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Math Help - I need help!

  1. #1
    Junior Member
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    I need help!

    Hey guys, lol I'm here again! with an integration question. I'm having a quiz on the topic this Monday and while doing revision, I couldn't quite get the following question:

    Evaluate (integrate; upper limit - 2, lower limit - 0) absolute (x^2 + 2x - 3) dx.

    I got 2/3 but the answer at the back says it's 4?
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  2. #2
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    Quote Originally Posted by margaritas
    Hey guys, lol I'm here again! with an integration question. I'm having a quiz on the topic this Monday and while doing revision, I couldn't quite get the following question:
    Evaluate (integrate; upper limit - 2, lower limit - 0) absolute (x^2 + 2x - 3) dx.
    I got 2/3 but the answer at the back says it's 4?
    Hello, margaritas,

    I've attached a diagram to show you what you have to calculate.

    1. Your function has a zero at x = 1. So you have to split the integral into:

    \int_{0}^{1}(-(x^2+2x-3))dx=\left[-\frac{1}{3}x^3-x^2+3x\right]_{0}^{1}=\frac{5}{3} and
    \int_{1}^{2}(x^2+2x-3)dx=\left[\frac{1}{3}x^3+x^2-3x\right]_{1}^{2}=\frac{7}{3}

    Add these partial values and you'll get 4.

    Greetings

    EB
    Attached Thumbnails Attached Thumbnails I need help!-flaeche_parab1.gif  
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  3. #3
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    That's fast, thanks for your help once again, EB!

    I'll take a look at your solution straight after I finish a report for a project. Yes I have loads of homework I know. lol
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