CHAPTER 5 The Integral 5.1 Approximating and Computing Area Preliminary Questions 1. The interval [2, 5] is divided into 6 subintervals in order to calculate R6 for some function. What are the right-endpoints of those subintervals? What are the left-endpoints? −2 2. If f (x) = x on [3, 7], which is larger: RN or L N ? 3. Which of the following pairs of sums are not equal? (a) 4 i, 4 i=1 =1 (b) 4 j 2, 5 k2 j=1 k=2 (c) 4 j, 5 (i − 1) j=1 i=2 4 ( + ), 5 ( − ) (d) i=1 i i 1 j=2 j 1 j 4. The interval [1, 5] is divided into 16 subintervals. (a) What are the left endpoints of the first and last subintervals? (b) What are the right endpoints of the first two subintervals? 5. True or False: (a) The right-endpoint rectangles lie below the graph of an increasing function. (b) If f is monotonic, then the area under the graph lies in between RN and L N . (c) If f is not monotonic, then L N and RN may converge to different limits as N →∞. (d) If f (x) is constant, then the right-endpoint rectangles all have the same height. Exercises 1. An athlete runs with velocity 4 mph for half an hour, 6 mph for the next hour, and 5 mph for another half-hour. Compute the total distance traveled and indicate on a graph how this quantity can be interpreted as an area. ( ) ( 1 ) + ( )( ) + ( ) ( 1 ) = 1 The total distance traveled is 4 2 6 1 5 2 10 2 miles. 1 2 Chapter 5 The Integral 6 5 4 mph 3 2 1 0 0.5 1 1.5 2 hours 2. Figure 1 shows the velocity of an object over a 3-minute interval. Determine the distance 3. traveledAssume overthat the the velocity intervals of[0 an, 3] objectand [1 is, 2 32.5t]ft/s.(remember Use Eq. to (?? convert) to determine from mph the to distance miles per minute).traveled by the object over the time intervals (in seconds) [0, 2] and [2, 5]. The total distance traveled is given by the area under the graph of v = 32t. [ , ] Figure1 ( 1 )( ) = During the interval 0 2 , the object travels 2 2 64 64 ft. [ , ] 1 ( )( − ) + ( )( ) = During the interval 2 5 , the object travels 2 3 160 64 3 64 336 ft. 160 140 120 100 80 60 40 20 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 4. Consider f (x) = 2x + 3on[0, 3]. 2 5. Let (a)f (x)Determine= x + x − the2. left- and right-endpoints if [0, 3] is divided into 6 subintervals. (a) Calculate R and L for the interval [2, 5]. (b) Compute3 R6 and3 L6. (b) Sketch(c) Find the the graph exact of areaf and using the geometry rectangles and that compute make up the each error of in the the approximations. two calculations of (b). Let f (x) = x 2 + x − 2andseta = 2, b = 5, n = 3, h = x = (b − a) /n = (5 − 2) /3 = 1. (a) Let xk = a + kh, k = 0, 1, 2, 3. Selecting left endpoints of subintervals, xk , k = 0, 1, 2, or {2, 3, 4},wehave 2 2 L3 = f (xk )x = h f (xk ) = (1) (4 + 10 + 18) = 32 k=0 k=0 Selecting right endpoints of subintervals, xk , k = 1, 2, 3, or {3, 4, 5},wehave 3 3 R3 = f (xk )x = h f (xk ) = (1) (10 + 18 + 28) = 56 k=1 k=1 (b) Here are figures of the three sets of rectangles that approximate the area under the curve f (x) over the interval [2, 5]. 