Find the Avg value of the function i=15(1-e^-t/2) from t=0 and t=4?

Question

i=15(1-e^-t/2) to prevent confusion its 15 times 1-e to the -t/2 power.

when i integrate i set 1-e^-t/2 to u so i get 15u and then i integrate, and get 7.5u^2. am i doing this right?? but i really am confused when i let t=4! someone plz help thanks.

Answer 1

not exactly, you also need to substitute what dt is in terms of du.

15∫(1 - e^-t/2)dt = 15(∫dt + ∫-e^-t/2dt)

let u = e^-t/2

du/dt = (-1/2)e^-t/2

2du = -e^-t/2dt

so by simply substituting 2du i get rid of everything:

∫-e^-t/2dt --> 2∫du = 2u

substitute back what u was

2e^-t/2

the other integral was easy, so total answer is

15( t + 2e^-t/2)

now evaluate at endpoints. Also, since you want the average of the function, divide by the length of the interval, in this case 4.

should get:

7.5(1 + e^-2)

Answer 2

First take the interval. Then divide by the length of the interval.

∫15{1 - e^(-t/2)}dt = 15∫dt - 15∫e^(-t/2)dt
= 15t - {15e^(-t/2)}/(-1/2) = 15t + 30e^(-t/2) eval over [0,4]
= {15*4 + 30e^(-4/2)} - {15*0 + 30e^(-0/2)}
= 60 + 30e^(-2) - 30 = 30 + 30e^(-2) = 30{1 + e^(-2)}

The average value over the domain is:

30{1 + e^(-2)}/(4 - 0) = 7.5{1 + e^(-2)}