Question

Find the average value of the function i = 15(1 - e1/2t) from t = 0 to t = 4.

A. 7.5(3 - e-2)
B. 7.5(1 + e-2)
C. 7.5(2 - e-2)
D. 7.5(2 + e-2)

Answer 1

Answer with explanation:

The average value of a function from a to b i.e. in the interval a to b is calculated by:

`Average=(1)/(b-a)\int\limits^a_b {f(x)} \, dx`

we are given a function f(t) as:

`f(t)=i=15(1-e^{(1)/(2)t})`

and a= 0 and b=4

Hence, the average value of the function is:

`=(1)/(4-0)\int\limits^4_0 {15(1-e^{(1)/(2)t)} \, dt\\ \\\\=(15)/(4)[[t-2e^{(1)/(2)t}]_(4)-[t-2e^{(1)/(2)t}]_(0)]`

since,

`\int\limits {1-e^{(1)/(2)t}} \, dx =t-2{e^{(1)/(2)t}`

Hence,

`Average\ value=(15)/(4)[4-2e^2-0+2e^0]\\\\\\Average\ value=(15)/(4)[4-2e^2+2]\\\\\\Average\ value=(15)/(4)[6-2e^2]\\\\\\Average\ value=7.5(3-e^2)`

The average value is:

`Average\ value=7.5(3-e^2)`

Answer 2

1) Function:


`i=15(1 - e^((1/2)t))`

2) Average = [value of the function a t = 0 + value of the function at t = 4 ] / 2

3) Value of the function at t = 0

15[1 - e^(0) ] = 15 [ 1 - 1] = 0

4) Value of the function at t = 4

15 [ 1 - e^ (4/2) ] = 15 [ 1 - e^(2) ]

5) Average

{ 0 + 15 [1 - e ^ (2) ] } / 2 = 7.5 [ 1 - e^(2)]

Given that all the options show e raised to negative 2, I guess you made a mistake on the function.