Answer:
a) A=0.86
b) A=2.81
c) A=2.88
Step-by-step explanation:
The average of a functions can be written as:
[tex]A=\frac{1}{b-a}\int^{b}_{a}f(t)dt[/tex] (1)
a.) The second hour means that the interval of time is from 0 to 2 hours, so a=0 and b=2. Using the equation (1) we can calculate the average intensity.
[tex]A=\frac{1}{2-0}\int^{2}_{0}te^{-0.1t}dt[/tex]
Using integration by parts we can solve it.
[tex]\int fdg=fg-\int gdf[/tex]
[tex]f=t[/tex] , [tex]df=dt[/tex]
[tex]dg=e^{-t/10}dt[/tex] , [tex]g=-10e^{-t/10}dt[/tex]
[tex]\int^{2}_{0}te^{-0.1t}dt=-10e^{-t/10}t|^{2}_{0}+10\int^{2}_{0} e^{-t/10}dt=-10e^{-t/10}t|^{2}_{0}-100e^{-t/10}|^{2}_{0}=1.75[/tex] (2)
Therefore, the average intensity at the second hour will be:
[tex]A=\frac{1}{2-0}\int^{2}_{0}te^{-0.1t}dt=\frac{1.75}{2}=0.86[/tex]
b) We can use the equation (2) to solve it. In this case the limits of integration will be a = 0 and b = 12 hours.
[tex]\int^{12}_{0}te^{-0.1t}dt=-10e^{-t/10}t|^{12}_{0}-100e^{-t/10}|^{12}_{0}=33.74[/tex]
Therefore, the average intensity at the twelfth hour will be:
[tex]A=\frac{1}{12-0}\int^{12}_{0}te^{-0.1t}dt=\frac{33.74}{12}=2.81[/tex]
c) Finally, here a = 0 and b = 24 hour.
[tex]\int^{24}_{0}te^{-0.1t}dt=-10e^{-t/10}t|^{24}_{0}-100e^{-t/10}|^{12}_{0}=69.16[/tex]
Therefore, the average intensity at the twenty-fourth hour will be:
[tex]A=\frac{1}{24-0}\int^{24}_{0}te^{-0.1t}dt=\frac{69.16}{24}=2.88[/tex]
I hope it helps you!
