Chapter 5 25

# Chapter 5 25 - SECTION 5.3 THE FUNDAMENTAL THEOREM 0F...

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Unformatted text preview: SECTION 5.3 THE FUNDAMENTAL THEOREM 0F CALCULUS U 457 26. f: w‘5 dm does not exist because the function ﬂat) 2 32‘5 has an inﬁnite discontinuity at a: = 0; that is. f is discontinuous on the interval [—2. 3]. 2 5 2 27. / —3 da: does not exist because the function f(m )— — —3 has an inﬁnite discontinuity at x— — 0 that is f 1s 5 m a" discontinuous on the interval [—5. 5]. 28. ff" cos6d6 = [sin6]:7r = si1127r — sin7r = 0 — 0 = 0 29. f02cc(2+a:5)dar:f02(2m+w6)dm= [\$132+ 13:7]0: (4+%) —(0+0):i76 4 1 4 1 331/2 4 30. — _ 42 -: _ = 21/2 :2 =4— 2—2 [ﬁdm— [an (id [1/2]1[\$ E \/_ 2f 31. fTr/4 see2 t dt: [tant 77/4 0 lo =tang—tan021—0:1 32- f01(3+\$\/§)d33:f01<3+x3/2)dm: [31’+%m5/2];: “3+9 _0] = 151 33. ff” csc2 6d6 does not exist because the function f(6) = csc2 6 has inﬁnite discontinuities at 6 : 7r and 6 : 27r; that is. f is discontinuous on the interval [7r, 2a]. 34. far/6 csc 6 cot 6 d6 does not exist because the function f(6) : csc 6 cot 6 has an inﬁnite discontinuity at 6 = 0; that IS, f 18 discontinuous on the interval [0. El‘ 23 2 1 x 35./101d\$:[10]::£1; i 0 O 9 9 1 1 1 9 35/ —dw:—/ Edw=%[ln|at|] :%(ln9*ln1)=éln9—0:1n91/2:ln3 1 1 1 In 10 In 10 In 10 In 10 ﬁ/2 6 ﬁ/z 1 37. / dt 6 dt 1/2 V 1 * t2 /2 V1 — t2 66=(-%):() 1 1 321/ 4 dt:4/ 1 dt=4[tan—1t];:4(1an’11~tan’10) :4(gi0) =7r 0 0 : 6[sin_1t]\/§/2 : 6[sin71( 1/2 ME )— sin_1( >l NIH II N. U ml: at: t2+1 1+12 39- f: 6““ du — [e eu+1]71 i 62 — 80 : e2 — 1 [or start with 6““ : euel] 4 2 2 2 4ll./2 +U2 duz/ (4u_3+u_1)du: [iui2+lnlul = ——2+lnu 1 1‘31 T2 1 “2 1 : (—é +ln2) — (72+ln1) : % +ln2 4‘- foszx = fol 1.4.1.3113”. =1éx51é+1éw6132<é~ 0) + (% ~%) 210.7 42. ff” f(.’17)d\$ : ffﬂxdani—f; sinmda: =[ — [cosa:]z)T = (0 ~ "7) — (cos71' — cos0) bob—1 2 W2 7T :—77(—1—1):2¥7 43. From the graph. it appears that the area is about 60. The actual area is 27 f0” Il/Bdw : [gal/3L 2711-81 — 0 = 9—3—3 : 60.75. This is g 0fthe area of the viewing rectangle. ...
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