1 Based on Stewart, James, Single Variable Calculus, Early Transcendentals, 7th edition, Brooks/Coles, 2012 Briggs, Cochran and Gillett: Calculus for Scientists and Engineers: Early Transcendentals, Pearson 2013 MATH1013 Calculus I Integration I (Chap. 5 ( 5.1, 5.2)) 1 Edmund Y. M. Chiang Department of Mathematics Hong Kong University of Science & Technology November 24, 2014
Definite integrals Riemann Sums Integrable functions
Area under curve We shall consider continuous functions defined on a closed interval only. The aim is to develop a theory that can be used to find area of the region under the given function.
Example We consider the problem of finding the area under the straight line f (x) = x for the interval 0 x 1. We divide the interval [0, 1] into five subintervals of equal width. By a partition of [0, 1] with five points: {x 0, x 1, x 2, x 3, x 4, x 5 } of [0, 1]. So we have the subintervals: [x 0, x 1 ] = [0, 1/5] [x 1, x 2 ] = [1/5, 2/5] [x 2, x 3 ] = [2/5, 3/5] [x 3, x 4 ] = [3/5, 4/5] [x 4, x 5 ] = [4/5, 5/5]
Right Riemann sum The f attains maximum values in each of the above intervals: f (x 1 ) = f (1/5) = 1/5, f (x 2 ) = f (2/5) = 2/5, f (x 3 ) = f (3/5) = 3/5, f (x 4 ) = f (4/5) = 4/5, f (x 5 ) = f (5/5) = 5/5 = 1. Let us sum the areas of the five rectangles with height at the right-end point of [x i 1, x i ] and with base 1/n. Thus we have ( ) ( ) ( ) ( ) ( ) 1 1 2 1 3 1 4 1 5 1 S 5 = f 5 5 + f 5 5 + f 5 5 + f 5 5 + f 5 5 = 1 ( ) 15 1 + 2 + 3 + 4 + 5 = 25 25 = 3 25, is called an right Riemann sum of f over [0, 1] with respect to the above partition. The approximation 3/25 is larger than the actual area.
Left Riemann sum The f attains minimum values in each of the above intervals: f (x 0 ) = f (0/5) = 0/5, f (x 1 ) = f (1/5) = 1/5, f (x 2 ) = f (2/5) = 2/5, f (x 3 ) = f (3/5) = 3/5, f (x 4 ) = f (4/5) = 4/5. Let us sum the areas of the five rectangles height at the left-end point in [x i 1, x i ] and with base 1/n. Thus we have ( ) ( ) ( ) ( ) ( ) 0 1 1 1 2 1 3 1 4 1 S 5 = f 5 5 + f 5 5 + f 5 5 + f 5 5 + f 5 5 = 1 ( ) 10 0 + 1 + 2 + 3 + 4 = 25 25 = 2 5, is called an left Riemann sum of f over [0, 1] with respect to the above partition. The approximation 2/25 is smaller than the actual area.
Left and Right Riemann sums figures Suppose the value of the area that we are to find is A. Then we clearly see from the figure below that the following inequalities must hold: 2 5 = S 5 A S 5 = 3 5.
Left and Right Riemann sums: general case It is also clear that the above argument that we divide [0, 1] into five equal intervals is nothing special. Hence the same argument applies for any number of intervals. Hence we must have S n A S n, (2) where n is any positive integer. We partition [0, 1] into n equal subintervals: {x 0, x 1, x 2,..., x n 1, x n } = {0, 1 n, 2 n,..., n 1 n, n n }. and the width of each of the subintervals of 1/n.
Hence the upper sum is S n = f ( 1 n Right Riemann sum: general case ) 1 n + f ( 2 n ) 1 n + + f ( n 1 n = 1 1 n n + 2 1 n n + + n 1 1 n n + n 1 n n = 1 ( ) 1 + 2 + 3 + + (n 1) + n n 2 = 1 n(n + 1) n 2 2 = 1 ( 1 ) 1 +. 2 n ) 1 n + f ( n ) 1 n n Note that we have used the fundamental formula that 1 + 2 + 3 + + (n 1) + n = n(n + 1) 2
Hence the lower sum is S n = f ( 0 ) 1 n n + f ( 1 n = 1 n Left Riemann sum: general case ) 1 n + + f ( n 2 n 0 n + 1 1 n n + + n 2 1 n n + n 1 1 n n = 1 ( ) 0 + 1 + 2 + + (n 2) + n 1 n 2 = 1 n 2 (n 1)n 2 We deduce that 1 2 ( 1 1 n = 1 2 ( 1 ) 1. n ) A 1 2 ) 1 n + f ( n 1 n ( 1 ) 1 +, for all n. n ) 1 n Letting n + gives that 1 2 A 1 2. That is A = 1/2 and lim S n = 1/2 = lim S n + n + n.
