{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 5 19 - 56 57 58 59 60 61 62 63 64 65 66 67 68...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. SECTION5.2 THEDEFlNITEINTEGRAL 3‘ 451 IfOSmS2.then0S:c3S8.sole3+1S9and1S\/m3+1S3.Thus. 1(2—0) gfoz \/m3+1dmg3(2—0) that is,2Sf02 x/x3-l—1da7S6. If? S m S §.then 1 S tans: S V350 1(§ — i) S fl/ftanxdm 3 WW? — g) or 1—"? S f://43tanmdm S 1—7r2‘/§- Let f(x) 2 m3 — 3x + 3 for 0 S :1: S 2. Then f’(m) 2 3:02 — 3 2 3(56 + 1)(a: ~ 1). so f is decreasing on (0.1) and increasing on (1. 2). f has the absolute minimum value f(1) 2 1. Since f(0) 2 3 and f(2) 2 5. the absolute maximum value off is f(2) 2 5. Thus. 1 S x3 — 311: + 3 S 5 form in [0. 2]. It follows from Property 8 that 1.(2—0)Sf02(m3—3:c+3)dmS5-(2—0);thatis.2Sf02(a:3~3$+3)d22S10. The only critical number of f(;£‘) 2 are‘m on [0. 2] is at 2 1. Since f(0) 2 0. f(1) 2 6T1 2 0.368, and f(2) 2 26‘2 2 0.271, we know that the absolute minimum value off on [0, 2] is 0. and the absolute maximum is 6‘1. By Property 8, 0 S 2706’z S e71 forO S :1: S 2 2 0(2 — 0) S f: ate—a” d9: S 64(2 2 0) 2 0 S [02 rte—I dm S 2/6. 37r/4 . Ifivr Sm S gm then 3? Ssinrc S land % S sin2m S 1.50 l(g71'—fi7r)S f sxn2mdm S 1(731’7r~ in); 2 7r/4 that is. ifl’ S [7374/4 siandm S $71: x/av‘1 + 1 2 \/$4 2 3:2.s0f13 \/l‘4 —l— 1da: 2 [13:32 d9: 2 %(33 — 13) 2 %. 0SsinatS1f0rOSmS§.so:vsin:cSm 2 fOW/2$sina:defO"/2mdm:é“ MI: 2 7.2 ) 202] : 3 Using a regular partition and right endpoints as in the proof of Property 2. we calculate 77. ffcflzv) d1: 2 lim f: cf(:c.,:)AasZ 2 lim Ci f(a:i)A$,-, 2 0 lim 2 f(a:i)A$i 20f: f(:c)dm. {:1 As in the proof of Property 2, we write fab f(m) drc 2 lim 2 flan) A22. Now f(:c,-) 2 0 and A2: 2 0. so n—voo 11:1 f (m,) Am 2 0 and therefore 2 f (33.) Ax Z 0. But the limit of nonnegative quantities is nonnegative by i=1 Theorem 2.32. so [5 f(:1:) day 2 0. Since — ]f($)| S f($) S ”(37) . it follows from Property 7 that ~ff|f(x>ld:c:f:f(m)dz:fflf(m>ldw => ]fff(m)drc]5fflf(w)ldm Note that the definite integral is a real number. and so the following property applies: —a S b S a 2 [b] S a for all real numbers b and nonnegative numbers a. ”02” f(.7:) sin 2x dx] S f027r |f(m) sin 2m] d3: [by Exercise 65] 2 02" |f(a:)] ]sin 21?] dx S f5" ]f(a:)] dw by Property 7., since lsin 255] S1 2 ]f(m)] ]sin 2x] S ]f(:v)] n , n . 4 . t4 , 1 z 1 4 11m — 2 11m — E — 2 3: dm n—>(>o n5 'nfioo 71 TL 0 i=1 1'21 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern