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5.2 THE DEFINITE INTEGRAL OF A CONTINUOUS FUNCTION The process we used to solve the two problems of Section 5.1 is called integration , and the end results of this process are called definite integrals . Our purpose here is to establish these notions more precisely. (5.2.1) By a partition of the closed interval [ a , b ] we mean a finite subset of [ a , b ] which contains the points a and b . We index the elements of a partition according to their natural order. Thus, if we write P = { x 0 , x 1 , x 2 , · · · , x n 1 , x n } is a partition of [ a , b ], you can conclude that a = x 0 < x 1 < · · · < x n = b . Example 1 The sets { 0, 1 } , { 0, 1 2 , 1 } , { 0, 1 4 , 1 2 , 1 } , { 0, 1 4 , 1 3 , 1 2 , 5 8 , 1 } are all partitions of the interval [0,1]. If P = { x 0 , x 1 , x 2 , · · · , x n 1 , x n } is a partition of [ a , b ], then P breaks up [ a , b ] into n subintervals [ x 0 , x 1 ], [ x 1 , x 2 ], . . . , [ x n 1 , x n ] of lengths x 1 , x 2 , . . . , x n , respectively. Suppose now that f is continuous on [ a , b ]. Then on each interval [ x i 1 , x i ] the function f takes on a maximum value, M i , and a minimum value, m i .
266 CHAPTER 5 INTEGRATION (5.2.2) The number U f ( P ) = M 1 x 1 + M 2 x 2 + · · · + M n x n is called the P upper sum for f , and the number L f ( P ) = m 1 x 1 + m 2 x 2 + · · · + m n x n is called the P lower sum for f . Example 2 The quadratic function f ( x ) = x 2 is continuous on [1, 3]. The partition P = { 1, 3 2 , 2, 3 } breaks up [1, 3] into three subintervals: [ x 0 , x 1 ] = [1, 3 2 ], [ x 1 , x 2 ] = [ 3 2 , 2], [ x 2 , x 3 ] = [2, 3] of lengths x 1 = 3 2 1 = 1 2 , x 2 = 2 3 2 = 1 2 , x 3 = 3 2 = 1, respectively. Since f is increasing on [1, 3], it takes on its maximum value at the right endpoint of each subinterval: M 1 = f ( 3 2 ) = 9 4 , M 2 = f (2) = 4, M 3 = f (3) = 9. The minimum values of f are taken on at the left endpoints of each subinterval: m 1 = f (1) = 1, m 2 = f ( 3 2 ) = 9 4 , m 3 = f (2) = 4. Thus, U f ( P ) = M 1 x 1 + M 2 x 2 + M 3 x 3 = 9 4 ( 1 2 ) + 4( 1 2 ) + 9(1) = 97 8 = 12. 125 (see Figure 5.2.1a) and L f ( P ) = m 1 x 1 + m 2 x 2 + m 3 x 3 = 1( 1 2 ) + 9 4 ( 1 2 ) + 4(1) = 45 8 = 5. 625. (see Figure 5.2.1b) x 3 2 1 Upper sum y 3 2 ( a ) ( b ) x 3 2 1 y 3 2 Lower sum Figure 5.2.1
5.2 THE DEFINITE INTEGRAL OF A CONTINUOUS FUNCTION 267 Example 3 The function f ( x ) = 5 x 2 is continuous on the interval [ 1, 3]. The partition P = {− 1, 1 2 , 3 2 , 2, 3 } breaks up [ 1, 3] into four subintervals [ x 0 , x 1 ] = [ 1, 1 2 ], [ x 1 , x 2 ] = [ 1 2 , 3 2 ], [ x 2 , x 3 ] = [ 3 2 , 2], [ x 3 , x 4 ] = [2, 3] of lengths x 1 = 1 2 ( 1) = 3 2 , x 2 = 3 2 1 2 = 1, x 3 = 2 3 2 = 1 2 , x 4 = 3 2 = 1, respectively. See Figure 5.2.2. y x 4 1 4 ( 1, 4) (0, 5) (3, 4) (2, 1) , 1 2 3 1 2 1 2 19 4 3 2 , 3 2 11 4 Figure 5.2.2 As you can check, the maximum value of f on each subinterval is given by: M 1 = f (0) = 5 on 1, 1 2 , M 2 = f ( 1 2 ) = 19 4 on 1 2 , 3 2 M 3 = f ( 3 2 ) = 11 4 on 3 2 , 2 , M 4 = f (2) = 1 on [2, 3]. The minimum value of f on each subinterval is given by m 1 = f ( 1) = 4 on 1, 1 2 , m 2 = f ( 3 2 ) = 11 4 on [ 1 2 , 3 2 ] m 3 = f (2) = 1 on 3 2 , 2 , m 4 = f (3) = − 4 on [2, 3]. Therefore, U f ( P ) = 5 ( 3 2 ) + 19 4 (1) + 11 4 ( 1 2 ) + 1(1) = 117 8 = 14. 625 and L f ( P ) = 4 ( 3 2 ) + 11 4 (1) + 1 ( 1 2 ) + ( 4)(1) = 21 4 = 5. 25 By an argument that we omit here (it appears in Appendix B.4), it can be proved that, with f continuous on [ a , b ], there is one and only one number I that satisfies the inequality L f ( P ) I < U f ( P ) for all partitions P of [ a , b ].

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