1 / 4

Hence

Example 3 Evaluate the lower Riemann sum L ( P,f ) and the upper Riemann sum U ( P,f ) where P = { 0, 1/4, 1/3, 1/2, 3/4, 4/5, 1 } and f ( x ) = arctan x ,

feleti
Download Presentation

Hence

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Example 3 Evaluate the lower Riemann sum L(P,f )and the upper Riemann sum U(P,f )where P = {0, 1/4, 1/3, 1/2, 3/4, 4/5, 1}and f(x) = arctan x, Solution To locate the minimum of f on the subintervals of [0,1] and compute L(P,f ), sketch the graph of f(x)= arctan x. Then t1*=0, the point where f has its minimum value on the subinterval [0,1/4], t2*=1/4, the point where f has its minimum value on [1/4,1/3], t3*=1/3, the point where f has its minimum value on the subinterval [1/3,1/2], t4*=1/2, the point where f has its minimum value on the subinterval [1/2,3/4], t5*=3/4, the point where f has its minimum value on the subinterval [3/4,4/5] and t6*=4/5, the point where f has its minimum value on the subinterval [4/5,1].

  2. t1*=0, t2*=1/4, t3*=1/3, t4*=1/2, t5*=3/4, t6*=4/5 P = {0, 1/4, 1/3, 1/2, 3/4, 4/5, 1} • Hence • L(P,f) = R(P,T*,f) = f(t1*)(x1-x0) + f(t2*)(x2-x1) + f(t3*)(x3-x2) + f(t4*)(x4-x3) + f(t5*)(x5-x4) + f(t6*)(x6-x5) • where x0=0, x1=1/4, x2=1/3, x3=1/2, x4=3/4, x5=4/5, x6=1: • L(P,f) = (arctan 0)(1/4 – 0) + (arctan 1/4)(1/3– ¼)+ (arctan 1/3)(1/2-1/3) + (arctan 1/2)(3/4-1/2) + (arctan 3/4)(4/5-3/4) + (arctan 4/5)(1-4/5) • (0)(.25) + (.245)(.083) + (.322)(.167) + (.464)(.250) + (.644)(.050) + (.675)(.200) • 0+ .020+ .054+ .116 + .032 +.135  0.357

  3. To locate the maximum of f on the subintervals of [0,1] and compute U(P,f ), sketch the graph of f(x)=arctan x. Then t1**=1/4, the point where f has its maximum value on the subinterval [0,1/4], t2**=1/3, the point where f has its maximum value on [1/4,1/3], t3**=1/2, the point where f has its maximum value on the subinterval [1/3,1/2], t4**=3/4, the point where f has its maximum value on the subinterval [1/2,3/4], t5**=4/5, the point where f has its maximum value on the subinterval [3/4,4/5] and t6**=1, the point where f has its maximum value on the subinterval [4/5,1].

  4. t1**=1/4, t2**=1/3, t3**=1/2, t4**=3/4, t5**=4/5, t6**=1 x0=0, x1=1/4, x2=1/3, x3=1/2, x4=3/4, x5=4/5, x6=1 • Hence • U(P,f) = R(P,T**,f) = f(t1**)(x1-x0) + f(t2**)(x2-x1) + f(t3**)(x3-x2) + f(t4**)(x4-x3) + f(t5**)(x5-x4) + f(t6**)(x6-x5). • = (arctan 1/4)(1/4 – 0) + (arctan 1/3)(1/3– ¼) + (arctan 1/2)(1/2-1/3) + (arctan 3/4)(3/4-1/2) + (arctan 4/5)(4/5-3/4) + (arctan 1)(1-4/5) • (.245)(.250) + (.322)(.083) + (.464)(.167) + (.644)(.250) + (.675)(.050) + (.785)(.200) • .061+ .027+ .076+ .161 + .034 +.157  0.516 Let A be the area of the region bounded by the graph of f(x) = arctan x, the x-axis and the line x=1. By Prop. 3.2.32, A  ½ [L(P,f)+U(P,f)] = ½ [ 0.357 + 0.516 ] = 0.437 with error less than ½[U(P,f)-L(P,f)] = ½ [ 0.516 – 0.357 ] = 0.080. Using the Riemann sum of Example 3.2C (3), the inequality L(P,f) R(P,T,f) U(P,f) is illustrated by: 0.357  0.437  0.516

More Related