the daily production level for a product is given by n(x) = 240 - 240e^(-0.4x) units, where x is the time in…

Explanation:

Step1: Recall average - value formula

The average value of a function $y = N(x)$ over the interval $[a,b]$ is given by $\bar{N}=\frac{1}{b - a}\int_{a}^{b}N(x)dx$. Here, $a = 0$, $b = 3$, and $N(x)=240 - 240e^{-0.4x}$. So, $\bar{N}=\frac{1}{3-0}\int_{0}^{3}(240 - 240e^{-0.4x})dx=\frac{1}{3}\int_{0}^{3}(240 - 240e^{-0.4x})dx$.

Step2: Integrate term - by - term

We know that $\int(240 - 240e^{-0.4x})dx=\int240dx-\int240e^{-0.4x}dx$. For $\int240dx = 240x+C_1$. For $\int240e^{-0.4x}dx$, let $u=-0.4x$, then $du=-0.4dx$ and $dx=-\frac{1}{0.4}du = -\frac{5}{2}du$. So, $\int240e^{-0.4x}dx=240\int e^{-0.4x}dx=240\times(-\frac{5}{2})e^{-0.4x}+C_2=-600e^{-0.4x}+C_2$. Then $\int(240 - 240e^{-0.4x})dx=240x + 600e^{-0.4x}+C$.

Step3: Evaluate the definite integral

$\frac{1}{3}\int_{0}^{3}(240 - 240e^{-0.4x})dx=\frac{1}{3}\left[240x + 600e^{-0.4x}\right]_{0}^{3}$. First, substitute $x = 3$: $240\times3+600e^{-0.4\times3}=720 + 600e^{-1.2}$. Then substitute $x = 0$: $240\times0+600e^{-0.4\times0}=600$. So, $\frac{1}{3}\left[(720 + 600e^{-1.2})-600\right]=\frac{1}{3}(120 + 600e^{-1.2})$.

Step4: Calculate the value

$e^{-1.2}\approx0.3012$. $\frac{1}{3}(120 + 600\times0.3012)=\frac{1}{3}(120+180.72)=\frac{1}{3}\times300.72 = 100.24\approx100$.

Answer:

100