# Integration

• Oct 25th 2006, 07:44 PM
samsum
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
• Oct 26th 2006, 02:09 AM
CaptainBlack
Quote:

Originally Posted by samsum
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:

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

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
This last, do you mean:

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

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