Practice Exam 1 Solutions

Transcription

1 Practice Exam 1 Solutions

2 1a. Let S be the region bounded by y = x 3, y = 1, and x. Find the area of S. What is the volume of the solid obtained by rotating S about the line y = 1? Area A = Volume 1 1 x 3 dx A(x) = π(r 2 ) where r = 1 x 3 Volume = V = 1 b a A(x) dx (1 x 3 ) 2 dx Practice Exam 1 Solutions 2 / 72

3 1 b: Area and Volumes Let S be the region bounded by the curves y = x 2, y = x and x. Find the area of S. What is the volume of the solid obtained by rotating S about (i) the x-axis? (ii) the y-axis? Where do the curves meet? Set x 2 = x x 2 x = x(x 1) = Cross at x = and x = 1 Area = 1 x x 2 dx Rotating around the x-axis A(x) = π(r 2 r 2 ) b Volume = A(x) dx Volume = π a 1 (slicing method) x 2 x 4 dx Practice Exam 1 Solutions 3 / 72

4 1 b: Area and Volumes Let S be the region bounded by the curves y = x 2, y = x and x. Find the area of S. What is the volume of the solid obtained by rotating S about (ii) the y-axis? Rotating around the y-axis A(y) = π(r 2 r 2 ) R = y, r = y b Volume = A(y) dy a (slicing method) Volume = π 1 y y 2 dy Practice Exam 1 Solutions 4 / 72

5 1 c: Volumes by slicing Consider the region S bounded by the parabolas y = x 2 1 and y = 1 x 2. Find the volume of the solid with base S and cross-sections perpendicular to S that are squares with bases parallel to the y-axis. V = b a A(x)) dx Slicing Method The curves cross when x 2 1 = 1 x 2, that is, when x = ±1 Base of typical cross-section: b = (1 x 2 ) (x 2 1) Area of cross-section A(x) = b 2 = (2 2x 2 ) 2 V = 1 1 (2 2x 2 ) 2 dx Practice Exam 1 Solutions 5 / 72

6 1 d: Volumes by slicing Find the volume of the solid with the same base S as before and cross-sections that are equilateral triangles. V = 1 1 A(x)) dx Base: 2y = b = (2 2x 2 ) = y = (1 x 2 ) Height = h: h 2 + y 2 = (2y) 2 = h = 3y Area = 1 2 bh = A = 3 y 2 Area of cross-section A(x) = 3(1 x 2 ) 2 V = 1 1 3(1 x 2 ) 2 dx Practice Exam 1 Solutions 6 / 72

7 2 a: Length of a curve y y = f(x) Arc Length Formula a b x L = b a 1 + (f (x)) 2 dx 2a: Find the length of the curve y = 6x + 2 on the interval x 1. dy dx = f (x) = 6 L = dx = 37 dx Practice Exam 1 Solutions 7 / 72

8 2b: Let y = x2 1 x e 2. 4 ln x 2 and find the length of y on the interval dy dx = x 2 1 2x 1 + ( x 2 1 2x )2 = 1 + ( x2 4 2(1 4 ) + 1 4x ) 2 L = = e 2 1 ( x (1 4 ) + 1 4x 2 ) = 1 + (f (x)) 2 dx = e 2 1 ( x x )2 = x x x x dx = 1 4 (e4 7) Practice Exam 1 Solutions 8 / 72

9 2c: Find the length of the curve y = 4x 3 2 on the interval x 2. f(x) = 4x 3 2 = f (x) = 6x 1 2 L = (6x 1 2 ) 2 dx = x dx Arc Length Formula L = b a 1 + (f (x)) 2 dx Practice Exam 1 Solutions 9 / 72

10 3. Work If a constant force F displaces an object a distance d in the direction of the force, the work done is given by W = F d Work = constant force distance When the force F (x) is not constant we approximate the work done over small intervals of length x and add the n approximations: Work F ( x k ) x k=1 Practice Exam 1 Solutions 1 / 72

11 3 a: True or False The work required to stretch a spring 2 inches beyond its natural length is twice that required to stretch it 1 inch. Hooke s Law F (x) = kx Work: W = b a F (x) dx W = W = 1 2 kx dx = k x2 1 2 = k 2 kx dx = k x2 2 2 = 4k 2 Stretch of 1 inch Stretch of 2 inches False: The work required to stretch a spring 2 inches beyond its natural length is 4 times that required to stretch it 1 inch. Practice Exam 1 Solutions 11 / 72

12 3 d It takes 16 J of work to compress a spring.8 m from its equilibrium position. How much work is required to compress it an additional.4 m? Assumption: Set up the coordinate system so a positive force compresses the spring (this is opposite of what we usually do). Work: W =.8 b a F (x) dx where F (x) = kx kx dx = 1 2 kx2.8 = 1 2 k(.8)2 = 16 = k = 5 Work: W = x dx = 5 2 x Practice Exam 1 Solutions 12 / 72

13 3 b True or False: 3 3 π(9 x 2 ) dx represents the volume of a sphere of radius 3. Slice: Disk of radius x where x 2 + y 2 = 9 V = b a A(y) dy A(y) = πx 2 = π( 9 y 2 ) 2 V = 3 3 π(9 y 2 ) dy Practice Exam 1 Solutions 13 / 72

14 3 b True or False: Another viewpoint 3 3 π(9 x 2 ) dx represents the volume of a sphere of radius 3. The half circle of radius r = 3 bounded by y = 9 x 2 when revolved around the x-axis produces a sphere of radius 3. Use the slicing method A typical slice is a disk of radius r = y = 9 x 2 Partition the interval [ 3, 3] The area of a cross-section is A(x) = πr 2 = π( 9 x 2 ) 2 V = 3 3 A(x) dx = V = 3 3 π(9 x 2 ) dx Practice Exam 1 Solutions 14 / 72

