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LECTURE 11: APPROXIMATING AREAS UNDER CURVES MINGFENG ZHAO January 28, 2015 Approximating areas by Riemann sums Figure 1. Approximation of the area Definition 1. Suppose [a, b] is a closed interval, for any positive integer n, a partition with n subintervals is a set of n + 1 points x0 , x1 , · · · , xn such that x0 = a < x1 < · · · < xn−1 < xn = b. The points x0 , x1 , · · · , xn−1 , xn are called the grid points of this partition. This partition contains n subintervals: [x0 , x1 ], [x1 , x2 ], · · · , [xn−1 , xn ], with ∆xi = xi − xi−1 for all i = 1, · · · , n. b−a for all i = 1, · · · , n, this partition is called the regular partition of the interval [a, b]. In this case, If ∆xi = n b−a ∆x = is called the subinterval length, and the k-th grid point is: n xk = a + k∆x, for k = 0, 1, 2, · · · , n. 1 2 MINGFENG ZHAO Example 1. A regular partition with subinterval length 1 of [1, 5] is: [1, 2], [2, 3], [3, 4], [4, 5]. That is, x0 = 1, x1 = 2, x2 = 3, x3 = 4, x4 = 5. This regular partition has 5 grid points. Example 2. Find the regular partition with 10 grid points of [2, 20]. By the assumption, we have n = 10 − 1 = 9, then the length of each small interval is: ∆x = 20 − 2 18 = = 2. 9 9 So we have xk = 2 + k∆x = 2 + 2k, for all k = 0, 1, · · · , 9. Then this regular partition is: [2, 4], [4, 6], [6, 8], [8, 10], [10, 12], [12, 14], [14, 16], [16, 18], [18, 20]. Definition 2. Let f be a function on the closed interval [a, b], and a = x0 < x1 < · · · < xn = b be a partition of [a, b], for any k = 0, 1, · · · , n, we choose any point x∗k in k-th interval [xk−1 , xk ], the the sum f (x∗1 )∆x1 + f (x∗2 )∆x2 + · · · + f (x∗n )∆xn is called a general Riemann sum of f on [a, b]. If a partition is regular, that is, ∆x1 = ∆x2 = · · · = ∆xn = b−a , the n general Riemann sum is called the Riemann sum with n subintervals, then • If x∗k = xk−1 for any k = 1, · · · , n, the sum is called a left endpoint Riemann sum. • If x∗k = xk for any k = 1, · · · , n, the sum is called a right endpoint Riemann sum. xk−1 + xk ∆x = xk−1 + for any k = 1, · · · , n, the sum is called a midpoint endpoint Riemann sum. • If x∗k = 2 2 Example 3. Calculate the left Riemann sum for f (x) = cos(x) on the interval [0, π] using n = 4. The length of the subinterval for the regular partition of [0, π] with n = 4 will be: ∆x = π−0 π = . 4 4 Then x0 = 0, x1 = 0 + ∆x = π π 3π , x2 = 0 + 2∆x = , x3 = 0 + 3∆x = , x4 = π. 4 2 4 LECTURE 11: APPROXIMATING AREAS UNDER CURVES 3 So the left Riemann sum is: f (x0 )∆x + f (x1 )∆x + f (x2 )∆x + f (x3 )∆x π π π π π 3π π · + cos · + cos = cos(0) · + cos · 4 4 4 2 4 4 4 √ √ π 2 π 2 π π = 1· + · +0· − · 4 2 4 4 2 4 π = . 4 Example 4. Estimate the area A under the graph of f on the interval [0, 2] using the left and right Riemann sums with n = 4, where f is continuous but known only at the points in the following table. Since [0, 2] has n = 4 subintervals, then ∆x = x0 = 0, x f (x) 0 1 0.5 3 1.0 4.5 1.5 5.5 2.0 6.0 2−0 = 0.5 and 4 x1 = 0.5, x2 = 1, x3 = 1.5, and x4 = 2. So • The left Riemann sum is: f (x0 )∆x + f (x1 )∆x + f (x2 )∆x + f (x3 )∆x = 1 · 0.5 + 3 · 0.5 + 4.5 · 0.5 + 4.5 · 0.5 + 5.5 · 0.5 = 7.0. • The right Riemann sum is: f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + f (x4 )∆x = 3 · 0.5 + 4.5 · 0.5 + 4.5 · 0.5 + 5.5 · 0.5 + 6.0 · 0.5 = 9.5. Sigma notation Definition 3. For a sequence of numbers, x1 , x2 ,· · · , xn , the sum x1 + x2 + · · · + xn is denoted by n X i=1 called the sigma notation. xi , which is 4 MINGFENG ZHAO Remark 1. In general, for integers m ∈ Z and n ≥ m, for a sequence of numbers, xm , xm+1 , · · · , xn , we use the representation: xm + xm+1 + · · · + xn = n X xi . i=m Example 5. For a regular partition x0 = a < x1 < · · · < xn = b of [a, b], then the Riemann sum can be written: f (x∗1 )∆x + f (x∗2 )∆x + · · · + f (x∗n )∆x = n X f (x∗i )∆x. i=1 For the choices x∗i ’s, we have a. For the left Riemann sum, we have x∗i = xi−1 = a + (i − 1)∆x. b. For the right Riemann sum, we have x∗i = xi = a + i∆x. a + (i − 1)∆x + a + i∆x 1 xi−1 + xi ∗ = =a+ i− ∆x. c. For the midpoint Riemann sum, we have xi = 2 2 2 Example 6. Compute 1 + 2 + 3 + · · · + 100. In fact, we have 1 + 2 + 3 + · · · + 100 = 100 X i. i=1 Let A = 1 + 2 + 3 + · · · + 100, then we have A = = 1 + 2 + 3 + · · · + 100 100 + 99 + 98 + · · · + 1. So we get 2A = 1 + 2 + 3 + · · · + 100 + 100 + 99 + 98 + · · · + 1 = 100 ∗ 101. That is, A= 100 ∗ 101 = 50 ∗ 101 = 5050. 2 Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C. Canada V6T 1Z2 E-mail address: [email protected]