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Math Help - integral subdivision rule

  1. #1
    Junior Member
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    integral subdivision rule

    Graph of f consistes of linesegments and semicircles.Use it to geometrically evaluate the following integrals.
    (1) integral (0 to 14) f(x) dx

    (2) integral (0 to 10) f(x) dx
    (3) integral (3 to 12) f(x) dx

    For diagram please see the doc fileattached.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by gracy View Post
    Graph of f consistes of linesegments and semicircles.Use it to geometrically evaluate the following integrals.
    (1) integral (0 to 14) f(x) dx

    (2) integral (0 to 10) f(x) dx
    (3) integral (3 to 12) f(x) dx

    For diagram please see the doc fileattached.
    Integral means you have to find the area between graph and x axis.
    If graph is above x-axis, tha area is positive
    if graph is below x axis, area is negative
    integral (0 to 6) f(x) dx=6*2 +area of semicicle= 12+\pi *3^2*\frac{1}{2}= 12+4.5\pi
    integral (6 to 8) f(x) dx=area of triangle of base length 2 and heght 2=2
    integral (8 to 10) f(x) dx=-2
    integral (0 to 6) f(x) dx=-(4*2 +area of semicicle)= -(8+\pi *2^2*\frac{1}{2})= -(8+2\pi)

    hence answers are
    1) 4+2.5\pi
    2) 12+4.5\pi
    for 3) you have to find the area between 3 and 6
    It will be equal to half of area between 0 and 6 doue to symmetry.
    3) -2-2.5\pi

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    Malay
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  3. #3
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    Hello, Gracy!

    Graph of f(x) consists of line segments and semicircles.
    Use it to geometrically evaluate the following integrals.

    (1)\;\int^{14}_0\!\!f(x)\,dx \qquad (2)\;\int^{10}_0\!\!f(x)\,dx \qquad (3)\;\int^{12}_3\!\!f(x)\,dx
    Code:
                * *
             *       *
           *    9π/2   *
          *             *
          *-------------*
          |             | *
          |     12      |   *
          |             |  2  *        10         14
        --+------+------+-------*-------+---------+
          0             6       8 * -2  |         |
                                    *   |   -8    |
                                      * |         |
                                        *---------*
                                        *   -2π   *
                                          *      *
                                            * *

    At the far left, there is a semicircle of radius 3.
    . . Its area is: . \frac{1}{2}\pi(3^2) \:=\:\frac{9\pi}{2}
    Below it is a 6-by-2 rectangle; its area is 12

    There are two isoceles right triangles, each with area 2.

    At the far left, there is a semicircle of radius 2.
    . . Its area is: . \frac{1}{2}\pi(2^2) \:=\:2\pi
    Above it is a 4-by-2 rectangle; its area is 8.

    The regions below the x-axis have "negative areas".


    (1)\;\int^{14}_9f(x)\,dx\;=\;\frac{9\pi}{2} + 12 + 2 - 2 - 2\pi^2 - 8 \;=\;\frac{\pi}{4} + 4

    (2)\;\int^{10}_0f(x)\,dx\;=\;\frac{9\pi}{2} + 12 + 2 - 2 \;=\;\frac{9\pi}{2} + 12

    (3)\;\int^{12}_3f(x)\,dx\;=\;\frac{1}{2}\left(\fra  c{9\pi}{2} + 12\right) + 2 - 2 - \frac{1}{2}\left(2\pi + 8\right) \;= \;\frac{5\pi}{4} + 2

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