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Math Help - Using Integrals :/

  1. #1
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    Using Integrals :/

    I need somewhere to start this problem. Were working on integrals in class and I need help with this Application.

    Question: Water flows into an empty storage tank at a rate of V'(t)= 40e^(-0.01t) liters per minute. How much water is in the storage tank after 20 minutes?


    What I did to set it up was I setup the integral from 0 to 20 with the integrand being 40e^(-0.01t). Now from here do I just take the anti-derivative and then plug in my times and subract them from eachother?
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    Re: Using Integrals :/

    Yes, you would have:

    V(20)-V(0)=\int_0^{20}V'(t)\,dt

    Since the tank is initially empty, i.e. V(0)=0 we then have:

    V(20)=40\int_0^{20}e^{-0.01t}\,dt
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    Re: Using Integrals :/

    Okay I got everything you said. The only thing is I try to do U substitution but I'm not sure if that would work. Im getting U= -.01t, DU= -.01dt. So I dont know if i multiply the 40 by the reciprocal of -.01 or what?
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    Re: Using Integrals :/

    Okay I got it to be 40(e^-.01(20)) and got it to be 792.03986. To the nearest thousandth would be 792.039 liters
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    Re: Using Integrals :/

    I did something wrong. Damn it.
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    Re: Using Integrals :/

    Your u substitution is what I would use:

    u=-0.01t\,\therefore\,du=-0.01\,dt

    and so we now have:

    V(20)=40(-100)\int_{u(0)}^{u(20)}e^{u}\,(-0.01\,dt)=4000\int_{-\frac{1}{5}}^0 e^u\,du

    Note: For the last step I used the rule -\int_a^b f(x)\,dx=\int_b^a f(x)\,dx
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    Re: Using Integrals :/

    For the answer I got -8.008 liters? Did I do something wrong?

    4000*[e^0 - e^(-.01)*(-1/5)] and got -8.008 liters.
    Last edited by illicitkush; November 26th 2012 at 09:16 PM.
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    Re: Using Integrals :/

    In liters, we have:

    V(20)=4000\int_{-\frac{1}{5}}^0 e^u\,du=4000\left[e^u \right]_{-\frac{1}{5}}^0=4000\left(e^0-e^{-\frac{1}{5}} \right)=

    4000\left(1-e^{-\frac{1}{5}} \right)\approx725.0769876880727
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  9. #9
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    Re: Using Integrals :/

    What happens to the -.01 inside the e function?
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  10. #10
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    Re: Using Integrals :/

    We did away with that when we substituted. We rewrote the definite integral in terms of u.
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