line.
    Code:
            |
          1 +       **              o * * * *
            |     * :   *           :       :
            |   *   :     *         :       :
            | *     :      *        :       :
          - * - - - + - - - * - - - + - - - + -
            |      π/2     2π*    3π/2     2π
            |                 *     :
            |                   *   :
         -1 +                      **
            |
    (b) F(x) is not differentiable at x = \frac{\pi}{2} and x = \frac{3\pi}{2}

    It has two different slopes at x = \frac{\pi}{2}

    It is discontinuous at x = \frac{3\pi}{2}


    (c) No calculus needed!

    The triangle has base \frac{\pi}{2} and height 1.
    . . . Its area is: A_1 \:= \:\frac{1}{2}\left(\frac{\pi}{2}\right)(1)\:=\; \frac{\pi}{4}

    The sine curve has a region above and a region below the x-axis.
    . . . The two regions have equal area and will cancel out.
    Its net area is: A_2\:=\:0

    The rectangle has base \frac{\pi}{2} and height 1.
    . . . Its area is: A_3\:=\:\left(\frac{\pi}{2}\right)(1)\:=\: \frac{\pi}{2}


    The value of the integral is the sum of these areas:
    . . . A_1 + A_2 + A_3 \;= \;\frac{\pi}{4} + 0 + \frac{\pi}{2}\;=\;\frac{3\pi}{4}
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