Riemann sums We consider a closed interval [a, b] and let P = {x 0, x 1, x 2,..., x n 1, x n }, x i = x i x i 1 = b a n, i = 1,, n to be any points lying inside [a, b]. Let f (x) be a continuous function defined on [a, b]. Then we define a Riemann sum of f (x) over [a, b] with respect to partition P to be the sum f (x 1 ) x 1 + f (x 2 ) x 2 + + f (x n) x n where xi is an arbitrary point lying in [x i 1, x i ], i = 1, 2,, n. Depending on the choices of xi, we have 1. Left Riemann sum if xi = x i 1 ; 2. Right Riemann sum if x i = x i ; 3. Mid-point Riemann sum if x i = (x i 1 + x i )/2.
Riemann sum figure Figure: (Briggs, et al, Figure 5.8)
Left Riemann sum figure Figure: (Briggs, et al, Figure 5.9)
Right Riemann sum figure Figure: (Briggs, et al, Figure 5.10)
Mid-point Riemann sum figure Figure: (Briggs, et al, Figure 5.11)
So 0.863 we have 1.003 1.125. Riemann sum of Sine We compute various Riemann sums under the sine curve from x = 0 to x = π/2 with six intervals. So we have x = b a n Left Riemann sum gives = π/2 0 6 = π 12. f (x 1 ) x 1 + f (x 2 ) x 2 + + f (x n) x n 0.863 Right Riemann sum gives f (x 1 ) x 1 + f (x 2 ) x 2 + + f (x n) x n 1.125 Mid-point Riemann sum gives f (x 1 ) x 1 + f (x 2 ) x 2 + + f (x n) x n 1.003
Partition of Riemann sum of Sine Figure: (Briggs, et al, Figure 5.2)
Partition of Riemann sum of Sine Figure: (Briggs, et al, Theorem 5.1)
Partition 50 Figure: (Briggs, et al, Figure 5.15)
Example (Briggs, et al, p. 372) After choosing a partition that divides [0, 2] into 50 subintervals: x k = x k x k 1 = 2 0 50 = 1 25 = 0.04. The Right Riemann Sum is given by 50 k=1 50 = f (xk ) x 50 k = f (x k ) (x k x k 1 ) k=1 k=1 f ( k 50 25 ) (0.04) = k=1 [( k ) 3 ] + 1 (0.04) 25 = [ 1 (50 51) 2 ] + 50 (0.04) = 6.1616. 25 3 2
Example (Briggs, et al, p. 372) II The Left Riemann Sum is given by 49 k=0 49 = f (xk ) x 49 k+1 = f (x k ) (x k+1 x k ) k=0 k=0 f ( k 49 25 ) (0.04) = k=0 [( k ) 3 ] + 1 (0.04) 25 = [ 1 (49 50) 2 ] + 50 (0.04) = 5.8416. 25 3 2
Example (Briggs, et al, p. 388) III After choosing a partition that divides [0, 2] into n subintervals: x k = x k x k 1 = 2 0 n The Right Riemann Sum is given by = 2 n. n f (xk ) x k = k=1 as n. n f (x k ) 2 n k=1 = 2 n [( 2k ) 3 ] + 1 = 2 ( 2 3 n n n n 3 k=1 = 2 ( 2 3 n n 3 n2 (n + 1) 2 ) + n 4 [ ( = 2 2 1 + 1 ) 2 ] + 1 6, n n k 3 + k=1 n k=1 ) 1
Example (Briggs, et al, p. 389) III The Left Riemann Sum is given by n 1 f (xk ) x n 1 k+1 = f (x k ) 2 n k=0 k=0 = 2 n 1 [( 2k ) 3 ] + 1 = 2 ( 2 3 n 1 n 1 ) n n n n 3 k 3 + 1 k=0 k=0 k=0 = 2 ( 2 3 n n 3 n2 (n 1) 2 ) [ ( + n = 2 2 1 1 ) 2 ] + 1 6, 4 n In fact, we have ( 4 1 1 ) 2 ( + 2 A 4 1 + 1 ) 2 + 2 n n to hold for every integer n. So A = 6.
Example (Stewart, p. 362) Let f (x) = x 2 and find the area A under f over the interval [0, 1]. Choose the partition P = {0, 1 n, 2 n,, n 1 n, n n = 1}. Show that the following inequalities 1 3 ( 1 )( 1 1 1 n 2n ) A 1 3 ( 1 )( 1 ) 1 + 1 +. n 2n by computing the left Riemann sum and right Riemann sum. Then show that the area is A = 1 3. We omit the details.
