Answer :
1) Function:
[tex]i=15(1 - e^{(1/2)t})[/tex]
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.
[tex]i=15(1 - e^{(1/2)t})[/tex]
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.
Answer with explanation:
The average value of a function from a to b i.e. in the interval a to b is calculated by:
[tex]Average=\dfrac{1}{b-a}\int\limits^a_b {f(x)} \, dx[/tex]
we are given a function f(t) as:
[tex]f(t)=i=15(1-e^{\dfrac{1}{2}t})[/tex]
and a= 0 and b=4
Hence, the average value of the function is:
[tex]=\dfrac{1}{4-0}\int\limits^4_0 {15(1-e^{\dfrac{1}{2}t)} \, dt\\ \\\\=\dfrac{15}{4}[[t-2e^{\dfrac{1}{2}t}]_{4}-[t-2e^{\dfrac{1}{2}t}]_{0}][/tex]
since,
[tex]\int\limits {1-e^{\dfrac{1}{2}t}} \, dx =t-2{e^{\dfrac{1}{2}t}[/tex]
Hence,
[tex]Average\ value=\dfrac{15}{4}[4-2e^2-0+2e^0]\\\\\\Average\ value=\dfrac{15}{4}[4-2e^2+2]\\\\\\Average\ value=\dfrac{15}{4}[6-2e^2]\\\\\\Average\ value=7.5(3-e^2)[/tex]
The average value is:
[tex]Average\ value=7.5(3-e^2)[/tex]