Integration Solutions 1. Let the volume of the solid of revolution be V. (ax 2) a V= = 2 0 a ( x 2 2) 2 dx (M1) (a2x2 + 4ax + 4 – x4 – 4x2 –4)dx 0 (M1) a 1 4 1 = a 2 x 3 2ax 2 x 5 x 3 5 3 0 3 2 2 = a 5 a 3 units3 3 15 (M1) (A1) 2a 3 π 2 (a + 5) 15 = (C4) Note: The last line is not required [4] 3. (a) y 2=k –1 0 1 2 k 2 1 3 x –1 –2 3 Notes: Award (A1) for the correct intercepts (A1) for graphing over the correct interval (A1) for the correct x-coordinate of the maximum point. 1 (b) Required area = k ( x x 2 )dx – ( x x 2 )dx 0 1 1 k x x x x = 3 0 2 3 1 2 2 3 (M1) 2 3 1 k2 k3 1 1 k3 k2 6 2 3 6 3 3 2 1 = (2 + 2k3 – 3k2) 6 = (A1) (M1) (A1) OR k Required Area = ( x x )dx 2 (M1) 1 k x2 x3 = 3 1 2 (M1)(A1) k3 k2 1 3 2 6 (A1) = 4 [7] 1 a2 1 dx = 0.22 a 1 x2 2 [arctan x ] aa Then = 0.22 4. If (M1) 2 arctan a – arctan a – 0.22 = 0 (A1) a = 2.04 or a = 2.62 (A1) (C3) Notes: Award final (A1) only if both correct answers are shown. If no working is shown and only one answer is correct, award (C1). GDC example: finding solutions from a graph. [3] 5. (a) y = ln x5 – 3x2 (i) 2.5 –0.5 0.5 1 1.5 2 –2.5 –5 –7.5 –10 –12.5 asymptote asymptote (G2) Note: Award (G1) for correct shape, including three zeros, and (G1) for both asymptotes (ii) (b) f (x) = 0 for x = 0.599, 1.35, 1.51 f (x) is undefined for (x5 – 3x2) = 0 x2(x3 – 3) = 0 Therefore, x = 0 or x = 31/3 f (x) = (c) (G1)(G1)(G1) 5 (M1) (A2) 5x 4 6 x 5x 3 6 or x 5 3x 2 x 4 3x 3 (M1)(A1) f (x) is undefined at x = 0 and x = 31/3 (d) For the x-coordinate of the local maximum of f (x), where 0 < x < 1.5 put f (x) = 0 5x3 – 6 = 0 (A1) 3 (R1) (M1) 1 6 3 x= 5 (e) (A1) 3 (A2) 2 The required area is 1.35 A= f ( x)dx 0.599 Note: Award (A1) for each correct limit. [16] 6. Let dv ln xdx u dx dx where u = lnx and Then du 1 and v = x. dx x dv =1 dx (M1) Using integration by parts, 2 1 ln xdx x ln x x x dx (A1) = x ln x – x + C (A1) [3] 7. x-intercepts are = π, 2π, 3π. Area required = 2π π (A1) sin x dx x 3π sin 2π x x dx (M1) = 0.4338 + 0.2566 = 0.690 units2 (G1) (C3) [3] The curves meet when x = –1.5247 and x = 0.74757. x 0.74757 2 3 dx e The required area = 1.5247 1 x 2 = 1.22. (G1) 9. (G1) (M1) (C3) [3] 10. (a) = (b) x3 ln x – 3 x2 ln x dx = x3 1 3 x dx (M1)(A1)(A1) x3 x3 ln x – (Constant of integration not required.) 3 9 2 1 x 2 ln x dx = 1.07 (A1) (C4) 7 8 or ln 2 – 9 3 (A2) (C2) [6] 11. Using integration by parts u = du = d => cos d = sin – sind => cos d = sin + cos v = sin θ dv = cos d Therefore, => ( cos – )d = sin + cos – Note: (M1) (M1)(A1) (A1) 2 +c 2 Award (C5) for sin + cos – (A2) (C6) 2 , ie 2 penalize omission of + c by [1 mark]. [6] 3 12. (a) (A1)(A1) 2 6 4 A 2 g ( x) –4 –3 –2 –1 1 2 3 4 5 6 –2 –4 f(x) –6 Note: Award (A1) for showing the basic shape of f (x). Award (A1) for showing both the vertical asymptote and the basic shape of g (x). (b) (c) (d) (e) (i) x = –3 is the vertical asymptote. (A1) (ii) x-intercept: x = 4.39 ( = e2 – 3) y-intercept: y = –0.901 ( = ln 3 – 2) (G1) (G1) 3 (G1)(G1) 2 f (x) = g (x) x = –1.34 or x = 3.05 (i) See graph (ii) Area of A = (iii) Area of A = 10.6 4 – 1 – x 3.05 2 0 – (ln (x + 3) – 2)dx (M1)(A1) (G1) 4 y = f (x) – g (x) y = 5 + 2x – x2 – ln(x + 3) dy 1 2 – 2x – dx x3 (M1) Maximum occurs when 2 – 2x = dy =0 dx 1 x3 5 – 4x – 2x2 = 0 x = 0.871 y = 4.63 OR Vertical distance is the difference f (x) – g (x). Maximum of f (x) – g (x) occurs at x = 0.871. The maximum value is 4.63. (A1) (A1) (M1) (G1) (G1) 3 [14] 14. (a) R cos = 1, R sin = π R = 2, = 3 cosx + 3 sinx = R cos cosx + R sin sinx (M1) 3 (A1)(A1) Note: 3 Award (M1)(A1)(A0) if degrees used instead of radians. 