Find the average value of the function $f(x) = x^2 - 19$ on $[0,9]$.
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The definite integral of f(x) on the interval [0,9] is given by: ∫[0,9] (x^2 - 19) dx = [1/3 * x^3 - 19x] evaluated from 0 to 9 = [1/3 * 9^3 - 19*9] - [1/3 * 0^3 - 19*0] = [243/3 - 171] - [0 - 0] = 81 - 171 = -90 Show more…
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