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what is the average value of 14−6x214-6x^2 on the interval [−1,3][-1,3]?

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Full solution

Q. what is the average value of 14−6x214-6x^2 on the interval [−1,3][-1,3]?
  1. Use Average Value Formula: To find the average value of a function f(x)f(x) on the interval [a,b][a, b], we use the formula:\newlineAverage value = 1(b−a)∫abf(x) dx\frac{1}{(b-a)} \int_{a}^{b} f(x) \, dx.\newlineHere, f(x)=14−6x2f(x) = 14 - 6x^2, a=−1a = -1, and b=3b = 3.
  2. Calculate Interval Width: Calculate the width of the interval [a,b][a, b] by subtracting aa from bb.\newlineWidth = b−a=3−(−1)=3+1=4b - a = 3 - (-1) = 3 + 1 = 4.
  3. Integrate Function: Now, we need to integrate the function f(x)=14−6x2f(x) = 14 - 6x^2 from −1-1 to 33.∫−13(14−6x2) dx\int_{-1}^{3} (14 - 6x^2) \, dx.
  4. Find Antiderivative: Find the antiderivative of f(x)f(x). The antiderivative of 1414 is 14x14x, and the antiderivative of −6x2-6x^2 is −6⋅(x33)=−2x3-6\cdot\left(\frac{x^3}{3}\right) = -2x^3. So, the antiderivative of f(x)f(x) is F(x)=14x−2x3F(x) = 14x - 2x^3.
  5. Evaluate Antiderivative: Evaluate the antiderivative F(x)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(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.F(-1) = 14(-1) - 2(-1)^3 = -14 - 2(-1) = -14 + 2 = -12.
  6. Subtract Limits: Subtract F(−1)F(-1) from F(3)F(3).F(3)−F(−1)=(−12)−(−12)=−12+12=0F(3) - F(-1) = (-12) - (-12) = -12 + 12 = 0.

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