(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.

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nuhulawal20 nuhulawal20 · Answer 1 · 2021-01-30T05:25:28+01:00

Answer:

a) the average value of e' over the time interval is [[tex]e^{54}[/tex] - 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 [tex]\int\limits^b_a f({x}) \, dx[/tex]

so we substitute

Av = 1/(54-0) ⁵⁴∫₀ [tex]e^{t}[/tex] dx

= 1/54 [[tex]e^{t}[/tex]]₀⁵⁴ dx

= 1/54 [[tex]e^{54}[/tex] - [tex]e^{0}[/tex] ]

= 1/54 [[tex]e^{54}[/tex] - 1 ]

= [[tex]e^{54}[/tex] - 1] / 54

Therefore, the average value of e' over the time interval is [[tex]e^{54}[/tex] - 1] / 54

b)

given that; [tex]e^{t}[/tex] = [[tex]e^{54}[/tex] - 1] / 54

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

ln ([tex]e^{t}[/tex]) = ln ([[tex]e^{54}[/tex] - 1] / 54)

t( ln(e) = ln [[tex]e^{54}[/tex] - 1] - ln (54)

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

t = 54 - 3.9889

t = 50.0111

Therefore, the value of t is 50.0111