Some left/right Riemann sums Figure: (Stewart, Figures 5.1.8-9)
Exercises (Briggs, et al: p. 376) Write the following sum in summation notation: 4 + 9 + 14 + + 44. (Stewart: p. 369, Q. 4) Estimate the area under f (x) = x over [0, 4] using four approximating rectangles. 4 (Briggs, et al, p. 377) Given that f (1 + k) 1 is a k=1 Riemann sum of a certain function f over an interval [a, b] with a partition of n subdivisions. Identify the f, [a, b] and n.
Upper/Lower Riemann sums Riemann originally considered for a partition P n = {x 0,, x n } Area = lim n R n = [f (x 1 ) x + f (x 2 ) x + f (x n ) x] Area = lim n L n = [f (x 0 ) x + f (x 1 ) x + f (x n 1 ) x] and more generally, the upper and lower sums defined respectively by Area = lim n U n = lim n [f (x 1 ) x + f (x 2 ) x + f (x n) x] where f (xk ) attains the largest value in the k th interval [x k 1, x k ]; Area = lim n L n = lim n [f (x 1 ) x + f (x 2 ) x + f (x n ) x] where f (x k ) attains the smallest value in the k th interval [x k 1, x k ]
Upper Riemann sum figure Figure: (Stewart, Figures 5.1.13)
Upper/lower Riemann figures Figure: (Stewart, Figures 5.1.14)
Definite Integrals definition Figure: (Stewart Figure p. 372)
Definite Integral notation Figure: (Publisher Figure 5.21)
Integrable functions Theorem (Compare Stewart p. 373) Let f be a continuous function except on a finite number of discontinuities over the interval [a, b]. Then f is integrable on [a, b]. That is, lim δx 0 k=1 n f (xk ) x k = b a f (x) dx, exists irrespective to the x k and the partition [x k 1, x k ] chosen. So Since f (x) = x 2 is continuous over [0, 1] so it is integrable and 1 0 x 2 dx = 1 2 according to a previous calculation. Since f (x) = x 3 is continuous over [0, 1] so it is integrable and 1 0 x 3 dx = 1 according to a previous calculation. 3
Special partition and sample points Figure: (Stewart p. 374)
Piecewise continuous functions The following function has a finite number of discontinuities and so is integrable. However, we note that part 2 of the area is negative: Figure: (Briggs, et al, Figure 5.23)
Negative area Figure: (Briggs, et al, Figure 5.18)
Negative area Figure: (Briggs, et al, Figure 5.17)
Figure: (Briggs, et al, Figure 5.20) Net area
Figure: (Briggs, et al, Figure 5.24) Recognizing integral
Computing net area Figure: (Briggs, et al, Figure 5.31)
Exercises 1. Write down the right Riemann sum for 2. Interpret the sum lim integral. 3. Let n k=1 f (x) = 2 0 4 x 2 dx; n 3 n(1 + 3k as a certain Riemann n ) { 2x 2, if x 2; x + 4, if x > 2. Compute both the net area and actual area of 5 0 f (t) dt.
One can analyze the the sum as n k=1 3 n(1 + 3k n ) = = Hints to Exercises n k=1 n k=1 1 (1 + 3 k n ) 3 n 1 (1 + k 3 n ) 3 n = n f (k 3 n ) 3 n k=1 which represents a right Riemann sum for f (x) = 1, and for 1 + x the integration range from a = 0 and b = 3. But then lim n 0 k=1 3 3 n(1 + 3k n ) = 0 1 1 + x dx = ln 1 + x = (ln 4 ln 1) = ln 4. 3 0
Hints to Exercises The net area of the last example is given by 2 0 (2x 2) dx + 5 2 ( x + 4) dx = (x 2 2x) 2 0 + ( x 2 /2 + 4x) 5 2 = (2 2 2 2) + 1 2 (22 5 2 ) + (20 8) = 0 + 1 2 21 + 12 = 3 2. The actual area is given by 1 2 4 (2x 2) dx + (2x 2) dx + 5 ( x + 4) dx + ( x + 4) dx 0 1 2 4 = (x 2 2x) 1 0 + (x 2 2x) 2 1 + ( x 2 /2 + 4x) 4 ( x 2 + 2 /2 + 4x) 5 4 = 1 + 1 + 2 + 1 2 = 9 2.
Properties of Definite Integral Figure: (Briggs, et al, Table 5.4)
Sum of integrals Figure: (Briggs, et al, Figure 5.29)
Comparison properties of Definite Integral Figure: (Stewart, p. 381)