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 $f(x)=6x^{2}+8$. So we need to calculate $\frac{1}{5 - 3}\int_{3}^{5}(6x^{2}+8)dx=\frac{1}{2}\int_{3}^{5}(6x^{2}+8)dx$.

Step2: Integrate term - by - term

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

$\frac{1}{2}\left[(2x^{3}+8x)\big|_{3}^{5}\right]=\frac{1}{2}\left[(2\times5^{3}+8\times5)-(2\times3^{3}+8\times3)\right]$. First, calculate $2\times5^{3}+8\times5=2\times125 + 40=250+40 = 290$. Second, calculate $2\times3^{3}+8\times3=2\times27+24 = 54 + 24=78$. Then, $\frac{1}{2}(290 - 78)=\frac{1}{2}\times212 = 106$.

Answer:

$106$