what is the average value of $6x^{2}+8$ on the interval $3,5$?

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 3$, $b = 5$, and $f(x)=6x^{2}+8$.

Step2: Calculate the integral $\int(6x^{2}+8)dx$

Using the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int(6x^{2}+8)dx=6\times\frac{x^{3}}{3}+8x + C=2x^{3}+8x + C$.

Step3: Evaluate the definite integral $\int_{3}^{5}(6x^{2}+8)dx$

[ \begin{align*} \int_{3}^{5}(6x^{2}+8)dx&=\left[2x^{3}+8x\right]_{3}^{5}\ &=(2\times5^{3}+8\times5)-(2\times3^{3}+8\times3)\ &=(2\times125 + 40)-(2\times27+24)\ &=(250 + 40)-(54 + 24)\ &=290 - 78\ &=212 \end{align*} ]

Step4: Calculate the average value

The average value is $\frac{1}{5 - 3}\int_{3}^{5}(6x^{2}+8)dx$. Since $\int_{3}^{5}(6x^{2}+8)dx = 212$ and $b - a=2$, the average value is $\frac{212}{2}=106$.

Answer:

106