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Evaluating the integral of absolute values

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data

    ∫(0 to 3pi/2) -7|sinx|dx

    2. Relevant equations



    3. The attempt at a solution

    I am not sure how to treat it as it has an absolute value

    i assumed that you could remove the -7 to get

    -7∫|sinx| dx

    then integrate sinx into -cosx but since there is absolute value i tried to change -cosx to cosx which ended up as

    -7cosx

    and then doing -7cos(3pi/2)-(-7cos(0))

    but this was incorrect. Can anyone give me any hints?

    thanks
     
  2. jcsd
  3. Jan 15, 2013 #2

    Zondrina

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    Homework Helper

    Usually for absolute values you have to break things into cases depending on what values of x you're integrating over.

    Notice that from 0 to 3π/2, x≥0? If x≥0, then |sinx| = sinx.

    Consider integrating from x=-2 to x=-1, x<0. If x<0, then |sinx| = -sinx.
     
  4. Jan 15, 2013 #3

    CAF123

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    Gold Member

    Sketch the graph of y = sinx from 0 to 3pi/2. From this, can you see what g = |sinx| would look like? Do you then see why integrating sinx to simply -cosx is wrong?
     
  5. Jan 15, 2013 #4

    Mark44

    Staff: Mentor

    This is NOT true. If 0 ≤ x ≤ π, then |sinx| = sinx, but for π ≤ x ≤ 2π, sin(x) ≤ 0.
    This is not true, either. There are infinitely many intervals for which x < 0 but sin(x) ≥ 0.
     
  6. Jan 15, 2013 #5
  7. Jan 15, 2013 #6

    CAF123

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    Gold Member

    It is still not right. Can you use |sinx| = sinx if sinx ≥ 0 and |sinx| = -sinx if sinx < 0? This is key to solving the problem.
     
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