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Math Help - Some reassurance

  1. #1
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    Some reassurance

    Hello,

    I would be most grateful if someone could check my work.

    <br />
f(x)= (x-3)(e^\frac{-x}{4}-1)<br />

    Find the area enclosed by the graph of this function and the x axis between x=0 and x=6.

    As the graph crosses the x-axis at x=0 and x=3 all values of x are positive between x=0 and x=3 all values of x>3 are negative.

    It is necessary to to find the positive area bounded by x=0 and x=3 and subtract it from the negative area bounded by x=3 and x=6.

    So


    <br />
\int^{3}_{0}(x-3)(e^\frac{-x}{4}-1)dx -\int^{6}_{3}(x-3)(e^\frac{-x}{4}-1)dx<br />

    After hopefully performing the integration correctly I get an answer of 4.1319
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  2. #2
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    Everything checks out to me, including your final answer.
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  3. #3
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    Also, if you're allowed to do these problems on the calculator I would suggest using the absolute value button. It's for you to note that total area means positive and negative area, not just the integral, but if you need a quick calculation, I know on the TI models you can enter \int |f(x)|dx
    Last edited by Jameson; July 3rd 2006 at 04:17 PM.
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  4. #4
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    Thanks for for that Jameson
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  5. #5
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    Quote Originally Posted by Jameson
    Also, if you're allowed to do these problems on the calculator I would suggest using the absolute value button. It's for you to note that total area means positive and negative area, not just the integral, but if you need a quick calculation, I know on the TI models you can enter |\int f(x)dx|
    No, Jameson it is.
    \int_a^b|f(x)|dx
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