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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 f defined below on the interval x=2 x=-2 to x=1 x=1 . Express your answer in simplest form.\newlinef(x)={3x28amp; for amp;x13x2amp; for amp;xgt;1 f(x)=\left\{\begin{array}{lll} 3 x^{2}-8 &amp; \text { for } &amp; x \leq-1 \\ -3 x^{2} &amp; \text { for } &amp; x&gt;-1 \end{array}\right.

Full solution

Q. Determine the average value of the piecewise function f f defined below on the interval x=2 x=-2 to x=1 x=1 . Express your answer in simplest form.\newlinef(x)={3x28 for x13x2 for x>1 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.
  1. Split Interval: To find the average value of the function on the interval from x=2x = -2 to x=1x = 1, we need to integrate the function over this interval and then divide by the length of the interval.
  2. First Interval Integration: First, we need to split the interval at the point where the function changes, which is at x=1x = -1. This gives us two intervals: [2,1][-2, -1] and (1,1](-1, 1].
  3. Second Interval Integration: On the first interval [2,1][-2, -1], the function is defined as f(x)=3x28f(x) = 3x^2 - 8. We need to integrate this function from 2-2 to 1-1.
  4. Total Area Calculation: The integral of 3x283x^2 - 8 from 2-2 to 1-1 is calculated as follows:\newline21(3x28)dx=[x38x]21=((1)38(1))((2)38(2))=(1+8)(8+16)=78=1.\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.
  5. Interval Length Calculation: On the second interval (1,1] (-1, 1] , the function is defined as f(x)=3x2 f(x) = -3x^2 . We need to integrate this function from 1 -1 to 1 1 .
  6. Average Value Calculation: The integral of 3x2-3x^2 from 1-1 to 11 is calculated as follows:\newline11(3x2)dx=[x3]11=(13)((1)3)=1(1)=2.\int_{-1}^{1}(-3x^2) dx = [-x^3]_{-1}^{1} = (1^3) - ((-1)^3) = 1 - (-1) = 2.
  7. Average Value Calculation: The integral of 3x2-3x^2 from 1-1 to 11 is calculated as follows:\newline11(3x2)dx=[x3]11=(13)((1)3)=1(1)=2.\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=2x = -2 to x=1x = 1.\newlineTotal area = 1+2=1-1 + 2 = 1.
  8. Average Value Calculation: The integral of 3x2-3x^2 from 1-1 to 11 is calculated as follows:\newline11(3x2)dx=[x3]11=(13)((1)3)=1(1)=2.\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=2x = -2 to x=1x = 1.\newlineTotal area = 1+2=1-1 + 2 = 1.The length of the interval from x=2x = -2 to x=1x = 1 is 1(2)=31 - (-2) = 3.
  9. Average Value Calculation: The integral of 3x2-3x^2 from 1-1 to 11 is calculated as follows:\newline11(3x2)dx=[x3]11=(13)((1)3)=1(1)=2.\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=2x = -2 to x=1x = 1.\newlineTotal area = 1+2=1-1 + 2 = 1.The length of the interval from x=2x = -2 to x=1x = 1 is 1(2)=31 - (-2) = 3.Finally, we divide the total area by the length of the interval to find the average value of the function.\newlineAverage value = Total area / Length of interval = 1/31 / 3.

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