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$Determine the average value of the piecewise function f defined below on the interval x=-2 to x=1. Express your answer in simplest form. f(x)={[3x^(2)-8," for ",x <= -1],[-3x^(2)," for ",x > -1]:}$

Determine the average value of the piecewise function $f$ defined below on the interval $x=-2$ to $x=1$ . Express your answer in simplest form. $\newline$ $f(x)=\left\{\begin{array}{lll} 3 x^{2}-8 & \text { for } & x \leq-1 \\ -3 x^{2} & \text { for } & x>-1 \end{array}\right.$

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Calculus

Intermediate Value Theorem

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Q. Determine the average value of the piecewise function

f

defined below on the interval

x=-2

x=1

. Express your answer in simplest form.

\newline

f(x)=\left\{\begin{array}{lll} 3 x^{2}-8 & \text { for } & x \leq-1 \\ -3 x^{2} & \text { for } & x>-1 \end{array}\right.

Split Interval: To find the average value of the function on the interval from $x = -2$ to $x = 1$ , we need to integrate the function over this interval and then divide by the length of the interval.
First Interval Integration: First, we need to split the interval at the point where the function changes, which is at $x = -1$ . This gives us two intervals: $[-2, -1]$ and $(-1, 1]$ .
Second Interval Integration: On the first interval $[-2, -1]$ , the function is defined as $f(x) = 3x^2 - 8$ . We need to integrate this function from $-2$ to $-1$ .
Total Area Calculation: The integral of $3x^2 - 8$ from $-2$ to $-1$ is calculated as follows: $\newline$ $\int_{-2}^{-1} (3x^2 - 8) \, dx = [x^3 - 8x]_{-2}^{-1} = ((-1)^3 - 8(-1)) - ((-2)^3 - 8(-2)) = (-1 + 8) - (-8 + 16) = 7 - 8 = -1.$
Interval Length Calculation: On the second interval $(-1, 1]$ , the function is defined as $f(x) = -3x^2$ . We need to integrate this function from $-1$ to $1$ .
Average Value Calculation: The integral of $-3x^2$ from $-1$ to $1$ is calculated as follows: $\newline$ $\int_{-1}^{1}(-3x^2) dx = [-x^3]_{-1}^{1} = (1^3) - ((-1)^3) = 1 - (-1) = 2.$
Average Value Calculation: The integral of $-3x^2$ from $-1$ to $1$ is calculated as follows: $\newline$ $\int_{-1}^{1}(-3x^2) dx = [-x^3]_{-1}^{1} = (1^3) - ((-1)^3) = 1 - (-1) = 2.$ Now, we add the results of the integrals from both intervals to get the total area under the curve from $x = -2$ to $x = 1$ . $\newline$ Total area = $-1 + 2 = 1$ .
Average Value Calculation: The integral of $-3x^2$ from $-1$ to $1$ is calculated as follows: $\newline$ $\int_{-1}^{1}(-3x^2) dx = [-x^3]_{-1}^{1} = (1^3) - ((-1)^3) = 1 - (-1) = 2.$ Now, we add the results of the integrals from both intervals to get the total area under the curve from $x = -2$ to $x = 1$ . $\newline$ Total area = $-1 + 2 = 1$ .The length of the interval from $x = -2$ to $x = 1$ is $1 - (-2) = 3$ .
Average Value Calculation: The integral of $-3x^2$ from $-1$ to $1$ is calculated as follows: $\newline$ $\int_{-1}^{1}(-3x^2) dx = [-x^3]_{-1}^{1} = (1^3) - ((-1)^3) = 1 - (-1) = 2.$ Now, we add the results of the integrals from both intervals to get the total area under the curve from $x = -2$ to $x = 1$ . $\newline$ Total area = $-1 + 2 = 1$ .The length of the interval from $x = -2$ to $x = 1$ is $1 - (-2) = 3$ .Finally, we divide the total area by the length of the interval to find the average value of the function. $\newline$ Average value = Total area / Length of interval = $1 / 3$ .