An amusement company maintains records for each video game it installs in an arcade. Suppose that $\mathrm{C}(\mathrm{t})$ and $\mathrm{R}(\mathrm{t})$ represent the total accumulated costs and revenues (in thousands of dollars), respectively, $\mathrm{t}$ years after a particular game has been installed and the derivatives $C^{\prime}(t)$ and $R^{\prime}(t)$ are the following functions. \[ C^{\prime}(t)=2 \quad \text { and } \quad R^{\prime}(t)=8 e^{-0.3 t} \text {. } \] Find the area between the graphs of $C^{\prime}$ and $R^{\prime}$ over the interval on the $t$-axis from 0 to the useful life of the game and interpret the results. What is the useful life of the game? $t=\square$ years (Round to the nearest tenth as needed.)
Solution
Step 1 ：First, we need to find the functions \(R(t)\) and \(C(t)\) by integrating \(R'(t)\) and \(C'(t)\).
Step 2 ：The integral of \(C'(t) = 2\) is \(C(t) = 2t + C_1\), where \(C_1\) is the constant of integration. Since the cost at time 0 is usually 0, we can assume \(C_1 = 0\), so \(C(t) = 2t\).
Step 3 ：The integral of \(R'(t) = 8e^{-0.3t}\) is \(R(t) = -\frac{8}{0.3}e^{-0.3t} + R_1\), where \(R_1\) is the constant of integration. Since the revenue at time 0 is usually 0, we can assume \(R_1 = 0\), so \(R(t) = -\frac{8}{0.3}e^{-0.3t}\).
Step 4 ：Setting \(R(t) = C(t)\), we get \(-\frac{8}{0.3}e^{-0.3t} = 2t\).
Step 5 ：Solving this equation for \(t\) gives us the useful life of the game. This is a transcendental equation and cannot be solved exactly, but it can be solved numerically.
Step 6 ：Using a calculator or a computer, we find that the solution is approximately \(t \approx 7.6\) years.
Step 7 ：\(\boxed{\text{So, the useful life of the game is approximately 7.6 years.}}\)
