Question

Drug Reaction The intensity of the reaction to a certain drug, in appropriate units, is given by

Answer 1

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:

$A=\frac{1}{b-a}\int^{b}_{a}f(t)dt$ (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.

$A=\frac{1}{2-0}\int^{2}_{0}te^{-0.1t}dt$

Using integration by parts we can solve it.

$\int fdg=fg-\int gdf$

$f=t$ , $df=dt$

$dg=e^{-t/10}dt$ , $g=-10e^{-t/10}dt$

$\int^{2}_{0}te^{-0.1t}dt=-10e^{-t/10}t|^{2}_{0}+10\int^{2}_{0} e^{-t/10}dt=-10e^{-t/10}t|^{2}_{0}-100e^{-t/10}|^{2}_{0}=1.75$ (2)

Therefore, the average intensity at the second hour will be:

$A=\frac{1}{2-0}\int^{2}_{0}te^{-0.1t}dt=\frac{1.75}{2}=0.86$

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.  

$\int^{12}_{0}te^{-0.1t}dt=-10e^{-t/10}t|^{12}_{0}-100e^{-t/10}|^{12}_{0}=33.74$

Therefore, the average intensity at the twelfth hour will be:

$A=\frac{1}{12-0}\int^{12}_{0}te^{-0.1t}dt=\frac{33.74}{12}=2.81$

c) Finally, here a = 0 and b = 24 hour.

$\int^{24}_{0}te^{-0.1t}dt=-10e^{-t/10}t|^{24}_{0}-100e^{-t/10}|^{12}_{0}=69.16$

Therefore, the average intensity at the twenty-fourth hour will be:

$A=\frac{1}{24-0}\int^{24}_{0}te^{-0.1t}dt=\frac{69.16}{24}=2.88$

I hope it helps you!