Question

Please check my steps:

Answer 1

\[
\begin{align*}
\int_0^\infty tf(t)dt &= \lambda\int_0^\infty te^{-\lambda t}dt\\
&=\lambda\left[\left(-\frac{1}{\lambda}te^{-\lambda t}\right)_0^\infty-\int_{0}^{\infty}-\frac{1}{\lambda} e^{-\lambda t}dt\right]\\
&=\lambda\left[-\frac{1}{\lambda}\left(te^{-\lambda t}\right)_0^\infty + \frac{1}{\lambda}\int_{0}^{\infty}e^{-\lambda t}dt \right]\\
&=\lambda\left[-\frac{1}{\lambda}\left(\lim_{t\to\infty}\left(\frac{t}{e^{\lambda t}}\right)-0\right)-\frac{1}{\lambda^2}\left(e^{-\lambda t}\right)_0^\infty\right]\\
&=-\lim_{t\to\infty}\left(\frac{t}{e^{\lambda t}}\right)-\frac{1}{\lambda}\left(\lim_{t\to\infty}\left(e^{-\lambda t}\right)-1\right)\\
&=-\lim_{t\to\infty}\left(-\frac{1}{\lambda e^{-\lambda t}}\right)+\frac{1}{\lambda}\\
&=\frac{1}{\lambda}
\end{align*}
\]

Answer 2

in 2nd last step,
that \(\Large e^{-\lambda t }\)
should be in numerator.
rest all steps are correct :)

Answer 3

Ah! The second last step is wrong! Is this correct though?
\[
\begin{align*}
-\lim_{t\to\infty}\left(\frac{t}{e^{\lambda t}}\right)-\frac{1}{\lambda}\left(\lim_{t\to\infty}\left(e^{-\lambda t}\right)-1\right)&=-\lim_{t\to\infty}\frac{1}{\lambda e^{\lambda t}}+\frac{1}{\lambda}\\
&=\frac{1}{\lambda}
\end{align*}
\]

Answer 4

oh, you used L'Hopital's rule, right ??

then there is only sign error in 2nd last step,
should be
\(λe^{λt}\) in the denominator

Answer 5

yep, everything elase is correct :)

Answer 6

One last check please:
\[
\begin{align*}
\int_0^\infty tf(t)dt &= \lambda\int_0^\infty te^{-\lambda t}dt\\
&=\lambda\left[\left(-\frac{1}{\lambda}te^{-\lambda t}\right)_0^\infty-\int_{0}^{\infty}-\frac{1}{\lambda} e^{-\lambda t}dt\right]\\
&=\lambda\left[-\frac{1}{\lambda}\left(te^{-\lambda t}\right)_0^\infty + \frac{1}{\lambda}\int_{0}^{\infty}e^{-\lambda t}dt \right]\\
&=\lambda\left[-\frac{1}{\lambda}\left(\lim_{t\to\infty}\left(\frac{t}{e^{\lambda t}}\right)-0\right)-\frac{1}{\lambda^2}\left(e^{-\lambda t}\right)_0^\infty\right]\\
&=-\lim_{t\to\infty}\left(\frac{t}{e^{\lambda t}}\right)-\frac{1}{\lambda}\left(\lim_{t\to\infty}\left(e^{-\lambda t}\right)-1\right)\\
&=-\lim_{t\to\infty}\left(\frac{1}{\lambda e^{\lambda t}}\right)+\frac{1}{\lambda} &\text{L'Hôpital's rule}\\
&=\frac{1}{\lambda}
\end{align*}
\]

Answer 7

yep. all are correct, good work! :)