Question

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

R(t)=te-0.1t,
where t is time (in hours) after the drug is administered. Find the average intensity during the following hours.
a. second hour
b. Tewlth hour
c. Twenty-fourth hour.

Answer 1

a) A=0.86

b) A=2.81

c) A=2.88

Explanation:

The average of a functions can be written as:

`A=(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=(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=(1)/(2-0)\int^(2)_(0)te^(-0.1t)dt=(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=(1)/(12-0)\int^(12)_(0)te^(-0.1t)dt=(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=(1)/(24-0)\int^(24)_(0)te^(-0.1t)dt=(69.16)/(24)=2.88`

I hope it helps you!