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Math Help - Integration

  1. #1
    Junior Member samsum's Avatar
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    Integration

    i think i have to do this


    given, x = 0.2e^-0.1t

    solve: A = integral (from 0 to tmax) of xdt



    given y = 0.0909 + 0.00002x - 0.000000001x^2

    solve: B = m*integral (from 100 to 500) of ydT

    SORRY FOR INCONVENIENT TYPING
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by samsum View Post
    i think i have to do this


    given, x = 0.2e^-0.1t

    solve: A = integral (from 0 to tmax) of xdt
    Do you mean:

    <br />
\int_0^{t_{max}} 0.2 e^{-0.1 t} dt<br />

    If so then:

    <br />
\int_0^{t_{max}} 0.2 e^{-0.1 t} dt = [-2 e^{-0.1 t}]_0^{t_{max}}<br />

    ............... =2[1-e^{-0.1\ t_{max}}]


    given y = 0.0909 + 0.00002x - 0.000000001x^2

    solve: B = m*integral (from 100 to 500) of ydT

    SORRY FOR INCONVENIENT TYPING
    This last, do you mean:

    <br />
\int_{100}^{500} 0.0909 + 0.00002x - 0.000000001x^2 dx<br />

    you seem to have a confusing set of variables in what you have written.

    <br />
\int_{100}^{500} 0.0909 + 0.00002x - 0.000000001x^2 dx=<br />

    ............ <br />
[0.0909x + 0.00002x^2/2 - 0.000000001x^3/3]_{100}^{500}<br />

    RonL
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