what is the average value of 14 - 6x² on the interval -1,3?

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $\bar{y}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a=-1$, $b = 3$, and $f(x)=14-6x^{2}$. So, $\bar{y}=\frac{1}{3-(-1)}\int_{-1}^{3}(14 - 6x^{2})dx=\frac{1}{4}\int_{-1}^{3}(14 - 6x^{2})dx$.

Step2: Integrate term - by - term

We know that $\int(14 - 6x^{2})dx=\int 14dx-\int 6x^{2}dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$) and $\int kdx=kx + C$ ($k$ is a constant), we have $\int 14dx=14x$ and $\int 6x^{2}dx=6\times\frac{x^{3}}{3}=2x^{3}$. So, $\int(14 - 6x^{2})dx=14x-2x^{3}+C$.

Step3: Evaluate the definite integral

$\int_{-1}^{3}(14 - 6x^{2})dx=\left[14x-2x^{3}\right]{-1}^{3}=(14\times3-2\times3^{3})-(14\times(-1)-2\times(-1)^{3})$. First, calculate $(14\times3-2\times3^{3})=(42 - 54)=-12$. Second, calculate $(14\times(-1)-2\times(-1)^{3})=(-14 + 2)=-12$. Then, $\int{-1}^{3}(14 - 6x^{2})dx=-12-(-12)=-12 + 12 = 0$.

Step4: Calculate the average value

$\bar{y}=\frac{1}{4}\times0 = 0$.

Answer:

$0$