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find the average value of the function ff over the indicated interval [a,b][a, b]: f(x)=4−x2f(x)=4-x^2, [−2,3][-2,3]

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Full solution

Q. find the average value of the function ff over the indicated interval [a,b][a, b]: f(x)=4−x2f(x)=4-x^2, [−2,3][-2,3]
  1. Calculate b−ab - a: To find the average value of a function over an interval [a,b][a, b], we use the formula: Average value = 1(b−a)×∫abf(x) dx\frac{1}{(b-a)} \times \int_{a}^{b} f(x) \, dx.
  2. Set up integral: First, let's calculate b−ab - a for our interval [−2,3][-2, 3].b−a=3−(−2)=3+2=5b - a = 3 - (-2) = 3 + 2 = 5.
  3. Find antiderivative: Now, we need to set up the integral of f(x)f(x) from −2-2 to 33.∫−23(4−x2) dx\int_{-2}^{3} (4 - x^2) \, dx.
  4. Evaluate antiderivative: Let's find the antiderivative of f(x)f(x).\newlineThe antiderivative of 44 is 4x4x, and the antiderivative of −x2-x^2 is −x33-\frac{x^3}{3}.\newlineSo, the antiderivative of f(x)f(x) is 4x−x334x - \frac{x^3}{3}.
  5. Subtract evaluations: Now we evaluate the antiderivative from −2-2 to 33.\newlinePlug in the upper limit: 4(3)−(3)3/3=12−9=34(3) - (3)^3/3 = 12 - 9 = 3.\newlinePlug in the lower limit: 4(−2)−(−2)3/3=−8−(−8/3)=−8+8/3=−24/3+8/3=−16/34(-2) - (-2)^3/3 = -8 - (-8/3) = -8 + 8/3 = -24/3 + 8/3 = -16/3.
  6. Divide for average value: Subtract the lower limit evaluation from the upper limit evaluation.\newline3−(−163)=3+163=93+163=2533 - (-\frac{16}{3}) = 3 + \frac{16}{3} = \frac{9}{3} + \frac{16}{3} = \frac{25}{3}.
  7. Divide for average value: Subtract the lower limit evaluation from the upper limit evaluation. 3−(−163)=3+163=93+163=2533 - (-\frac{16}{3}) = 3 + \frac{16}{3} = \frac{9}{3} + \frac{16}{3} = \frac{25}{3}.Finally, divide the result by (b−a)(b - a) to find the average value. Average value = (15)∗(253)=2515=53(\frac{1}{5}) * (\frac{25}{3}) = \frac{25}{15} = \frac{5}{3}.

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