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Math Help - integral abs sinx [0,3pi/2]

  1. #1
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    integral abs sinx [0,3pi/2]

    integral abs sinx [0,3pi/2]

    would this be -cosx(abs sinx/sinx)?
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  2. #2
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    Quote Originally Posted by kensington View Post
    integral abs sinx [0,3pi/2]

    would this be -cosx(abs sinx/sinx)?
    Draw the graph, then set up and evaluate the required integrals.
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  3. #3
    Junior Member autumn's Avatar
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    You cannot ignore the absolute value.
    Break up the integral into the parts where sin x is positive and negative and flip the negative part over.
    Then you can integrate.
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    what do you mean flip the negative part over?
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    Quote Originally Posted by kensington View Post
    what do you mean flip the negative part over?
    This means that [LaTeX ERROR: Convert failed]
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    Quote Originally Posted by Vlasev View Post
    This means that [LaTeX ERROR: Convert failed]
    soo then is it

    abs sinx= sinx if x=> 3pi/2
    =-sinx if x< 3pi/2

    so then integral from 0 to pi sinx+ integral from pi to 3pi/2 -sinx
    which is then [-cosx]+ [cosx]
    [(-cos0)-(-cospi)]+ [(cospi)-(cos3pi/2)]
    [-1-(1)] + [-1-(-1/2)]
    -2-1/2
    =-2.5??

    Would that be correct?
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  7. #7
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    This may surprise you but: \int_0^{\frac{{3\pi }}<br />
{2}} {\left| {\sin (x)} \right|dx}  = 3\int_0^{\frac{\pi }<br />
{2}} {\sin (x)dx} .

    Think about it after looking at the graph.
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  8. #8
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    Quote Originally Posted by Plato View Post
    This may surprise you but: \int_0^{\frac{{3\pi }}<br />
{2}} {\left| {\sin (x)} \right|dx}  = 3\int_0^{\frac{\pi }<br />
{2}} {\sin (x)dx} .

    Think about it after looking at the graph.
    so then would it be
    3 [[(-cos0)-(-cospi/2)]+ [(coso)-(cospi/2)]]?
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  9. #9
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    No. The answer is 3.
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  10. #10
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Plato View Post
    No. The answer is 3.
    and I thought every answer was 42, well at least 3 times 14 is 42.
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