# Sinh x0 1 x x 1 x2 x0 2x cosh x0 tanh x0 coth

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Unformatted text preview: 1 √ ( x + x/4)dx = (2x3/2 /3 + x2 /8) 0 4 = 22/3 0 2 2 (y − 1/y 2 )dy = (y 2 /2 + 1/y ) 3. A = 1 =1 1 2 2 (2 − y + y )dy = (2y − y /3 + y /2) 2 4. A = 3 2 0 = 10/3 0 4 16 (4x − x2 )dx = 32/3 5. (a) A = (b) A = 0 √ ( y − y/4)dy = 32/3 0 y (4, 16) y = 4x y = x2 5 x 1 6. Eliminate x to get y 2 = 4(y + 4)/2, y 2 − 2y − 8 = 0, (y − 4)(y + 2) = 0; y = −2, 4 with corresponding values of x = 1, 4. 1√ 4√ √ (a) A = [2 x − (−2 x)]dx + [2 x − (2x − 4)]dx 0 1 √ 0 y = 2x – 4 x 1 1 (1, -2) 4 −2 1 7. A = [(y/2 + 2) − y /4]dy = 9 2 (b) A = (4, 4) y2 = 4x (2 x − 2x + 4)dx = 8/3 + 19/3 = 9 4 xdx + = √ 4 y √ ( x − x2 )dx = 49/192 y 1/4 y = √x (1, 1) y = x2 x 1 4 256 257 Chapter 8 π /2 2 [0 − (x3 − 4x)]dx 8. A = (0 − cos 2x)dx 9. A = π/4 0 2 π /2 (4x − x3 )dx = 4 = =− cos 2x dx = 1/2 0 π/4 y y 1 x y = cos 2x 2 x 3 y = 2x 3 – 4x 6 -1 3π/4 10. Equate sec2 x and 2 to get sec2 x = 2, 11. A = sin y dy = √ 2 π/4 y y 2 (# , 2) (3 , 2) y = sec 2 x 1 x = sin y x 9 √ sec x = ± 2, x = ±π/4 3 π /4 A= −π/4 (2 − sec2 x)dx = π − 2 x 2 ln 2 [(x + 2) − x ]dx = 9/2 2 12. A = −1 13. A = e 2x −e x dx = 0 = 1/2 y y (2, 4) y = e 2x 4 y= (–1, 1) x2 x 2 y = ex x=y–2 x ln 2 e 14. A = 1 dy = ln y y y e =1 1 e 1 x 1/e 1 1 2x e − ex 2 ln 2 0 Exercise Set 8.1 258 3 − x, x ≤ 1 , 1 + x, x ≥ 1 15. y = 2 + |x − 1| = 1 y (–5, 8) y = –1x+7 5 1 − x + 7 − (3 − x) dx 5 A= −5 y= 1+x 5 1 − x + 7 − (1 + x) dx 5 + 1 1 4 x + 4 dx + 5 = −5 (5, 6) y = 3–x 5 1 x 6 6 − x dx 5 = 72/5 + 48/5 = 24 2/5 1 (4x − x)dx 16. A = (x3 − 4x2 + 3x)dx 17. A = 0 0 1 3 (−x + 2 − x)dx + [−(x3 − 4x2 + 3x)]dx + 2/5 1 2/5 (2 − 2x)dx = 3/5 3x dx + = = 5/12 + 32/12 = 37/12 1 4 2/5 0 ( y 2 5 , 8 5 ) -1 y = −x + 2 y = 4x 4 (1, 1) x -8 y=x 9 18. Equate y = x3 − 2x2 and y = 2x2 − 3x to get x3 − 4x2 + 3x = 0, x(x − 1)(x − 3) = 0; x = 0, 1, 3 with corresponding values of y = 0, −1.9. 1 [(x3 − 2x2 ) − (2x2 − 3x)]dx A= -1 0 3 -2 3 [(2x3 − 3x) − (x3 − 2x2 )]dx + 1 1 3 (x3 − 4x2 + 3x)dx + = 0 (−x3 + 4x2 − 3x)dx 1 8 37 5 += = 12 3 12 19. From the symmetry of the region 5π/4 A=2 1 √ (sin x − cos x)dx = 4 2 π/4 0 -1 o 259 Chapter 8 0 20. The region is symmetric about the origin so −1 −2 (y 3 − y )dy + −(y 3 − y )dy 0 = 1/2 0 A=2 1 21. A = (x3 − 4x)dx = 8 1 3.1 -1 -3 1 3 -1 -3.1 1 4.1 y 3 − 4y 2 + 3y − (y 2 − y ) dy 22. A = 0 4 y 2 − y − (y 3 − 4y 2 + 3y ) dy + 1 = 7/12 + 45/4 = 71/6 -2.2 12.1 0 23. Solve 3 − 2x = x6 +2x5 − 3x4 + x2 to ﬁnd the real roots x = −3, 1; from a plot it is seen that the line 1 is above the polynomial when −3 &lt; x &lt; 1, so A = −3 (3 − 2x − (x6 +2x5 − 3x4 + x2 )) dx = 9152/105 1 24. Solve x5 − 2x3 − 3x = x3 to ﬁnd the roots x = 0, ± 2 √ √ (6+2 21)/2 27 + (x3 − (x5 − 2x3 − 3x)) dx = A=2 4 0 k 25. √ 2 ydy = 9 √ 2 ydy k x2 dx k 1 13 k = (8 − k 3 ) 3 3 y 1/2 dy k 0 2 x2 dx = 0 9 y 1/2 dy = 7√ 21 4 k 26. k 0 √ 6 + 2 21. Thus, by symmetry, k3 = 4 √ k= 34 2 2 3/2 k = (27 − k 3/2 ) 3 3 k 3/2 = 27/2 2/3 k = (27/2) √ 3 y x = √y = 9/ 4 y y=9 y=k x 2 x=k x Exercise Set 8.1 260 2 (2x − x2 )dx = 4/3 27. (a) A = 0 (b) y = mx intersects y = 2x − x2 where mx = 2x − x2 , x2 + (m − 2)x = 0, x(x + m − 2) = 0 so x = 0 or x = 2 − m. The area below the curve and above the line is 2−m 2−m 1 1 (2 − m)x2 − x3 2 3 0 0 √ 3 so (2 − m)3 /6 = (1/2)(4/3) = 2/3, (2 − m)3 = 4, m = 2 − 4. (2x − x2 − mx)dx = 28. The line through (0, 0) and (5π/6, 1/2) is y = 5π/6 A= 0 2−m [(2 − m)x − x2 ]dx = = 0 1 (2 − m)3 6 y 3 x; 5π y = sin x 1 √ 5 3 3 x dx = − π+1 sin x − 5π 2 24 5c 6 ( , c 1 2 ) x 29. (a) It gives the area of the region that is between f and g when f (x) &gt; g (x) minus the area of the region between f and g when f (x) &lt; g (x), for a ≤ x ≤ b. (b) It gives the area of the region that is between f and g for a ≤ x ≤ b. 1 30. (b) (x1/n − x) dx = lim lim n→+∞ n→+∞ 0 x2 n x(n+1)/n − n+1 2 1 1 n − n+1 2 = lim 0 n→+∞ = 1/2 31. The curves intersect at x = 0 and, by Newton’s Method, at x ≈ 2.595739080 = b, so b A≈ (sin x − 0.2x)dx = − cos x + 0.1x2 0 b 0 ≈ 1.180898334 32. By Newton’s Method, the points of intersection are at x ≈ ±0.824132312, so with b b (cos x − x2 )dx = 2(sin x − x3 /3) b = 0.824132312 we have A ≈ 2 0 33. distance = ≈ 1.094753609 0 |v | dt, so 60 (3t − t2 /20) dt = 1800 ft. (a) distance = 0 T (b) If T ≤ 60 then distance = (3t − t2 /20) dt = 0 32 1 T − T 3 ft. 2 60 T (a2 (t) − a1 (t)) dt = v2 (T ) − v1 (T ) is the diﬀerence in the velocities 34. Since a1 (0) = a2 (0) = 0, A = of the two cars at time T . 0 y 35. Solve x1/2 + y 1/2 = a1/2 for y to get y = (a1/2 − x1/2 )2 = a − 2a1/2 x1/2 + x a a (a − 2a1/2 x1/2 + x)dx = a2 /6 A= 0 x a 261 Chapter 8 √ 36. Solve for y to get y = (b/a) a2 − x2 for the upper half of the ellipse; make use of symmetry to a 4b a 4b 1 2 b get A...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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