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Math Help - Can somebody find my error in solving this integral?

  1. #1
    Member Em Yeu Anh's Avatar
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    Question Can somebody find my error in solving this integral?

    Evaluate \int_0^{\pi}f(x)dx
    where f(x) = 2sin(x), 0 \leq x< \frac{\pi}{2}
    7cos(x), \frac{\pi}{2} \leq x < \pi
    (Sorry I can't get the LaTex for piecewise to work properly, hope it's not too hard to read)

    \int_0^{\pi}f(x)dx = \int_0^{\frac{\pi}{2}}f(x)dx + \int_{\frac{\pi}{2}}^{\pi}f(x)dx

    = \int_0^{\frac{\pi}{2}}2sin(x)dx + \int_{\frac{\pi}{2}}^{\pi}7cos(x)dx

    =  [-2cos(x)]_0^{\frac{\pi}{2}} + [7sin(x)]_{\frac{\pi}{2}}^{\pi}

    = 2 + 7 = 9

    I believe the correct answer is -2 + 7 = 5.
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  2. #2
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    Quote Originally Posted by Em Yeu Anh View Post
    Evaluate \int_0^{\pi}f(x)dx
    where f(x) = 2sin(x), 0 \leq x< \frac{\pi}{2}
    7cos(x), \frac{\pi}{2} \leq x < \pi
    (Sorry I can't get the LaTex for piecewise to work properly, hope it's not too hard to read)

    \int_0^{\pi}f(x)dx = \int_0^{\frac{\pi}{2}}f(x)dx + \int_{\frac{\pi}{2}}^{\pi}f(x)dx

    = \int_0^{\frac{\pi}{2}}2sin(x)dx + \int_{\frac{\pi}{2}}^{\pi}7cos(x)dx

    =  [-2cos(x)]_0^{\frac{\pi}{2}} + [7sin(x)]_{\frac{\pi}{2}}^{\pi}

    = 2 + 7 = 9

    I believe the correct answer is -2 + 7 = 5.

    If you draw the function it must be clear there's much more area under the x-axis than over it, so the integral will be negative:

     [-2cos(x)]_0^{\frac{\pi}{2}} + [7sin(x)]_{\frac{\pi}{2}}^{\pi} =-2(0-1) + 7(0-1)=2-7=-5

    Now, if you were asked to evaluate the (geometrical) AREA between the graph of the function and the x-axis then you have to change the sign of the second integral to make it positive and then the area is 9.

    Tonio
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