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Riemann Sums. Part 2 Lecture 30 Riemann Sums. Part 2 Objectives ......................................... ...................... 2 Theareaunderaparabola ................................ .................... 3 Theareaunderaparabola ................................ .................... 4 Theareaunderaparabola ................................ .................... 5 Theareaunderaparabola ................................ .................... 6 Theareaunderaparabola ................................ .................... 7 Approximating an Integral by Riemann sums . ....................... 8 Theintegralasasignedarea............................. ...................... 9 The geometric interpretation of the left Riemann sum. ........................ 10 The geometric interpretation of the right Riemann sum. ....................... 11 LeftandrightRiemannsums............................. ..................... 12 GeneralleftandrightRiemannsums ....................... ..................... 13 GeneralleftandrightRiemannsums ....................... ..................... 14 The integral as the limit of Riemann sums. ....................... 15 Calculationoflimits................................ ........................ 16 Calculationoflimits................................ ........................ 17 Summary............................................. .................. 18 Comprehension checkpoint ............................. ...................... 19 1 Objectives In this lecture we will continue to work with the definite integral as the limit of Riemann sums. Remember that for a piece-wise continuous function f(x), the definite integral may be calculated as the limit of a Riemann sum Ln or Rn : b f(x) dx = lim Ln = lim Rn, where a n→∞ n→∞ Z n−1 Ln = f(xi)∆x is the left Riemann sum, i=0 Pn Rn = f(xi)∆x is the right Riemann sum, i=1 Pb − a ∆x = are n subintervals of [a,b], n xi = a + i∆x are the points subdividing [a,b]. Riemann sums have a simple geometric interpretation. In this lecture, we will do some elementary calculations with Riemann sums. 2 / 19 The area under a parabola 1 2 Problem. 1. Approximate x dx by L2 and R2 . 0 Z 2. Calculate the Riemann sums Ln and Rn , where n is a positive integer. 1 2 3. Calculate x dx explicitly as the limit of Ln and Rn . 0 Z Solution. The integrand is f(x)= x2 , the interval is [0, 1] . b − a 1 1. To calculate L2 and R2 , we subdivide [0, 1] into two equal parts of length ∆x = = : n 2 1 1 2 2 The subdivision points are 1 0 1 1 x0 = 0, x1 = , x2 = 1 . 2 2 3 / 19 2 The area under a parabola 1 2 The integral x dx represents the area under the parabola y = x2 between x = 0 and x = 1 : 0 Z y y = x2 1 1 4 1 x 0 2 1 1 1 1 1 1 1 1 L2 = f(x0)∆x + f(x1)∆x = f(0) · + f( ) · = 0 · + · = . 2 2 2 2 4 2 8 1 1 1 1 1 1 5 R2 = f(x1)∆x + f(x2)∆x = f( ) · + f(1) · = · + 1 · = . 2 2 2 4 2 2 8 1 1 2 1 2 5 x dx ≈ L2 = (underestimate), x dx ≈ R2 = (overestimate) 0 8 0 8 Z Z 4 / 19 The area under a parabola 2. To calculate the general Riemann sums Ln , Rn , b − a 1 we subdivide [0, 1] into n equal parts of length ∆x = = . n n 1 i The points of the subdivision are x = x0 + i∆x =0+ i = , where i = 0, 1,...,n . i n n 2 n−1 n−1 i i n−1 i 1 1 n−1 L = f(x )∆x = f = = i2, n i n n n n n3 i=0 i=0 i=0 i=0 P P P 2 P n n i i n i 1 1 n R = f(x )∆x = f = = i2 . n i n n n n n3 i=1 i=1 i=1 i=1 P P P P As we know from Algebra (or can prove by induction), n n(n + 1)(2n + 1) i2 = 12 + 22 + 32 + · · · + n2 = i=1 6 P 5 / 19 3 The area under a parabola Therefore, n−1 1 2 (n − 1)n(2n − 1) Ln = 3 i = 3 and n i=0 6n Pn 1 2 n(n + 1)(2n + 1) Rn = 3 i = 3 . n i=1 6n 1 P 2 The Riemann sums Ln and Rn approximate x dx and, 0 since f(x)= x2 is increasing on [0, 1], we haveZ 1 2 Ln ≤ x dx ≤ Rn for any positive integer n . Calculate the limits: 0 Z 1 1 (n − 1)n(2n − 1) 1 − n · 1 · 2 − n 1 · 1 · 2 1 lim Ln = lim = lim = = , n→∞ n→∞ n3 n→∞ 6 6 6 3 1 1 n(n + 1)(2n + 1) 1 · 1+ n · 2+ n 1 · 1 · 2 1 lim Rn = lim = lim = = . n→∞ n→∞ n3 n→∞ 6 6 6 3 6 / 19 The area under a parabola In the inequality for the integral, let n →∞: 1 2 Ln ≤ x dx ≤ Rn 0 1 Z 2 1 = Therefore, x dx = . n →∞ n →∞ 0 3 1 Z 3 y y = x2 1 2 1 x dx = lim Ln = 0 n→∞ 3 Z 1 2 1 x dx = lim Rn = 0 n→∞ 3 Z x 0 1 7 / 19 4 Approximating an Integral by Riemann sums 2 3 Problem. 