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Math Help - Yet another integration question.

  1. #1
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    Yet another integration question.

    Hi guys, yes it's me again.

    Ok I can't quite get to the answer of this question despite it being similar to the last one I sought help for, anyway here's the question:

    Find (integrate; upper limit 3, lower limit -4) absolute (e^x - e^-x) dx, giving your answer to 3 significant figures.

    Answer: 70.8

    Thanks if you could help!
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  2. #2
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    Quote Originally Posted by margaritas
    Hi guys, yes it's me again.

    Ok I can't quite get to the answer of this question despite it being similar to the last one I sought help for, anyway here's the question:

    Find (integrate; upper limit 3, lower limit -4) absolute (e^x - e^-x) dx, giving your answer to 3 significant figures.

    Answer: 70.8

    Thanks if you could help!
    Okay you got,
    \int_{-4}^3 |e^x-e^{-x}|dx
    Since you are working with absolute value, you need to divide the the intervals were the function is below the x-axis and above the x-axis. (Look at graph below without absolute value)
    Therefore, the point where this change happens is at zero.
    Thus you can set up your integral as,
    -\int_{-3}^0 e^x-e^{-x}dx+\int_0^4 e^x-e^{-x}dx
    (Note the negative sign in front of the integral because since it is below the intregral is going to be negative and you need it to be positive cuz you have absolute value).
    Attached Thumbnails Attached Thumbnails Yet another integration question.-picture5.gif  
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  3. #3
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    Thanks! Lol careless me.
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