Respuesta :
The average weight of the colony over the interval [1,4] is 47.83 newtons
The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.
Given,
[tex]w(t)=12.5e^{0.5t}[/tex]
The average value of the given function
[tex]w_{avg} =\frac{1}{b-a}\int\limits^a_b {w(t)} \, dt[/tex]
Here the interval is [1,4]
Therefore, a=1 and b= 4.
Substitute the values of w(t),a and b in the equation
[tex]w_{avg}= \frac{1}{4-1}\int\limits^4_1 {12.5e^{0.5t} } \, dx \\w_{avg}= \frac{1}{3}\int\limits^4_1 {\frac{25e^{0.5t} }{2} } \, dx[/tex]
Consider,
[tex]u=\frac{t}{2}\\ \frac{du}{dt}=\frac{1}{2}\\dt=2du[/tex]
Then,
[tex]w_{avg}= \frac{25}{3}\int\limits^4_1 {e^{u} } \, du\\ w_{avg}= \frac{25}{3}[e^{4/2}-e^{1/2}]\\ w_{avg}= \frac{25}{3}[e^{2}-\sqrt{e}]\\ w_{avg}= \frac{25}{3}[5.74]\\ w_{avg}=47.83[/tex]
Hence, the average weight of the colony over the interval [1,4] is 47.83 newtons
Learn more about average value here
brainly.com/question/14402983
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