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Lesson 6-5

Lesson 6-5. Average Value of a Function. x = 2. ∫ x² √x³ + 1 dx. x = 0. Icebreaker. Evaluate Ponder: If the slope of the secant represents the average rate of change over an interval, what corresponds, graphically, to the “Average Value” of the function over the interval?. Objectives.

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Lesson 6-5

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  1. Lesson 6-5 Average Value of a Function

  2. x = 2 ∫ x²√x³ + 1 dx x = 0 Icebreaker • Evaluate • Ponder: If the slope of the secant represents the average rate of change over an interval, what corresponds, graphically, to the “Average Value” of the function over the interval?

  3. Objectives • Find the average value of a function

  4. Vocabulary • Average value – average height (in the rectangle) associated with the area under the curve between two points

  5. x = b 1/(b-a)∫ f(x) dx = f(c) for a ≤ c ≤ b x = a Average Value Average value of a function, f(x), on [a,b] is The integral represents the area under the curve and f(c) represents the height of rectangle that is equal to the area under the rectangle. Formerly called the Mean Value Theorem for Integrals f(x) c b a

  6. x = b average value of f(x) = 1/(b-a)∫ f(x) dx = f(c) for a ≤ c ≤ b x = a x = 4 x = 4 x = 4 ∫ f(x) dx = ∫ (3x² - 2x) dx = x³ - x² | = (64 – 16) – (1 – 1) = 48 x = 1 x = 1 x = 1 6-5 Example 1 f(x) = 3x² – 2x on [1, 4] 1 1 ------ = ------ = ⅓ b – a 4 - 1 average value of f(x) = ⅓ • 48 = 16

  7. x = π x = π x = π ∫ f(x) dx = ∫ (sin x) dx = - cos x | = - (-1 – 1) = 2 x = 0 x = 0 x = 0 x = b average value of f(x) = 1/(b-a)∫ f(x) dx = f(c) for a ≤ c ≤ b x = a 2/π² π 0.69 2.45 6-5 Example 2 f(x) = sin x on [0, π] 1 1 ------ = ------ = 1/π b – a π - 0 average value of f(x) = 1/π • (2) = 2/π

  8. x = 4 x = 4 x = 4 ∫ f(x) dx = ∫ x² dx = ⅓ x³ | = ⅓ (64 – 1) = 21 x = 1 x = 1 x = 1 x = b average value of f(x) = 1/(b-a)∫ f(x) dx = f(c) for a ≤ c ≤ b x = a 6-5 Example 3 f(x) = x² on [1, 4] 1 1 ------ = ------ = ⅓ b – a 4 - 1 average value of f(x) = ⅓ • (21) = 7 f(c) = c² = 7 c = √7

  9. 6-5 Example 4 let u = x² + 2 du = 2x dx form: du/u x = √6 x = k x = √6 x = k x f(x) dx = --------- dx x² + 2 = ½ ln | x² + 2| = ½ (ln 8 – ln 2) = ½ ln 8/2 = ½ ln 4 = ln 2 x f(x) dx = --------- dx x² + 2 = ½ ln | x² + 2| = ½ ((ln k² + 2) – ln 2) = ½ln 2 = ln (k² + 2) = 2ln 2 = ln 2² k² + 2 = 4 k = √2 ∫ ∫ ∫ ∫ x = 0 x = 0 x = 0 x = 0 x = k x = √6 | | x = 0 x = 0 1 √6 -------- = ------ √6 – 0 6 average value of f(x) = (√6/6) ln 2

  10. Summary & Homework • Summary: • Average value is like the average slope (secant line) • Remember the FTC Part II • Homework: • pg 467 1, 3, 4, 7, 9, 10, 17, 18

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