What is the average value of $14-6 x^{2}$ on the interval $[-1,3]$ ?

Full solution

Q. What is the average value of

14-6 x^{2}

on the interval

[-1,3]

Calculate Interval Width: To find the average value of a function $f(x)$ on the interval $[a, b]$ , we use the formula: $\newline$ Average value = $\frac{1}{(b-a)} \cdot \int_{a}^{b} f(x) \, dx$ $\newline$ Here, $f(x) = 14 - 6x^2$ , $a = -1$ , and $b = 3$ .
Set Up Integral: First, calculate the width of the interval $[a, b]$ by subtracting $a$ from $b$ . $\newline$ Width = $b - a = 3 - (-1) = 3 + 1 = 4$
Calculate Integral: Now, set up the integral to find the average value. $\newline$ Average value = $(1/4) \times \int_{-1}^{3} (14 - 6x^2) \, dx$
Evaluate Integral Limits: Calculate the integral of the function $f(x) = 14 - 6x^2$ . $\int (14 - 6x^2) dx = 14x - \frac{6}{3}x^3 = 14x - 2x^3$
Subtract Integral Values: Evaluate the integral from $-1$ to $3$ . $\newline$ Plug in the upper limit of the integral: $\newline$ $14(3) - 2(3)^3 = 42 - 2(27) = 42 - 54 = -12$ $\newline$ Plug in the lower limit of the integral: $\newline$ $14(-1) - 2(-1)^3 = -14 - 2(-1) = -14 + 2 = -12$
Subtract Integral Values: Evaluate the integral from $-1$ to $3$ . $\newline$ Plug in the upper limit of the integral: $\newline$ $14(3) - 2(3)^3 = 42 - 2(27) = 42 - 54 = -12$ $\newline$ Plug in the lower limit of the integral: $\newline$ $14(-1) - 2(-1)^3 = -14 - 2(-1) = -14 + 2 = -12$ $\newline$ Now, subtract the value of the integral at the lower limit from the value at the upper limit. $\newline$ Integral from $-1$ to $3$ = $(-12) - (-12) = -12 + 12 = 0$