for the function f(x)= -2x² - 6, do the following. (a) find the average value of f(x) on the interval 0, 2…

Answer:

$-10$

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $f_{avg}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 0$, $b = 2$, and $f(x)=-2x^{2}-6$. So, $f_{avg}=\frac{1}{2 - 0}\int_{0}^{2}(-2x^{2}-6)dx=\frac{1}{2}\int_{0}^{2}(-2x^{2}-6)dx$.

Step2: Integrate term - by - term

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

Step3: Evaluate the definite integral

$\frac{1}{2}\left[-\frac{2}{3}x^{3}-6x\right]_{0}^{2}=\frac{1}{2}\left[\left(-\frac{2}{3}(2)^{3}-6(2)\right)-\left(-\frac{2}{3}(0)^{3}-6(0)\right)\right]$. First, calculate $-\frac{2}{3}(2)^{3}-6(2)=-\frac{16}{3}-12=-\frac{16 + 36}{3}=-\frac{52}{3}$. Then, $\frac{1}{2}\times\left(-\frac{52}{3}\right)=-\frac{26}{3}- 4=-\frac{26+14}{3}=-10$.