What is the average value of -x^(2)+3 on the interval [0,6] ?

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What is the average value of  -x^(2)+3 on the interval  [0,6] ?

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Set up integral: To find the average value of the function -x^2 + 3 on the interval [0,6], we need to integrate the function over the interval and then divide by the length of the interval.Find antiderivative: First, let's set up the integral for the function over the interval [0,6]. The integral of -x^2 + 3 from 0 to 6 is ∫(-x^2 + 3)dx from 0 to 6.Evaluate at limits: Now, we need to find the antiderivative of -x^2 + 3. The antiderivative of -x^2 is -x^3/3 and the antiderivative of 3 is 3x. So, the antiderivative of -x^2 + 3 is -x^3/3 + 3x.Calculate values: Next, we evaluate the antiderivative at the upper and lower limits of the interval and subtract. This gives us (-6^3/3 + 3*6) - (0^3/3 + 3*0).Simplify expression: Calculating the values, we get (-216/3 + 18) - (0 + 0). This simplifies to (-72 + 18) - 0.Find interval length: Now, we simplify the expression to get the definite integral value. -72 + 18 equals -54. So, the definite integral of the function over the interval [0,6] is -54.Divide for average value: The length of the interval [0,6] is 6 - 0, which is 6.Finally, to find the average value, we divide the definite integral value by the length of the interval. The average value is -54 / 6.Calculating the division, we get -54 / 6 equals -9. So, the average value of the function -x^2 + 3 on the interval [0,6] is -9.

What is the average value of $-x^{2}+3$ on the interval $[0,6]$ ?

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What is the average value of $-x^{2}+3$ on the interval $[0,6]$ ?