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  1. #1
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    help

    Piecewise function

    let f(x) = 2/pi*x [0.pi/2]
    = 2sinx (pi/2, 3pi/2]
    = 1 (3pi/2,2pi]

    a. find a continuous funciton, F(x) , satifying F(x) = integral of f(t)dt on [0,2pi].

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

    c. comupte integral of f(x)dx from 0 to 2pi.
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  2. #2
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    Hello, Nichelle14 !

    Didn't I solve this at another site?

    Let 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), satifying F(x) = \int^{2\pi}_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 the interval \left[0,\frac{\pi}{2}\right], the graph is a straight line.

    On the interval \left(\frac{\pi}{2},\,\frac{3\pi}{2}\right], we have a portion of the sine curve.

    On the interval \left(\frac{3\pi}{2},\,2\pi\right], the graph is a horizontal line.

    The graph looks like this:
    Code:
            |
           1+       *
            |     * :   *           o * * * *
            |   *   :     *         :
            | *     :      *        :
          - * - - - + - - - * - - - + - - - +
            |      π/2      π*     3π/2    2π
            |                 *     :
            |                   *   :
          -1+                       *
    You should be able to answer the questions now . . .
    .
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  3. #3
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    I understand the graph how that works. I did that too, but I really dont' know what to do next. I know the F(x) notation should ring a bell but it's not.

    Can you help some more? Thanks.

    I think you did do this before, but I don't see any thing else on it.
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  4. #4
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    Hello, Nichelle14 !

    Here's what I said:
    On the interval \left[0,\frac{\pi}{2}\right], the graph is a straight line.

    On the interval \left(\frac{\pi}{2},\,\frac{3\pi}{2}\right], we have a portion of the sine curve.

    On the interval \left(\frac{3\pi}{2},\,2\pi\right], the graph is a horizontal line.

    The graph looks like this:
    Code:
            |
           1+       *
            |     * :   *           o * * * *
            |   *   :     *         :
            | *     :      *        :
          - * - - - + - - - * - - - + - - - +
            |      π/2      π*     3π/2    2π
            |                 *     :
            |                   *   :
          -1+                       *
    So you got the graph, too . . . good for you!


    I'm not sure how they want part (a) answered.
    F(x)\:=\:\int^{2\pi}_0\!f(x)\,dx is the area under the graph on [0,\,2\pi]

    On \left[0,\,\frac{\pi}{2}\right], the area increases steadily.

    On \left(\frac{\pi}{2},\,\pi\right], the area still increases but more slowly, until it stops increasing at \pi.

    Then on \left(\pi,\,\frac{3\pi}{2}\right], the area decreases.

    Finally, on \left(\frac{3\pi}{2},\,2\pi\right], the area increases steadily.

    And I can't find a continuous function that behaves that way.

    After answering part (c), I found a way . . . but probably not what they're looking for.



    (c) The area under the curve can be found without calculus!

    First, we have a triangle with base \frac{\pi}{2} and height 1.
    . . Its area is: . \frac{1}{2}\left(\frac{\pi}{2}\right)(1)\:=\:\frac  {\pi}{4}

    Then we have two equal regions of the sine curve, above and below the x-axis.
    . . These will "cancel out" . . . net area: 0.

    Finally we have a rectangle with base \frac{\pi}{2} and height 1.
    . . Its area is: . \left(\frac{\pi}{2}\right)(1)\:=\:\frac{\pi}{2}

    The total area under the graph is: \frac{\pi}{4} + 0 + \frac{\pi}{2}\;=\;\frac{3\pi}{4}


    My interpretation is: . F(x)\;=\;\int^{2\pi}_0\!f(x)\,dt \;= \;\frac{3\pi}{4}

    That is: any function which satifies that equation should be satisfactory.


    So we can use: f(x)\:=\:\frac{3}{8} . . . because: \int^{2\pi}_0\frac{3}{8}\,dx\;=\;\frac{3\pi}{4}


    Or we can use: f(x) \,= \,\frac{3}{4\pi}x . . . because: \int^{2\pi}_0\!\frac{3}{4\pi}x\,dx\;=\;\frac{3\pi}  {4}

    See what I mean?
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    Hello, Nichelle14 !

    I'm not sure how they want part (a) answered.
    F(x)\:=\:\int^{2\pi}_0\!f(x)\,dx is the area under the graph on [0,\,2\pi]
    Hello Sorobon,
    Good Effort.
    I note something which I want to tell you:
    1) We have to find F(x), not f(x)
    2) If you see the above equation (in quote), you will see that RHS is a constant. Hence F(x) will be a constant function and a constant function is continuous.
    Am I right?
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  6. #6
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    Hello, malaygoel!

    I'm certain that you're right.

    Right now my brain is fried from all that (unnecessary) Thinking.
    If it took two brain cells to get to China, I couldn't get out of sight . . .
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  7. #7
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    Hello, malaygoel!

    I'm certain that you're right.

    Right now my brain is fried from all that (unnecessary) Thinking.
    If it took two brain cells to get to China, I couldn't get out of sight . . .
    sorry, but I don't clearly understand you
    what do you consider unnecessary here
    and what do you want to convey through your last line?
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