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18 September, 00:03

How to find average value of a function over a given interval?

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  1. 18 September, 03:50
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    f (x) = 8x-6 f (x) = 8x-6, [0,3] [0,3]

    The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined. (-∞,∞) (-∞,∞) x∈R x∈ℝ f (x) f (x) is continuous on [0,3] [0,3]. f (x) f (x) is continuousThe average value of function ff over the interval [a, b] [a, b] is defined as A (x) = 1 b-a ∫ ba f (x) dx A (x) = 1 b-a ∫ab f (x) dx. A (x) = 1 b-a ∫ ba f (x) dx A (x) = 1 b-a ∫ab f (x) dx Substitute the actual values into the formula for the average value of a function. A (x) = 1 3-0 (∫ 30 8x-6dx) A (x) = 1 3-0 (∫03 8x-6dx) Since integration is linear, the integral of 8x-6 8x-6 with respect to xx is ∫ 30 8xdx + ∫ 30 - 6dx ∫03 8xdx + ∫03 - 6dx. A (x) = 1 3-0 (∫ 30 8xdx + ∫ 30 - 6dx) A (x) = 1 3-0 (∫03 8xdx + ∫03 - 6dx) Since 88 is constant with respect to xx, the integral of 8x 8x with respect to xx is 8 ∫ 30 xdx 8 ∫03 xdx. A (x) = 1 3-0 (8 ∫ 30 xdx + ∫ 30 - 6dx) A (x) = 1 3-0 (8 ∫03 xdx + ∫03 - 6dx) By the Power Rule, the integral of xx with respect to xx is 12 x2 12 x2. A (x) = 1 3-0 (8 (12 x2 ] 30) + ∫ 30 - 6dx)
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