What is the average value of $8-2 x^{3}$ on the interval $[2,4]$ ?

Full solution

Q. What is the average value of

8-2 x^{3}

on the interval

[2,4]

Set up integral: To find the average value of the function $8 - 2x^3$ on the interval $[2,4]$ , we need to integrate the function over the interval and then divide by the length of the interval.
Calculate integral: First, let's set up the integral for the function over the interval $[2,4]$ . The average value formula is given by: $\newline$ Average value = $\frac{1}{(b-a)} \cdot \int_{a}^{b} f(x) \, dx$ $\newline$ Here, $a = 2$ , $b = 4$ , and $f(x) = 8 - 2x^3$ .
Evaluate antiderivative: Now, we calculate the integral of $f(x)$ from $2$ to $4$ . $\int_{2}^{4} (8 - 2x^3) \,dx$ $= [8x - (\frac{1}{2}) \cdot 2x^4]_{2}^{4}$ $= [8x - x^4]_{2}^{4}$
Divide by interval length: Next, we evaluate the antiderivative at the upper and lower limits of the interval. $\newline$ $= (8\times4 - 4^4) - (8\times2 - 2^4)$ $\newline$ $= (32 - 256) - (16 - 16)$ $\newline$ $= -224 - 0$ $\newline$ $= -224$
Divide by interval length: Next, we evaluate the antiderivative at the upper and lower limits of the interval. $\newline$ $= (8\cdot4 - 4^4) - (8\cdot2 - 2^4)$ $\newline$ $= (32 - 256) - (16 - 16)$ $\newline$ $= -224 - 0$ $\newline$ $= -224$ Now, we divide the result of the integral by the length of the interval, which is $b - a = 4 - 2 = 2$ . $\newline$ Average value $= (-224) / (4 - 2)$ $\newline$ Average value $= (-224) / 2$ $\newline$ Average value $= -112$