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Thread: area enclosed

  1. #1
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    area enclosed

    area enclosed-win_20170318_11_02_19_pro.jpg


    the question is to find the area enclosed between the y axis, the line y=10 and the exponential curve using integration.

    so first i need to switch the function around. f(x)=2e^x y=2e^x y/2 =e^x im not sure what i do next is correct. i get ln of both sides ln y/2 = ln e^x so

    ln y/2 =x ln e ln y/2 =x (1) ln y/2 =x. i dont know how to integrate ln y/2 after this though i would find the definite integral of from 10 to 2.
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  2. #2
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    Re: area enclosed

    You're making this more difficult than it isn't ...

    The curve intersects the line at $x=\ln{5}$, the upper limit of integration.

    Area is just the integral of the upper function minus the lower function ...

    $\displaystyle A = \int_0^{\ln{5}} 10-2e^x \, dx$

    Evaluate ... you should get $A=10\ln{5}-8$
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  3. #3
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    Re: area enclosed

    i dont see you get that. i get 10ln5-2e^ln5 =10ln5-10
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  4. #4
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    Re: area enclosed

    $\bigg[10x-2e^x\bigg]_0^{\ln{5}}$

    $\left(10\ln{5}-2e^{\ln{5}}\right)-\left(0-2e^0\right)$

    note, $e^{\ln{5}}=5$ and $e^0=1$ ...
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