current objective find the average value of a function over an interval question find the average value of…

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ over the interval $[a,b]$ is given by $\bar{y}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a=-3$, $b = 2$, and $f(x)=-5x + 2$.

Step2: Calculate the integral

First, find $\int_{-3}^{2}(-5x + 2)dx$. Using the power - rule of integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(-5x + 2)dx=-\frac{5}{2}x^{2}+2x+C$. Then, $\int_{-3}^{2}(-5x + 2)dx=\left(-\frac{5}{2}x^{2}+2x\right)\big|_{-3}^{2}=\left(-\frac{5}{2}(2)^{2}+2(2)\right)-\left(-\frac{5}{2}(-3)^{2}+2(-3)\right)$. [ \begin{align*} &=\left(-\frac{5}{2}\times4 + 4\right)-\left(-\frac{5}{2}\times9-6\right)\ &=(-10 + 4)-\left(-\frac{45}{2}-6\right)\ &=-6-\left(-\frac{45 + 12}{2}\right)\ &=-6+\frac{57}{2}\ &=\frac{-12 + 57}{2}=\frac{45}{2} \end{align*} ]

Step3: Calculate the average value

Now, $b - a=2-(-3)=5$. The average value $\bar{y}=\frac{1}{b - a}\int_{a}^{b}f(x)dx=\frac{1}{5}\times\frac{45}{2}=\frac{9}{2}$.

Answer:

$\frac{9}{2}$