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.

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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

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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

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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

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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

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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

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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

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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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8 Calculation of limits n−1 n − i Example. Calculate the limit lim→∞ 2 . n i=0 n Solution. We calculate the limit byP interpreting it as the limit of a Riemann sum, and expressing the limit of the Riemann sum as an integral. n−1 n − i n−1 The sum 2 looks like the left Riemann sum f(xi)∆x . i=0 n i=0 P n − i P i 1 That would make = f(x )∆x , which we can write as 1 − = f(x )∆x. This gives n2 i n n i b − a 1 i ∆x = = , x = , f(x ) = 1 − x . n n i n i i 0 It follows that b − a = 1 , a = x0 = = 0 , f(x) = 1 − x . n Putting it all together, a = 0, b = 1, f(x) = 1 − x and n−1 n − i 1 lim 2 = (1 − x)dx . n→∞ =0 n 0 i Z P 16 / 19

Calculation of limits 1 Now we evaluate the integral (1 − x)dx . 0 Z y 1 1 y = 1 − x 1 (1 − x)dx = Area of triangle = 0 2 Z x 0 1

n−1 n − i 1 1 Finally, lim 2 = (1 − x)dx = . n→∞ =0 n 0 2 i Z P 17 / 19

9 Summary Riemann sums may be used for the approximate calculation of deﬁnite integrals. Remember the following formulas: b f(x) dx = lim Ln = lim Rn, where n→∞ n→∞ Za 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 P b − a ∆x = are subintervals, n xi = a + i∆x are the subdivision points of [a, b].

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Comprehension checkpoint 1 x2 • Approximate the integral e dx by the Riemann sums L2,R2 and L4,R4 . Use a calculator! 0 Z Give geometric interpretations of these Riemann sums. Are the approximations underestimates or overestimates?

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