Find the average value over the given interval.
\[
f(x)=x^{2}+x-8 ;[0,16]
\]
The average value of $f(x)=x^{2}+x-8$ on $[0,16]$ is (Type an integer or a simplified fraction.)
Answer
The average value of $f(x)=x^{2}+x-8$ on $[0,16]$ is \(\boxed{\frac{256}{3}}\).
Step 1 ：The average value of a function $f(x)$ over the interval $[a, b]$ is given by the formula: \[\frac{1}{b-a} \int_{a}^{b} f(x) dx\]
Step 2 ：In this case, $f(x) = x^{2} + x - 8$, $a = 0$, and $b = 16$. So we need to calculate the integral of $f(x)$ from $0$ to $16$, and then divide by $16 - 0 = 16$.
Step 3 ：Calculate the integral of $f(x)$ from $0$ to $16$.
Step 4 ：Divide the result by $16$ to get the average value.
Step 5 ：The average value of $f(x)=x^{2}+x-8$ on $[0,16]$ is \(\boxed{\frac{256}{3}}\).
