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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- Oct 25th 2006, 07:44 PMsamsumIntegration
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 - Oct 26th 2006, 02:09 AMCaptainBlack
Do you mean:

$\displaystyle

\int_0^{t_{max}} 0.2 e^{-0.1 t} dt

$

If so then:

$\displaystyle

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

$

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

Quote:

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

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

SORRY FOR INCONVENIENT TYPING

$\displaystyle

\int_{100}^{500} 0.0909 + 0.00002x - 0.000000001x^2 dx

$

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

$\displaystyle

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

$

............$\displaystyle

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

$

RonL