Answer
$\color{blue}{\$50,\!000}$
Work Step by Step
\begin{align*}
E(Q(Y)) &= E(\underbrace{2(1-e^{-2Y}}_{\times 10^5})) \\
&= \int_{\mathbb{R}} \underbrace{2(1-e^{-2y})}_{\times 10^5}\cdot f_Y(y)\ dy \\
&= \int_0^\infty \underbrace{2(1-e^{-2y})}_{\times 10^5}\cdot(6e^{-6y})\ dy \qquad \left[\ \text{since}\ f_Y(y)=6e^{-6y},\ y \gt 0\ \right] \\
&= (12\times 10^5) \int_0^\infty (e^{-6y} - e^{-8y})\ dy \\
&= (12\times 10^5)\left( \frac{e^{-6y}}{-6} - \frac{e^{-8y}}{-8}\right)_0^\infty \\
&= (12\times 10^5)\left[\left( \frac{e^{-\infty}}{-6} - \frac{e^{-\infty}}{-8}\right) - \left( \frac{e^{-0}}{-6} - \frac{e^{-0}}{-8}\right)\right] \\
&= (12\times 10^5)\left[\left( \frac{0}{-6} - \frac{0}{-8}\right) - \left( \frac{1}{-6} - \frac{1}{-8}\right)\right] \\
&= (12\times 10^5)\left[\left(0 - 0\right) + \frac{1}{6} - \frac{1}{8}\right] \\
&= (12\times 10^5)\left(\frac{8-6}{48}\right) \\
&= (10^5)\left(\frac{24}{48}\right) \\
&= \frac{10^5}{2} \\
\color{blue}{E(Q(Y))}\ &\color{blue}{=50,\!000}
\end{align*}
The company's expected profit is $\color{blue}{\$50,\!000}$.
