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
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}}
$
...............$\displaystyle =2[1-e^{-0.1\ t_{max}}]$
This last, do you mean: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