Exercise 8.4.11

Question

\begin{enumerate}[label=(\alph*)] 
\item Evaluate $\int_0^b t e^{-\alpha t}$ using the integration-by-parts formula from Exercise 7.5.6. The result will be an expression in $\alpha$ and $b$.
\item Now compute $\int_0^\infty t e^{-\alpha t} dt $ and verify equation (4).
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item
$$\int_0^b te^{-\alpha t} dt = -\frac{b e^{-\alpha b}}{\alpha} + \frac{1}{\alpha} \int_0^b e^{-\alpha t} =-\frac{b e^{-\alpha b}}{\alpha} + \frac{1}{\alpha^2} (1 - e^{-\alpha b})$$
\item $\lim_{b \to \infty} -\frac{b e^{-\alpha b}}{\alpha} + \frac{1}{\alpha^2} (1 - e^{-\alpha b}) = \frac{1}{\alpha^2}$
 \end{enumerate}

Exercise 8.4.11

Answers

Comments

Exercise 8.4.11

Answers

Comments

Add answer