Question

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

Answer 1

Answer:

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