4 (b) π , 3 π when x fmax = 2 3 ; fmin = 1 (when x = 0) Since f (x) = 2 cos x – (i) Range is [1, 2] (ii) (c) (A1)(A1) (A1) Inverse does not exist because f is not 1:1 (R2) Notes: Award (R2) for a correct answer with a valid reason. Award (R1) for a correct answer with an attempt at a valid reason, eg horizontal line test. Award (R0) for just saying inverse does not exist, without any reason. 2 π = 3 2 π π x– = 3 4 π x= 12 2 cos x – f (x) = OR f (x) = 5 (M1) (A1) (A1) 2 (M1) x = 0.262 (G1) π 12 x= (A1) 3 π (d) I= 1 2 π sec x – dx 0 2 3 (M1) π 1 π π 2 = ln sec x – tan x – 2 3 3 0 1 2 1 3 3 ln = 2 2– 3 1 3 2 3 = 1 ln (3 + 2 3 ). ln = 2 2 – 3 2 3 2 Note: (A1) (A1)(A1) (M1)(AG) 5 Award zero marks for any work using GDC. [16] 5 15. y = ex – e y = ln x Curves intersect at x = 0.233 and x = 1 Area = 1 0.233 (G1) (G1) (ln x e x e)dx (M1)(A1) = 0.201 (G2) (C6) [6] 17. METHOD 1 Region required is given by y (2, 5) (4, 7) x (–3, 0) from gdc outer intersections are at x = –3 and x = 4 Area = 4 (A1)(A1) y1 – y 2 dx –3 (M1) = 101.75 (A3) (C6) METHOD 2 From gdc intersections are at x = –3, x = 2, x = 4 Area = x 2 3 3 – 3 x 2 – 9 x 27 – ( x 3) dx = 101.75 x 3 – ( x 4 2 (A1)(A1)(A1) 3 – 3x 2 – 9 x 27 ) dx (M1)(M1) (A1) (C6) [6] 6 18. Using the chain rule f (x) = 2 cos 5 x 5 2 (a) = 10 cos 5 x 2 (b) f (x) = = (M1) A1 2 f ( x) dx 2 π cos 5 x + c 5 2 A1 π π Substituting to find c, f π = – 2 cos 5 + c = 1 5 2 2 2 c = 1 + 2 cos 2 = 1 + 2 = 7 5 5 5 M1 (A1) f (x) = – 2 cos 5 x + 7 2 5 5 A1 N2 4 [6] 19. Substituting u = x + 2 u – 2 = x, du = dx (M1) (u 2) 3 x3 d x du ( x 2) 2 u2 A1 3 2 u 6u 2 12u 8 du u A1 u du (6) du 12 du 8u 2 du u A1 2 u 6u 12 ln u 8u 1 c 2 ( x 2) 2 8 6( x 2) 12 ln x 2 c 2 x2 A1 A1 N0 [6] 20. e x cos x dx e x cos x e x sin x dx = e x cos x e x sin x e x cos x dx 2 e x cos x dx e x cos x sin x c x e cos x dx ex cos x sin x k 2 Note: (M1)(A1) (M1)(A1) (M1) (A1) (C6) Do not penalize for missing integration constants. [6] 21. Attempting to find point of intersection Intersection at x = 2 Note: Award M1A1 if x = 2 is seen as upper limit of an integral. Using appropriate definite integrals Area = 1.66 (M1) (A1) M2 A2 N2 7 [6] +c=1 M1 c = 1 + 2 2 7 2 5 x 7 2 5 22. 5 5 5 5 e 2x METHOD 1 sin x dx e 2 x cos x 2 e 2 x cos x dx = e 2 x cos x 2 e 2 x sin x 2 e 2 x sin x dx (M1)A1 (M1)A1 = e 2 x cos x 2 e 2 x sin x 4 e 2 x sin x dx 5 e 2 x sin x dx e 2 x 2 sin x cos x e 2 x sin x dx (M1) e2x 2 sin x cos x C 5 A1 N0 METHOD 2 e 2x 1 1 2x sin x dx e 2 x sin x e cos x dx 2 2 = 1 2x 1 1 e sin x e 2 x cos x e 2 x sin x dx 2 4 4 5 2x 1 1 e sin x dx e 2 x sin x e 2 x cos x 4 2 4 e2x e sin x dx 5 2 sin x cos x C 2x Note: 23. (a) (M1)A1 (M1) A1 Do not penalize the absence of constants of integration. [6] 1 f x ln x x 1 x (M1) = ln x (b) (M1)A1 A1 N2 Using integration by parts METHOD 1 ln x 2 dx ln x x x 2 = x ln x 2 2 2 ln x dx x ln xdx = x (ln x)2 2(x ln x x) + C A1A1 (A1) A1 (= x (ln x)2 2x ln x + 2x + C) METHOD 2 ln x 2 dx x ln x x ln x 2 ln x 1dx = x (ln x)2 x ln x (x ln x x x) + C A1A1A1 A1 (= x (ln x)2 2 x ln x + 2x + C) Note: Do not penalize the absence of + C. [6] 8 9