f(x)=40*e^(-0.5x)

2. Originally Posted by chevy900ss
f(x)=40*e^(-0.5x)
It should be $-80e^{\frac{-1}{2}x}+C$. When you u-substitute for -1/2x.

3. Originally Posted by chevy900ss
f(x)=40*e^(-0.5x)
Let $u=-0.5x$, then $du=-0.5dx\Rightarrow\int{40}e^{-0.5}dx=\int\frac{40}{-0.5}e^udu=-80\int{e}^udu=-80e^{-0.5}+C$

4. To determine how much of a drug is absorbed into the body, researchers measure the difference between the dosage H and the amount of the drug excreted from the body. The total amount excreted is found by integrating the excretion rate r(t) from 0 to infinity. Therefore, the amount of the drug absorbed by the body is
H Integral (r(t)dt) from 0 to infinity_
If the initial dose is H = 200 milligrams (mg), and the excretion rate is r(t)=40e^(-0.5t)mg per hour, find the amount of the drug absorbed by the body.

5. Originally Posted by chevy900ss
To determine how much of a drug is absorbed into the body, researchers measure the difference between the dosage H and the amount of the drug excreted from the body. The total amount excreted is found by integrating the excretion rate r(t) from 0 to infinity. Therefore, the amount of the drug absorbed by the body is
H Integral (r(t)dt) from 0 to infinity_
If the initial dose is H = 200 milligrams (mg), and the excretion rate is r(t)=40e^(-0.5t)mg per hour, find the amount of the drug absorbed by the body.