Question

integrate 6e^(-.2t) from 0 to t with respect to t

the book says the answer is 30[1-e^(-.2t)] but I cant seem to get there

Answer 1

integral by u-sub
let u = -0.2t
du = -0.2 dt
dt = -5 du

see the integral becomes :
int [0,t] 6(-5e^u) du
= int [0,t] -30 e^u du
= [-30e^u] [0,t]
substitute back that u = -0.2t, that means
= [-30e^(-0.2t)] [0,t]
now, can you evaluate for the intervals of integral ?

Answer 2

I got it, thank you!