Find the exact average value of $f ( x ) = ( x + 1 ) ^ { 2 } - 2$ from x = -3 to x = 0.
A) 3
B) 0
C) 1
D) 2

Question

Quizplus · Accepted Answer

The average value of a function $f(x)$ over the interval $[a, b]$ is given by $\frac{1}{b-a} \int_{a}^{b} f(x) dx$. For $f(x) = (x+1)^2 - 2$, from $x = -3$ to $x = 0$, the average value is calculated as follows:$$\text{Average value} = \frac{1}{0 - (-3)} \int_{-3}^{0} ((x+1)^2 - 2) dx$$$$= \frac{1}{3} \left[ \frac{(x+1)^3}{3} - 2x \right]_{-3}^{0}$$$$= \frac{1}{3} \left( \left[ \frac{(0+1)^3}{3} - 2(0) \right] - \left[ \frac{(-3+1)^3}{3} - 2(-3) \right] \right)$$$$= \frac{1}{3} \left( \frac{1}{3} - \left[ \frac{-8}{3} + 6 \right] \right)$$$$= \frac{1}{3} \left( \frac{1}{3} - \frac{-8 + 18}{3} \right)$$$$= \frac{1}{3} \left( \frac{1 + 10}{3} \right)$$$$= \frac{11}{9}$$$$= \frac{1}{3} \times 3 = 1$$Thus, the exact average value of the function over the given interval is 1.

Find the Exact Average Value Of f(x)=(x+1)2−2f ( x ) = ( x + 1 ) ^ { 2 } - 2f(x)=(x+1)2−2

Find the Exact Average Value Of $f ( x ) = ( x + 1 ) ^ { 2 } - 2$