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Ch 5 solutions

# A x1 3xdxket chapter 5 1sheet2number 24 page number

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Unformatted text preview: 2π /4) π /8 √2 − 2)π /8 (a) + 2 1)(3/ /4) ( (0)(π π /2) ( ( 2/2)( 4) 3( 2)( − 2) = 3( 2. (a) ( 2.2/2)(π /2) +/(−1)(3π /4) + (0)(π /4) +(0)(π / π /4) = 3( 2 − 2)π /8 2/ 2. (a) ( π /4 2)(π /2) + (−1)(3π /4) + (0)(π /2) + ( 2/2)(π /4) = 3( 2 − 2)π /8 138 (b) 3 Chapter 5 (b) 3π /4 (b) 3π /4 (b) 3π /4 3. n (a) (−9/4)(1) + (3)(2) + (63/16)(1) + (−5)(3) = + 117/16 n n 3. (a) (−9/4)(1) (63/16)(1) + (−5)(3) − (−5)(3) 3. (a) (−9/4)(1) + (3)(2) + + (3)(2) + (63/16)(1) = −117/16= −117/16 3. (a)a (−b /) = (a+− b ) ++a − b ) + · ·++−a − b=)−117/16 a + · · · + a ) − (b + b + · · · + b ) = (b) k 3 9k 4)(1) 1 (3)(2) ( (63/16)(1) · ( (5)(3) n = (a1 + 2 67. ( ak − bk . 2 2 n n 1 2 n (b) 3 1 (b) −3 (b) 3 k=1 k=1 k=1 4. (a) (−8)(2)(a)(0)(1) + (0)(1) + (8)(2) = 0 + (8)(2)(b)0 2 4. + (−8)(2) (0)(1) + (8)(2) = = (b) 2 4. (a) (−8)(2) + (0)(1) + + (0)(1) + (0)(1) 0 (b) 2 4. (a) (−8)(2) + (0)(1) + (0)(1) + (8)(2) = 0 (b) 2 Exercise Set 5.5 2 2 2 22 5. 2 x dx 2 5. x2 dx 1. 5. −1(42x dx +−1 /2)(1) + (4)(2) = 71/6. (a) x /dx 3)(1) (5 5. −1 −1 2 23 6. 2 x dx 6. 3 6. 1 (b) 2. x3x dx 6. dx 1 x3 dx 1 1 3. (a) 3 (−9/4)(1) + (3)(2) + (63/16)(1) + (−5)(3) = −117π/2 /16. (b) 3. 3 π /2 3 π /2 2 − 8. x 7. 2 3 4x(17. 3x)dx x(1 − 3x)dx π /2 sin 8.2 dx 4 8. 0 sin x dx sin x dx 7. − 4x(1 − 3x)dx 4x 8. sin2 x dx0 7. 2 3−3 (1 − 3x)dx −3 0 n n 5. x2 −3 dx 0 ∗ n n n∗ n xk x∗ 9. −1 (a) lim (a) n 2xlim xk ; a =21,∗b = 2 a = 1, b =(b) ∆ lim (b) n ∗lim ∗ ∆xk ; a =k0, b = 1 a = 0, b = 1 k∗ ∗xk 9.xk →0 xk ∆b k ; 2 x= 2(b)max ∆xk →0 x+ → 9. (a)max ∆ lim k=1 2xk ∆xk ; a = 1, lim k=1 xk∆∗ k1 0 ∆xkx∗ + 10, b k ; 1 ; a = ∆x = x 9. 16:01 max ∆xk →0 max ∗ ∆k →0 k=1 x ∆x (b) lim xk → max k +∆x =1 = 1k 3 vember 10, 2008 (a) maxlimk →0 ET2chapter; 5”= 1, b = 2 number 24 Pagemaxx∆→0 0 k=1∗xk blackk ; a k 0, b = 1 ”ISM nk=1 k xk a Sheet ∆x maxnumber 248 k + 1 ∆k x n =1n k 7. (a) x(1 − 3x)dxkET chapter 5” = 1,Sheet2number 24 Page number =1 4 vember 10, 2008 −3 248 black 10. 