1. Lesson 5-2R Riemann Sums

2. Objectives Understand Riemann Sums

3. 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)

4. Example 2e Use sums to describe the area of the region between the graph of y = x² + 1 and the x-axis from x = 0 to x = 2. Partition [0,2] into n intervals, the width of the intervals will be (2-0)/n = 2/n. Since the function is increasing on this interval, the left-hand (inscribed) heights will be f(x i-1 ) and the right-hand (circumscribed) heights will be f(x i ). RectangleInscribed AreaCircumscribed Area 1 2 3 4 5 i (2/n) f(0) (2/n) f(0+2/n) (2/n) f(0+1(2/n)) (2/n) f(0+2(2/n)) (2/n) f(0+3(2/n)) (2/n) f(0+4(2/n)) (2/n) f(0+5(2/n)) (2/n) f(0+(i-1)(2/n)) (2/n) f(0+(i)(2/n)) (2/n) (1 + (2/n)²) (2/n) (1 + (4/n)²) (2/n) (1 + (6/n)²) (2/n) (1 + (8/n)²) (2/n) (1 + (10/n)²) (2/n) (1 + (2i/n)²) Right -Hand

5. Example 3 Find the area bounded by the function f(x) = x² + 1 and the x-axis on the interval [0,2] using limits. y x 2 5 0 0 ∆x = (2-0)/n = 2/n f(x i ) = 1 + (2i/n)² = 1 + 4i²/n² A i = 2/n (1 + 4i²/n²) Lim ∑A i = Lim ∑f(x i )∆x n→∞ Lim ∑ (2/n + 8i²/n³) n→∞ Lim (2/n³) ∑ (n² + 4i²) n→∞ = Lim (2/n³) (n³ + 4(n³/3 + n²/2 + n/6)) n→∞ = Lim (2 + 8/3 + 4/n + 8/6n²) = 4.67 n→∞

6. 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 the sum f(c i )∆x i is called a Riemann Sum of f. Furthermore, if exists, lim f(c i )∆x i we say f is integrable on [a,b]. The definite integral, f(x)dx, is the area under the curve ∑ i=1 n n→∞ ∑ i=1 n ∫ b a

7. Definite Integral vs Riemann Sum Area = f(x) dx ∫ b a Area = Lim ∑A i = Lim ∑f(x i ) ∆x n→∞ i=1 i=n i=1 i=n ∆x = (b – a) / n Area = (3x – 8) dx ∫ 5 2 3i 3 Area = Lim ∑ [3(----- + 2) – 8] (---) n n n→∞ i=1 i=n 5-2=35-2=3 ∆x [f(x)] xixi a

8. Σ ca i = c Σ a i i = m i = n Operations: i = m i = n Σ (a i ± b i ) = Σ a i ± Σ b i i = m i = n i = m i = n i = m i = n constants factor out summations split across ± C is a constant, n is a positive integer, and a i and b i are dependent on i Formulas: C is a constant, n is a positive integer, and a i and b i are dependent on i Σ i = 1 i = n 1 = n Σ i =1 i = n Σ i = 1 i = n Σ i = 1 i = n Σ i = 1 i = n c = cn n(n + 1) n² + n i = ------------- = ---------- 2 2 n(n + 1)(2n + 1) 2n³ + 3n² + n i² = -------------------- = -------------------- 6 6 n(n + 1) ² n² (n² + 2n + 1) i³ = ----------- = --------------------- 2 4 Sigma Notation

9. Example 4 In the following summations, simplify in terms of n. 1. (5) = 2. (2i + 1) = 3. (6i² - 2i) = 4. (4i³ - 6i²) = Σ i = 1 i = n Σ i = 1 i = n Σ i = 1 i = n Σ i = 1 i = n 5n 2(n² + n) ------------- + n = n² + 2n 2 6(2n³ + 3n² + n) 2(n² + n) ---------------------- - ------------- = 2n³ + 2n² 6 2 4(n² (n² + 2n + 1)) 6(2n³ + 3n² + n) ------------------------ - ---------------------- 4 6 = n 4 + 2n³ + n² - 2n³ - 3n² - n = n 4 - 2n² - n

10. Example 5 Rewrite following summations as definite integrals. Σ i =1 i = n 3i --- n 2 3 --- n a) Lim n→∞ Σ i =1 i = n 2i --- n 3 2 --- n b) Lim n→∞ Σ i =1 i = n 4i 2i 1 + ---- + ---- n n 2 2 --- n e) Lim n→∞ ∫ 3 0 x² dx ∫ 2 0 x³ dx ∫ 2 0 (1 + 2x + x²) dx Σ i =1 i = n πi sin --- n π --- n c) Lim n→∞ Σ i =1 i = n 2i 2i 1 + ---- + ---- n n 2 2 --- n e) Lim n→∞ ∫ 2 0 (1 + x + x²) dx ∫ π 0 sin(x) dx

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