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img009xs3 - 166 Chapter 7 Integration Techniques...

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Unformatted text preview: 166 Chapter 7 Integration Techniques, L’Hépital's Rule, and Improper Integrals voke-kl 32(1 — 5"" + ) 1 32 _ , 32(1—e-k‘) . _h 69- :13; k - 113:, k + £51009 > + —kr =11“132(0 te ) . V0 + 11m — = 32! + v0 k—>O 1 k—)0 e’“ 70. Let N be a fixed value for n. Then . xN-1_. (N—1)xN-2_. (N—1)(N—2)x”‘3_ _, [(N~1)! El: 8" _}l§§c e" ‘31 e‘ _"‘_}1>"§c e" 71. Area of triangle: %(2x)(1 — cos x) = x — x cosx Shaded area: Area of rectangle — Area under curve X X 2x(1— cosx) — 2]. (1 —- cos t)d! = 2x(1— cosx) — 2|:t - 31ml) 0 = 2.x(1— cosx) — 2(x — sinx) = 25inx — lxcosx RatiO' lim x—xcosx _ lim l+xsinx—cosx 'x—mZSinx—Zxcosx xe02cosx+2xsinx—2cosx . 1+xsinx—cosx =11m——_— x—ro lxsmx 1imxcosx + sinx + sinx x—m 2xcosx + Zsinx 1imxcosx+251nx l/cosx xaoncosx+ZSinx l/cosx lim x+2tanx xaolx+2tanx 1. 1+Zseczx=§ 121112+Zsec2x 4 72. f(x) = "k; 1 k = , f(x) = x — 1 k= o1 f(x) = ”3.1—1 — 10(x°’— 1) k-OOl —x°m_1-l °°‘- '— 1 f(x) _ 0.01 00(X 1) _ k — 1 _ . xkfln ) 1:11-311‘ x k _ 111153+ 1 x lnx 73. f(x) = x3, g(x) = x2 + 1, [0, 1] 74- f(x) = %g(x) = x2 - M112] 1%) —f(a) = E f(2) -f(1) ’( ) gm — g(a) we) 3(2) — gm = 2ch) f0) - (0) = E —1/2 —1/c2 g(1) — 3(0) 26 —3— = 26 l = 3_C 1 1 1 2 "E 2 'fi c =§ 21:3 = = 0. (See Exercise 56) ...
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