What is the average value of 6x^(2)+8 on the interval [3,5] ?

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What is the average value of  6x^(2)+8 on the interval  [3,5] ?

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Calculate Interval Width: To find the average value of a function f(x) on the interval [a, b], we use the formula: Average value = (1/(b-a)) * ∫ from a to b of f(x) dx Here, f(x) = 6x^2 + 8, a = 3, and b = 5.Set Up Integral: First, we calculate b - a, which is the width of the interval. b - a = 5 - 3 = 2Find Antiderivative: Next, we set up the integral of the function f(x) from 3 to 5. ∫ from 3 to 5 of (6x^2 + 8) dxEvaluate Antiderivative: We need to find the antiderivative of f(x). The antiderivative of 6x^2 is 6*(x^3/3) and the antiderivative of 8 is 8x. Antiderivative of f(x) = 6*(x^3/3) + 8x Simplify the antiderivative: 2x^3 + 8xSubtract Values: Now we evaluate the antiderivative at the upper and lower limits of the interval and subtract. Plug in x = 5: 2*(5)^3 + 8*(5) = 2*125 + 40 = 250 + 40 = 290 Plug in x = 3: 2*(3)^3 + 8*(3) = 2*27 + 24 = 54 + 24 = 78Divide for Average: Subtract the value of the antiderivative at x = 3 from the value at x = 5. 290 - 78 = 212Finally, we divide the result by the width of the interval to find the average value. Average value = (1/2) * 212 = 106

What is the average value of $6 x^{2}+8$ on the interval $[3,5]$ ?

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What is the average value of $6 x^{2}+8$ on the interval $[3,5]$ ?