1. Define the PDF
The time $X$ follows an exponential distribution with a mean of 0.5 .
- The mean of an exponential distribution is $\frac{1}{\lambda}$.
- So, $\frac{1}{\lambda}=0.5$, which means the rate $\lambda=2$.
- The probability density function (PDF) for $X$ is $f(x)=\lambda e^{-\lambda x}=\mathbf{2 e}^{-2 \mathrm{x}}$ for $x \geq 0$.
2. Set Up the Expectation Integral
We need to find $E[R]$, which is $\int_0^{\infty} R(x) f(x) d x$. Since $R(x)$ is a piecewise function, we must split the integral at $x=1$.
$$
\begin{gathered}
E[R]=\int_0^1 R(x) f(x) d x+\int_1^{\infty} R(x) f(x) d x \\
E[R]=\int_0^1(10)\left(2 e^{-2 x}\right) d x+\int_1^{\infty}\left(10 x^2\right)\left(2 e^{-2 x}\right) d x \\
E[R]=20 \int_0^1 e^{-2 x} d x+20 \int_1^{\infty} x^2 e^{-2 x} d x
\end{gathered}
$$
3. Solve Part 1 (from 0 to 1 )
$$
\begin{gathered}
20 \int_0^1 e^{-2 x} d x=20\left[\frac{e^{-2 x}}{-2}\right]_0^1 \\
\quad=-10\left[e^{-2 x}\right]_0^1 \\
\quad=-10\left(e^{-2}-e^0\right) \\
\quad=-10\left(e^{-2}-1\right)
\end{gathered}
$$
Part 1 Result: $10\left(1-e^{-2}\right)=10-10 e^{-2}$
4. Solve Part 2 (from 1 to $\infty$ )
This requires integration by parts twice. We'll use the tabular method for $\int x^2 e^{-2 x} d x$ :
\[
\begin{array}{lll}
D\ (\text{Differentiate}) & I\ (\text{Integrate}) & \text{Sign} \\
x^{2} & e^{-2x} & + \\
2x & \dfrac{e^{-2x}}{-2} & - \\
2 & \dfrac{e^{-2x}}{4} & + \\
0 & \dfrac{e^{-2x}}{-8} & -
\end{array}
\]
The integral is: $20\left[-\frac{x^2 e^{-2 x}}{2}-\frac{2 x e^{-2 x}}{4}-\frac{2 e^{-2 x}}{8}\right]_1^{\infty} 20\left[-e^{-2 x}\left(\frac{x^2}{2}+\frac{x}{2}+\frac{1}{4}\right)\right]_1^{\infty}$
- At $x=\infty$ : The $e^{-2 x}$ term goes to 0 , so the entire upper limit is 0.
- At $x=1: 0-\left(20\left[-e^{-2(1)}\left(\frac{1^2}{2}+\frac{1}{2}+\frac{1}{4}\right)\right]\right)=20\left[e^{-2}\left(1+\frac{1}{4}\right)\right]=20\left[e^{-2}\left(\frac{5}{4}\right)\right]= \left(20 \cdot \frac{5}{4}\right) e^{-2}$
Part 2 Result: $25 e^{-2}$
$E[R]=($ Part 1 Result $)+($ Part 2 Result $)$
$$
\begin{gathered}
E[R]=\left(10-10 e^{-2}\right)+\left(25 e^{-2}\right) \\
E[R]=10+15 e^{-2}
\end{gathered}
$$
