26. spread of drug use\nin a study of the spread of illicit drug use from an enthusiastic user to a…

26. spread of drug use\nin a study of the spread of illicit drug use from an enthusiastic user to a population of ( n ) users, the authors model the number of expected new users by the equation\n gamma=int_{0}^{infty} \frac{c nleft(1-e^{-k t}\right)}{k} e^{-lambda t} d t \nwhere ( c, k ), and ( lambda ) are positive constants. evaluate this integral to express ( gamma ) in terms of ( c, n, k ), and ( lambda ).

26. spread of drug use\nin a study of the spread of illicit drug use from an enthusiastic user to a population of ( n ) users, the authors model the number of expected new users by the equation\n gamma=int_{0}^{infty} \frac{c nleft(1-e^{-k t}\right)}{k} e^{-lambda t} d t \nwhere ( c, k ), and ( lambda ) are positive constants. evaluate this integral to express ( gamma ) in terms of ( c, n, k ), and ( lambda ).

Answer

Explanation:

Step1: Expand the integrand

$$\gamma=\frac{cN}{k}\int_{0}^{\infty}(e^{-\lambda t}-e^{-(k + \lambda)t})dt$$

Step2: Integrate term - by - term

For the first integral $\int_{0}^{\infty}e^{-\lambda t}dt$, using the formula $\int_{0}^{\infty}e^{-at}dt=\frac{1}{a}$ ($a>0$), when $a = \lambda$, we have $\int_{0}^{\infty}e^{-\lambda t}dt=\frac{1}{\lambda}$. For the second integral $\int_{0}^{\infty}e^{-(k+\lambda)t}dt$, when $a=k + \lambda$, we have $\int_{0}^{\infty}e^{-(k+\lambda)t}dt=\frac{1}{k+\lambda}$.

Step3: Calculate the value of $\gamma$

$$\gamma=\frac{cN}{k}\left(\frac{1}{\lambda}-\frac{1}{k + \lambda}\right)$$ Simplify the expression: $$\gamma=\frac{cN}{k}\cdot\frac{k+\lambda-\lambda}{\lambda(k + \lambda)}=\frac{cN}{\lambda(k + \lambda)}$$

Answer:

$\gamma=\frac{cN}{\lambda(k + \lambda)}$