Lesson 5-2R

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(xi-1) and the right-hand (circumscribed) heights will be f(xi). Right -Hand (2/n) (1 + (2/n)²) (2/n) f(0) (2/n) f(0+2/n) (2/n) (1 + (4/n)²) (2/n) f(0+2(2/n)) (2/n) f(0+1(2/n)) (2/n) (1 + (6/n)²) (2/n) f(0+2(2/n)) (2/n) f(0+3(2/n)) (2/n) (1 + (8/n)²) (2/n) f(0+3(2/n)) (2/n) f(0+4(2/n)) (2/n) (1 + (10/n)²) (2/n) f(0+5(2/n)) (2/n) f(0+4(2/n)) (2/n) f(0+(i)(2/n)) (2/n) f(0+(i-1)(2/n)) (2/n) (1 + (2i/n)²)

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. Lim ∑Ai = Lim ∑f(xi)∆x n→∞ n→∞ ∆x = (2-0)/n = 2/n f(xi) = 1 + (2i/n)² = 1 + 4i²/n² y 5 Ai = 2/n (1 + 4i²/n²) Lim ∑ (2/n + 8i²/n³) n→∞ x 0 0 2 Lim (2/n³) ∑ (n² + 4i²) = Lim (2/n³) (n³ + 4(n³/3 + n²/2 + n/6)) = Lim (2 + 8/3 + 4/n + 8/6n²) = 4.67 n→∞ n→∞ 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 ∆xi is the width of the ith interval, ci, is any point in the subinterval, then the sum f(ci)∆xi is called a Riemann Sum of f. Furthermore, if exists, lim f(ci)∆xi we say f is integrable on [a,b]. The definite integral, f(x)dx , is the area under the curve n n n→∞ ∑ ∑ b ∫ i=1 i=1 a

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

8. i = n i = n i = n i = n i = n Σ Σ Σ Σ Σ i = 1 i = 1 i = 1 i = 1 i =1 Sigma Notation Operations: C is a constant, n is a positive integer, and ai and bi are dependent on i i = n i = n i = n i = n i = n Σ(ai ± bi) = Σ ai± Σ bi Σcai = c Σ ai i = m i = m i = m i = m i = m summations split across ± constants factor out Formulas: C is a constant, n is a positive integer, and ai and bi are dependent on i 1 = 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

9. i = n i = n i = n i = n Σ Σ Σ Σ i = 1 i = 1 i = 1 i = 1 Example 4 In the following summations, simplify in terms of n. 1. (5) = 2. (2i + 1) = 3. (6i² - 2i) = 4. (4i³ - 6i²) = 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 = n4 + 2n³ + n² - 2n³ - 3n² - n = n4 - 2n² - n

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

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