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

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Full solution

Q. what is the average value of

14-6x^2

on the interval

[-1,3]

?

Use Average Value Formula: To find the average value of a function $f(x)$ on the interval $[a, b]$ , we use the formula: $\newline$ Average value = $\frac{1}{(b-a)} \int_{a}^{b} f(x) \, dx$ . $\newline$ Here, $f(x) = 14 - 6x^2$ , $a = -1$ , and $b = 3$ .
Calculate Interval Width: Calculate the width of the interval $[a, b]$ by subtracting $a$ from $b$ . $\newline$ Width = $b - a = 3 - (-1) = 3 + 1 = 4$ .
Integrate Function: Now, we need to integrate the function $f(x) = 14 - 6x^2$ from $-1$ to $3$ . $\int_{-1}^{3} (14 - 6x^2) \, dx$ .
Find Antiderivative: Find the antiderivative of $f(x)$ . The antiderivative of $14$ is $14x$ , and the antiderivative of $-6x^2$ is $-6\cdot\left(\frac{x^3}{3}\right) = -2x^3$ . So, the antiderivative of $f(x)$ is $F(x) = 14x - 2x^3$ .
Evaluate Antiderivative: Evaluate the antiderivative $F(x)$ at the upper and lower limits of the interval and subtract. $F(3) = 14(3) - 2(3)^3 = 42 - 2(27) = 42 - 54 = -12.$ $F(-1) = 14(-1) - 2(-1)^3 = -14 - 2(-1) = -14 + 2 = -12.$
Subtract Limits: Subtract $F(-1)$ from $F(3)$ . $F(3) - F(-1) = (-12) - (-12) = -12 + 12 = 0$ .

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