what is the average value of $x^{3}-9x$ on the interval $-1leq xleq3$?

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, $f(x)=x^{3}-9x$, $a=-1$, and $b = 3$.

Step2: Calculate the integral

First, find $\int(x^{3}-9x)dx=\frac{x^{4}}{4}-\frac{9x^{2}}{2}+C$. Then, $\int_{-1}^{3}(x^{3}-9x)dx=\left[\frac{x^{4}}{4}-\frac{9x^{2}}{2}\right]_{-1}^{3}=\left(\frac{3^{4}}{4}-\frac{9\times3^{2}}{2}\right)-\left(\frac{(-1)^{4}}{4}-\frac{9\times(-1)^{2}}{2}\right)$.

• Calculate $\frac{3^{4}}{4}-\frac{9\times3^{2}}{2}=\frac{81}{4}-\frac{81}{2}=\frac{81 - 162}{4}=-\frac{81}{4}$.
• Calculate $\frac{(-1)^{4}}{4}-\frac{9\times(-1)^{2}}{2}=\frac{1}{4}-\frac{9}{2}=\frac{1 - 18}{4}=-\frac{17}{4}$.
• Then $\int_{-1}^{3}(x^{3}-9x)dx=-\frac{81}{4}-\left(-\frac{17}{4}\right)=-\frac{81 - 17}{4}=-\frac{64}{4}=-16$.

Step3: Calculate the average - value

Since $b - a=3-(-1)=4$, the average value $\bar{y}=\frac{1}{4}\times(-16)=-4$.

Answer:

$-4$