1. Approximate (x − 2x)dx by L6 and R6 . −1 Z 2. Calculate the Riemann sums Ln and Rn , where n is a positive integer. 2 3 3. Calculate (x − 2x)dx as the limit of Ln and Rn . −1 Z Solution. The integrand is f(x)= x3 − 2x , the interval is [−1, 2] . 1. To calculate L6 and R6 , we subdivide [−1, 2] into six equal parts b − a 2 − (−1) 3 1 of length ∆x = = = = : n 6 6 2 x0 x1 x2 x3 x4 x5 x6 −1 −.5 0 .5 1 1.5 2 The subdivision points are x0 = −1, x1 = −.5, x2 =0, x3 = .5, x4 =1, x5 =1.5, x6 =2. i In abbreviated form, xi = x + i∆x = −1+ , where i =0, 1,..., 6 . 0 2 8 / 19 The integral as a signed area 2 3 The integral (x − 2x)dx represents the signed area −1 Z 3 between the graph of y = x − 2x and the x− axis: y 3 y = x − 2x + + x −1 − 2 9 / 19 5 The geometric interpretation of the left Riemann sum 5 L6 = f(xi)∆x i=0 = fP(x0)∆x + f(x1)∆x + f(x2)∆x + f(x3)∆x + f(x4)∆x + f(x5)∆x 1 3 = f(−1) + f(−.5) + f(0) + f(.5) + f(1) + f(1.5) · = 2 16 y 3 y = x − 2x .5 1 x −1 −.5 1.5 2 10 / 19 The geometric interpretation of the right Riemann sum 6 R6 = f(xi)∆x i=1 = fP(x1)∆x + f(x2)∆x + f(x3)∆x + f(x4)∆x + f(x5)+ f(x6)∆x 1 27 = f(−.5) + f(0) + f(.5) + f(1) + f(1.5) + f(2) · = 2 16 3 y y = x − 2x .5 1 x −1 −.5 1.5 2 11 / 19 6 Left and right Riemann sums 3 27 We have calculated L6 = and R6 = . 16 16 Both left and right Riemann sums approximate the value of the integral: 3 3 3 3 3 27 (x − 2x)dx ≈ and (x − 2x)dx ≈ . −1 16 −1 16 Z Z The exact value of the integral, as we will calculate soon, is 2 3 3 12 (x − 2x)dx = = . −1 4 16 Z L6 Exact value R6 3 12 27 16 16 16 12 / 19 General left and right Riemann sums 2 3 To compute Ln and Rn for (x − 2x)dx , recall the formulas: −1 Z n−1 n Ln = f(xi)∆x and Rn = f(xi)∆x , where i=0 i=1 P P 2 − (−1) 3 3i ∆x = = and x = x0 + i∆x = −1+ , i = 0, 1, 2,...,n . n n i n 3 3i 3i 3i Since f(x )= f −1+ = −1+ − 2 −1+ , we find i n n n 3 n−1 n−1 3i 3i 3 L = f(x )∆x = −1+ − 2 −1+ and n i n n n i=0 i=0 " # P P 3 n n 3i 3i 3 R = f(x )∆x = −1+ − 2 −1+ . n i n n n i=1 i=1 " # P P 13 / 19 7 General left and right Riemann sums After algebraic manipulations, we get n−1 n−1 n−1 n−1 n−1 3 3 27 2 27 3 3 9 81 2 81 3 Ln = 1+ i − i + i = 1+ i − i + i n i n n2 n3 n i n2 i n3 i n4 i =0 =0 =0 =0 =0 P P P 2 P P (∗) 3 9 (n − 1)n 81 (n − 1)n(2n − 1) 81 (n − 1)n = n + − + n n2 2 n3 6 n4 2 9 1 27 1 1 81 1 2 =3+ 1 − − 1 − 1 2 − + 1 − 1 . 2 n 2 n n 4 n 9 81 3 So lim Ln =3+ − 27+ = . n→∞ 2 4 4 n 3 3 27 27 (∗) 3 Similarly, Rn = 1+ i − i2 + i3 , and lim Rn = . n i n n2 n3 n→∞ 4 =1 In our calculations (*)P , we used the following formulas: n n n(n + 1) n n(n + 1)(2n + 1) n n(n + 1) 2 1= n, i = , i2 = , i3 = i i 2 i 6 i 2 =1 =1 =1 =1 P P P P 14 / 19 The integral as the limit of Riemann sums b b Since f(x)dx = lim Ln or f(x)dx = lim Rn , we have n→∞ n→∞ Za Za 2 n−1 3 3 3 27 2 27 3 3 (x − 2x)dx = lim 1+ i − 2 i + 3 i = or −1 n→∞ n n n n 4 i=0 Z X 2 n 3 3 3 27 2 27 3 3 (x − 2x)dx = lim 1+ i − 2 i + 3 i = . −1 n→∞ n n n n 4 Z i=1 2 X 3 3 Finally, (x − 2x)dx = . −1 4 Z ☞ The calculation of definite integrals using Riemann sums is tedious. A more efficient method of calculation (using the Fundamental Theorem of Calculus) will be given in the next lecture.
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