16:01 10.”ISM n lim x∗ ∆xk , a x∗ b = k ∆ number 24 Page number Page black ber 10, 2008 10. (a) max ∆xk →0 chapter ∗5” xk , aSheet b k , a = 1, b = 2number 24 248 number 248 16:01 ”ISM16:01 max ∆∗x →0 limET =k1, x November 10, 2008 (a)k=1 limkk ∆ ET chapter = 2 ”ISM Sheet black x∆x , a = 1, b = 5” 10. (a) lim xk 2 max ∆xk →0 kk=1 max ∆xk →0n k=1 n k=1 n Chapter 5 x∗ n k n 9. (a) lim 2x∗ ∆xcos x=) 1, xk= a = −π /2, b = π /2 ; a ∗ ∆ b , 2. ∗ (b) lim k+ (b) max ∆xk →0 n (1limk lim (b) ∗ + 1 ∆xk ; a = 0, b = 1.Chapter 5 248 ∆ / a = −π→0 =1 k k/ (b)max ∆xk →0 kn max(1x+→0 ∗x∗ ) ∆xk , a xk )−πxk , max ∆x/2 2,kb =xπ /2 cosk (1 + cos b 248 Chapter 5 n 248 limk→0 (1 ∆ cos x ) k x , a = = π /2, 2, = =/π Chapter 5 ￿ x∗ (b) lim =1=1 + k k=1 k − b π2 max ∆xk ￿ k∆ k k=1 ∗ ∆x ; a = 1, b = 2 maxlimk →0 ∆x 9. (a) (b) lim k n n k=1 2xk ∗ + 1 ∆xk ; a = 0, b = 1 max ∆xk →n depends 0 11. Theorem 5.5.4(a)￿ ∗ 1 on ￿ fact that a constant canmax ∆xk →1 k=1 xkx∗ 1 sign, which by Deﬁnition 5.5.1 the move past￿ integral n an 10 n 9. (a) A = 1 k=1= xk ∆x ; a lim ￿ ∗ 9/2 2 =2 (b) −A = ￿1 xk∗ k = xn3 2 =k0, b 1 x k; 2) 11. possible 11.x (3)(3)constantkcan = 1,/b past a limit and/or maxlim− 0 (1)(1∗+= −∆￿/a +∗2) = = 3/2 ∗ 22 (b) ∗ Alim because A = (3)(3) = x = move summation−xk + x − = 0, ∆xk → 9. 11. (a)max ∆1 k(a)k2xk ∆xk ;2a = 1,2b9k ∆xk ; a = 1, b(b)2 (b) a −∆x= 1 k=1 sign.∆2)k=a−3/2 b = 1− ; a = 0, b = 1 (a) lim = →0 a=1 = 9/ limA k(b) (1)(1 + 1 ;(1)(1 ∗ 9. 2 (3)(3) lim = 2 A →0 (a) − 2k →0 2 −3/xk + 1 max ∆xk max ∆x → − (1)(1 +12) = max 2 11. (a) A = (3)(3) max ∆2 (b) −A k= 0 k=12xk +∆xk →0 2 =1 n = 9/ x k ￿￿ k=1 k=1 2y y2 y y x∗ ∆xk , a = 1, b = 2 10. (a) lim y y n y max ∆xk →n ￿ ￿ k 0 y x n –2 –1 k￿ ￿= x 10. (a) lim ￿ =1 x∗ ∆xk , a ￿ 1, b = 2 –2 x –1 k –2 –1 max ∆xk(a) x 10. (a) lim →0 k=1x∗lim k , a = 1, x∗= 2 k , a = 1, b = 2 b k ∆x 10. –2 A 1 – k ∆x n max max ∆xk →0 A ￿ ∆xk →0 k=1 A k=1 A Ak (b) lim A A n (1 + cos x∗ ) ∆xk , a = −π /2, b = π /2 x max ∆xk →n ￿ x A0 k n 3 (b) lim ￿ =1 (1 + x x x∗3 ∆xk , a = −π /2, b = π /2 cos ￿ ) max ∆xk(b) (1 3 lim x∗ ) ∆x(1 a = −x∗ ) ∆b k , π /2 −π /2, b = π /2 (b) lim →0 k=1 + cos k k k , + cos πk 2, x = a = / 1 1 3 max ∆xk →0 max ∆xk →0 13. (a) Ak= (3)(3) = 9/2. (b) −A = − (1)(1 + 2) = −3/2. k=1 =1 1 2 /2 (c) −A15.5.4(a) depends = =the1fact that /2constant canA1 + (d)= −A integral sign,2which by + A2 = −A + 8 on15− + 8 = 15a − 1A (d) − move past0an + A = 0 A2 11. Theorem A + A = 1 1 ++ 8 = 15/2 2 2 2 (c) − 1 (c) 2 − (d) −A1 + A2 = 01 (c) −A1...
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