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Lesson 5-2 The Definite Integral. Ice Breaker See handout questions 1 and 2.

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Presentation on theme: "Lesson 5-2 The Definite Integral. Ice Breaker See handout questions 1 and 2."— Presentation transcript:

1 Lesson 5-2 The Definite Integral

2 Ice Breaker See handout questions 1 and 2

3 Objectives Understand Riemann Sums Understand the properties of the Definite Integral

4 Vocabulary Riemann Sum – a summation of n rectangles used to estimate the area under curve; when used with a limit as n approached infinity, then the Riemann sum is the definite integral Definite Integral – the integral evaluated at an upper limit (b) minus it evaluated at a lower limit (a); gives the area under the curve (in two dimensions)

5 Riemann Sums Let f be a function that is defined on the closed interval [a,b]. If ∆ is a partition of [a,b] and ∆x i is the width of the ith interval, c i, is any point in the subinterval, then Area = f(x) dx ∫ b a Area = Lim ∑A i = Lim ∑f(c i ) ∆x n→∞ i=1 i=n i=1 i=n ∆x = (b – a) / n (width) f(c i ) (height)

6 Example 0 Sketch the curve on the given interval, find the area geometrically and then write the definite integral that expresses the area pictured: y = 3x on [0, 3] y x 5 10 0 0 Area = 3x dx ∫ 3 0 Geometrically: Area = ½ b h = ½ 3 (9) = 27/2 Area = h w = (9)(2) = 18 + 9 dx ∫ 5 3. Now add the line y = 9 from 3 to 5

7 Example 1 Sketch the curve on the given interval and then write the definite integral that expresses the area pictured: y =  x on [0, 4] y x 5 5 0 0 Area =  x dx ∫ 4 0

8 y x Sketch the curve on the given interval and then write the definite integral that expresses the area pictured: y = cos x on [-π/2, π/2] Area = (cos x) dx ∫ π/2 -π/2 SKIP Example 2

9 Example 3 Sketch the curve on the given interval and then write the definite integral that expresses the area pictured: x = y², y  [0, 2] y x Area = y² dy ∫ 2 0

10 Properties of Definite Integrals 1. f(x) dx = 0 (no area above a pt) 2. f(x) dx = - f(x) dx (negative area) 3.If f is integrable on three closed intervals defined by a, b, and c, then f(x) dx = f(x) dx + f(x) dx (total = sum of its parts) ∫ a a ∫ b a ∫ a b ∫ b a ∫ c a ∫ b c

11 Properties of Definite Integrals (cont) 4. c f(x) dx = c f(x) dx (constants factor out) 5. [f(x) ± g(x)] dx = f(x) dx ± g(x) dx (integral operations can cross ± ) ∫ b a ∫ b a ∫ a b ∫ b a b a ∫

12 Example 6 a) b) Suppose and, find: ∫ 1 0 f(x) dx = 4 ∫ 1 0 g(x) dx = -2 ∫ 1 0 3 f(x) dx = ∫ 1 0 [f(x) - g(x)] dx = c) ∫ 1 0 [3f(x) + 2g(x)] dx = ∫ 1 0 3 f(x) dx = 3(4) = 12 ∫ 1 0 (f(x) dx – g(x)) dx = 4 - (-2) = 6 ∫ 1 0 ∫ 1 0 3 (f(x) dx + 2 g(x)) dx = 3(4) + 2(-2) = 8 ∫ 1 0

13 Example 6 cont d) e) Suppose and, find: ∫ 1 0 f(x) dx = 4 ∫ 1 0 g(x) dx = -2 ∫ 0 1 2 f(x) dx = ∫ 1 0 [2g(x) – 3f(x)] dx = ∫ 1 0 2 (g(x) dx - 3 f(x)) dx = 2(-2) - 3(4) = -16 ∫ 1 0 ∫ 0 1 2 f(x) dx = -2 f(x) dx = -2(4) = -8 ∫ 1 0

14 Summary & Homework Summary: –Riemann Sums are Limits of Infinite sums –Riemann Sums give exact areas under the curve –Riemann Sum is equal to the definite integral Homework: –pg 390 - 393: 3, 5, 9, 17, 20, 33, 38


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