Question

(a) Find the average value of e' over the time interval 0 < t < 54

Average value =
(b) Find the value of t at which e^t takes this average value.
The answer is close to 65 because exponential functions get big so quickly that the average value is dominated by values near the end. Thus the average human population of the planet (which has been growing exponentially) over the last 2000 years occurred since 1800.

Answer 1

To find the average value of e' over the time interval and the value of t at which e^t takes this average value, we need to perform integration and solve an equation.

Step-by-step explanation:

To find the average value of e' over the time interval 0 < t < 54, we need to integrate e' with respect to t over this interval and then divide by the length of the interval. In this case, the length of the interval is 54 - 0 = 54. So, the average value is given by:

average value = (1/54) * integral of e' dt from 0 to 54

Now, to find the value of t at which e^t takes this average value, we need to solve the equation e^t = average value. Rearranging the equation, we have:

t = ln(average value)

Substituting the value of average value that we found earlier into this equation will give us the value of t.

Answer 2

a) the average value of e' over the time interval is [
`e^(54)` - 1] / 54

b) the value of t is 50.0111

Step-by-step explanation:

Given that;

time interval 0 < t < 54

let y = f(x)

average value of f(x) [a,b]

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

so we substitute

Av = 1/(54-0) ⁵⁴∫₀
`e^(t)` dx

= 1/54 [
`e^(t)`]₀⁵⁴ dx

= 1/54 [
`e^(54)` -
`e^(0)` ]

= 1/54 [
`e^(54)` - 1 ]

= [
`e^(54)` - 1] / 54

Therefore, the average value of e' over the time interval is [
`e^(54)` - 1] / 54

b)

given that;
`e^(t)` = [
`e^(54)` - 1] / 54

we apply natural logarithm (ln) on both RHS and LHS

ln (
`e^(t)`) = ln ([
`e^(54)` - 1] / 54)

t( ln(e) = ln [
`e^(54)` - 1] - ln (54)

t(1) = (54 - 0) - 3.9889

t = 54 - 3.9889

t = 50.0111

Therefore, the value of t is 50.0111