5.1 Approximating and Computing Area 3 x2+x−2 x2+x−2 30 30 25 25 20 20 15 15 10 10 5 5 2 2.5 3 3.5 4 4.5 5 2 2.5 3 3.5 4 4.5 5 x x 6. Let f (x) = cos x. 7. Consider f (x) = 5x on [0, 3]. [ , π ] (a) Calculate R4 and L4 for the interval 0 2 . (a) Find(b) Sketch a formula the graphfor RN of(usef and formula the rectangles (??)). that make up each of the approximations. (b) Is R bigger or smaller than the area under the graph? (c) NIs the area under the graph larger or smaller than R4?ThanL4? (c) Calculate limN→∞ RN . (d) Calculate the area under the graph using geometry and verify your answer to (c). Let f (x) = 5x on [0, 3].Letn be a (symbolic) positive integer and set a = 0, b = 3, h = x = (b − a) /n = (3 − 0) /n = 3/n. (a) Let xk = a + kh = 3k/n, k = 1, 2,... ,n be the right endpoints of the n subintervals of [0, 3].Then n 3 n 3k 45 n R = h f (x ) = 5 = k n k n n n2 k=1 k=1 k=1 2 = 45 n + n = 45 + 45 n2 2 2 2 2n (b) Since f (x) = 5x is increasing, the right endpoint approximation Rn is greater than the area under the graph. 45 45 45 (c) The area under the graph is lim Rn = lim + = = 22.5. n→∞ n→∞ 2 2n 2 = 1 = 1 ( )( ) = 45 (d) The area under the graph is the area of a triangle: A 2 BH 2 3 15 2 ,which agrees with the answer in (c). 8. Estimate R6 and L6 over [0, 1.5] for the function shown in Figure 2. 9. Estimate R6 and L6 for the graph in Figure 3. Figure 2 1 .5 0 1 2 1 4 5 2 3 3 3 3 Figure 3 ( ) [ , ] = = ( − )/ = 1 Let f x on 0 2 be given by the figure in the text. For n 6, h 2 0 6 3 , { }6 = , 1 , 2 , , 4 , 5 , xk k=0 0 3 3 1 3 3 2 . 4 Chapter 5 The Integral Therefore 1 5 1 1 5 7 9 3 L6 = f (xk ) = + + + + 1 + = 1.625 3 k=0 3 2 8 8 8 4 1 6 1 5 7 9 3 R6 = f (xk ) = + + + 1 + + 1 ≈ 1.792 3 k=1 3 8 8 8 4 In Exercises 10–19, calculate the approximation for the given function and interval. 10. R8,7− x, [3, 5] 2 11. R6,2x − x + 2, [1, 4] ( ) = 2 − + [ , ] = = ( − )/ = 1 Let f x 2x x 2on 1 4 .Forn 6, h 4 1 6 2 , { }6 = , 1 , , 1 , , 1 , xk k=0 1 1 2 2 2 2 3 3 2 4 . Therefore 1 6 = ( 2 − + ) = . R6 2xk xk 2 47 5 2 k=1 12. 2 L6,2x − x + 2, [1, 4] 3 13. R4, x − 2x + 5, [0, 1] ( ) = 3 − + [ , ] = = ( − )/ = 1 Let f x x 2x 5on 0 1 .Forn 4, h 1 0 4 4 , { }4 = , 1 , 1 , 3 , xk k=0 0 4 2 4 1 . Therefore 1 4 = ( 3 − + ) ≈ . R4 xk 2xk 5 4 140625 4 k=1 14. 3 R4, x − 2x + 5, [−2, 2] [ π , 3π ] 15. R5,sinx, 4 4 ( ) = [ π , 3π ] = = (3π/4−π/4) = π Let f x sin x on 4 4 .Forn 5, h 5 10 , { }5 = π , 7π , 9π , 11π , 13π , 3π xk k=0 4 20 20 20 20 4 . Therefore π 5 R5 = sin xk ≈ 1.402563 10 k=1 16. −1 L5, x , [1, 2] −2 17. L4, x , [1, 3] ( ) = −2 [ , ] = = ( − )/ = 1 { }4 = , 3 , , 5 , Let f x x on 1 3 .Forn 4, h 3 1 4 2 , xk k=0 1 2 2 2 3 . Therefore 1 3 = −2 ≈ . L4 xk 0 927222 2 k=0 18. [ π , π ] L4,cosx, 4 2 2 19. L5, x + 3|x|, [−3, 2] Let f (x) = x 2 + 3 |x| on [−3, 2].Forn = 5, h = (2 − (−3))/5 = 1, { }5 = {− , − , − , , , } xk k=0 3 2 1 0 1 2 . Therefore 4 = ( 2 + | |) = L5 1 xk 3 x 36 k=0 5.1 Approximating and Computing Area 5 20. R&W Compute the average of R5 and L5 to approximate the area under the graph of 21. Calculatef (x) = x the−1 over sums:[3.5, 5]. Explain why the average is more accurate than either endpoint 5 approximation. (a) 3 i=1 5 (b) 3 i=0 4 (c) k3 k=2 4 π (d) sin j j=3 2 When the number of summands is small, we may compute the sum directly. Where applicable (especially when the number of summands is large), we may use summation laws. For illustration, both ways are shown in this exercise. 5 (a) Do this directly: 3 = 3 + 3 + 3 + 3 + 3 = 15. Or use the law for summing a i=1 5 constant: 3 = (3)(5) = 15. i=1 5 5 (b) Here 3 = 3 + 3 + 3 + 3 + 3 + 3 = 18 or 3 = (3)(6) = 18. i=0 i=0 4 (c) Again, k3 = 23 + 33 + 43 = 99 or k=2 4 4 1 k3 = k3 − k3 = = = k 2 k 1 k 1 44 43 42 14 13 12 = + + − + + = 99. 4 2 4 4 2 4 4 jπ (d) Finally, sin = 1 + 0 + (−1) + 0 = 0. j=1 2 In Exercises 22–34, use the Linearity Rules and formulas (??)–(??) to rewrite and evaluate the sums. 22. 6 3 10 j 3 23. j4=1j j=1 10 3 = 10 3 = 104 + 103 + 102 = We have j=1 4 j 4 j=1 j 4 4 2 4 12100.