MATH 12002 - CALCULUS I 4.2: Riemann Sum Examples Professor Donald - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 4.2: Riemann Sum Examples Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 7

Riemann Sums Definition Let y = f ( x ) be a function defined on an interval [ a , b ]. Let n be a positive integer. Break the interval [ a , b ] into n equal subintervals with endpoints a = x 0 < x 1 < x 2 < · · · < x n − 1 < x n = b , so that each subinterval has length ∆ x = b − a n . For each i = 1 , . . . , n , choose a point x ∗ i in the i th subinterval [ x i − 1 , x i ]. The n th Riemann Sum for f on [ a , b ] with this choice of sample points is S n = f ( x ∗ 1 )∆ x + f ( x ∗ 2 )∆ x + · · · + f ( x ∗ n )∆ x . D.L. White (Kent State University) 2 / 7

Riemann Sums The n th right Riemann sum R n is obtained by letting x ∗ i = x i , the right endpoint of the i th subinterval [ x i − 1 , x i ]: R n = f ( x 1 )∆ x + f ( x 2 )∆ x + · · · + f ( x n )∆ x . The n th left Riemann sum L n is obtained by letting x ∗ i = x i − 1 , the left endpoint of the i th subinterval [ x i − 1 , x i ]: L n = f ( x 0 )∆ x + f ( x 1 )∆ x + · · · + f ( x n − 1 )∆ x . D.L. White (Kent State University) 3 / 7

Example Example Compute the sixth left Riemann sum L 6 and the sixth right Riemann sum R 6 for the function f ( x ) = x 3 − 5 on the interval [0 , 3] . D.L. White (Kent State University) 4 / 7

Example In the notation of the definition, n = 6, a = 0, b = 3, so ∆ x = 3 − 0 = 1 2 . 6 The partition of our interval [0 , 3] is 1 3 5 0 1 2 3 2 2 2 The left endpoints of the intervals are 0 , 1 2 , 1 , 3 2 , 2 , 5 2 , the right endpoints of the intervals are 1 2 , 1 , 3 2 , 2 , 5 2 , 3, and we have 0 3 − 5 = 0 − 5 = − 5 f (0) = 2 ) 3 − 5 = 1 f ( 1 ( 1 8 − 5 = − 39 2 ) = 8 1 3 − 5 = 1 − 5 = − 4 f (1) = 2 ) 3 − 5 = 27 f ( 3 ( 3 8 − 5 = − 13 2 ) = 8 2 3 − 5 = 8 − 5 = 3 f (2) = 2 ) 3 − 5 = 125 f ( 5 ( 5 8 − 5 = 85 2 ) = 8 3 3 − 5 = 27 − 5 = 22 . f (3) = D.L. White (Kent State University) 5 / 7

Example Therefore, we have f (0) · 1 2 + f ( 1 2 ) · 1 2 + f (1) · 1 2 + f ( 3 2 ) · 1 2 + f (2) · 1 2 + f ( 5 2 ) · 1 = L 6 2 ( − 5) · 1 2 + ( − 39 8 ) · 1 2 + ( − 4) · 1 2 + ( − 13 8 ) · 1 2 + (3) · 1 2 + ( 85 8 ) · 1 = 2 − 5 2 − 39 16 − 2 − 13 16 + 3 2 + 85 16 = − 15 = 16 = − 0 . 9375 and f ( 1 2 ) · 1 2 + f (1) · 1 2 + f ( 3 2 ) · 1 2 + f (2) · 1 2 + f ( 5 2 ) · 1 2 + f (3) · 1 = R 6 2 ( − 39 8 ) · 1 2 + ( − 4) · 1 2 + ( − 13 8 ) · 1 2 + (3) · 1 2 + ( 85 8 ) · 1 2 + (22) · 1 = 2 − 39 16 − 2 − 13 16 + 3 2 + 85 16 + 11 = 201 = 16 = 12 . 5625 . D.L. White (Kent State University) 6 / 7

Example Both L 6 = − 0 . 9375 and R 6 = 12 . 5625 are (poor) approximations for � 3 x 3 − 5 dx = lim n →∞ L n = lim n →∞ R n . 0 Since f ( x ) is increasing, for any n we have � 3 x 3 − 5 dx � R n , L n � 0 so we know � 3 x 3 − 5 dx � 12 . 5625 . − 0 . 9375 � 0 The larger n we use, the closer L n and R n will be to the integral. We will see later that the exact value is � 3 x 3 − 5 dx = 5 . 25 . 0 D.L. White (Kent State University) 7 / 7