15 3 c: Work A force of 3 N is required to maintain a spring stretched from its natural length of 12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to 2 cm? Newtons is the metric unit for force (units are kilograms, meters and seconds) 12 cm equals.12 m Hooke s Law F (x) = kx F (.3) = k(.3) = 3 k = 1, F (x) = 1x Work: W = W =.8 = 1 x2 2 b a 1x dx.8 F (x) dx = 5(.8) 2 = (5)(64) 1 Nm Practice Exam 1 Solutions 15 / 72

16 3 i: Work Using a rope, it takes 8 minutes to lift a 5-pound box of dirt from the ground to a height of 2 feet above the ground. The rope weighs 1.5 pounds per foot and dangles from a platform that is 3 feet above the ground. Find the total work done to lift the box of dirt to a height of 2 feet above the ground, taking into account the weight of the rope. We divide this problem into two parts: Find the work done in lifting the 5 lb box and the bottom 1 feet of rope. Both are lifted exactly 2 feet. The box weighs 5 lbs and the 1 feet of rope weighs = 15 lbs. Work W = ( ) 2 = 13 Then we find the work done in lifting the upper 2 feet of rope. Practice Exam 1 Solutions 16 / 72

17 Coordinate system: y = the bottom and y = 3 at the top Slice the top 2 feet of rope into n equal pieces of length y The weight of the k-th segment is 1.5 y lbs so the force required to lift it is 1.5 y lbs. It is lifted a distance of approximately 3 y k ft W k (1.5 y)(3 y k ) work lifting this segment n Work W (1.5 y) (3 y k ) W = 1.5 k=1 3 1 (3 y) dy Total work: W = 1.5 (work done in lifting the upper 2 feet of rope) 3 1 (3 y) dy + 13 Practice Exam 1 Solutions 17 / 72

18 Work done in pumping oil out of cone-shaped tank Oil depth is 8 W = π ρ g 8 Slice by horizontal planes into circular disk-like slices Cross-section through the coordinate y has radius r = 1 2 y Force on a slice (weight) F (y) = (πr 2 y) ρ g Distance lifted (1 y) m Work lifting the k-th slice W k F k (1 y k ) n Total work W = ( 1 2 y)2 (1 y) dy k=1 W k Practice Exam 1 Solutions 18 / 72

19 3 e A spherical tank of radius r = 12 meters is filled with water to a depth of h = 9 meters. Use the slicing method to find the volume of water. (i) Find an expression for the area A(x) of a cross-section by a horizontal plane through the coordinate x. Set coordinate system with an upward vertical x-axis and center of the sphere at x =. A horizontal circular cross-section through x is shown The radius γ of the circular cross-section satisfies x 2 + γ 2 = 12 2 γ 2 = 144 x 2 = A(x) = πγ 2 = π(144 x 2 ) Practice Exam 1 Solutions 19 / 72

20 Radius r = 12 meters and water depth is h = 9 meters. (ii) Find the volume of water in the tank. h = 9 m Vertical x-axis pointing up with x = corresponding to the center Water level 12 x 3 The volume of the k-th slice (shown) is V k A(x k ) x where A(x) = π(144 x 2 ), area of typical cross-section. Volume = 3 12 A(x) dx = π x 2 dx Practice Exam 1 Solutions 2 / 72

21 3 e Radius r = 12 meters and water depth is h = 9 meters. (ii) Find the volume of water in the tank. (iii) Find the work done in pumping the water out of the tank through a hole in the top of the tank. 9 m Water level 12 x 3 A(x) = π(144 x 2 ), area of typical cross-section V k A(x k ) x, volume of kth-slice F k ρ g A(x k ) x, weight of kth-slice d = 12 x k, distance kth-slice is lifted Work lifting one slice (12 x k )(ρ g)a(x k ) x Total Work = ρ g π 3 12 (12 x)(144 x 2 ) dx Practice Exam 1 Solutions 21 / 72

22 3 f: Find work required to pump the water out the spout. Water depth is 3 m 1. Set up a y-axis pointing up with bottom of the tank at y = 2. We slice the water into n horizontal layers of thickness y and approximate the work required to lift a typical layer. 3. The k-th cross-section is lifted a distance of 5 y k meters 5 y k y k y = 2x 4. We need to find the area A(y) of a typical cross-section. For this, use geometry to find the width. The length is fixed at 8 m. Practice Exam 1 Solutions 22 / 72

23 Two approaches to find the width w. The area of a typical slice is A(y) = 8w 5 y y = 2x k y k 3 y k 3 w y k 3 (1) The equation y = 2x = w = 2x k = y k V = 3 A(y) dy A(y) = 8y = V = OR Volume by slicing 3 8 y dy (2) Use similar triangles w y k = 3 3 = w = y k Practice Exam 1 Solutions 23 / 72

24 The weight of the k-th slice is ρ g V k = 8 ρ g y k y The work required to lift the k-th slice a distance of (5 y k ) is 5 y k y k y = 2x W k 8 ρ g y k y (5 y k ) n Total work W = W k W = 8 ρ g 3 k=1 y(5 y) dy n k=1 8 ρ g y k (5 y k ) y Practice Exam 1 Solutions 24 / 72

25 3f (b) Find the total hydrostatic force on the triangular-shaped end of the tank. 5 y y = 2x k yk 3 yk Area of k-th strip is A k = w y = y k y The depth is approximately 3 y k. Force on k-th strip P k area = [ρ g (3 y k )] y k y n Total Force ρ g (3 y k )y k y F = ρ g 3 k=1 (3 y)y dy (g = 9.81, ρ = 1 for water) Practice Exam 1 Solutions 25 / 72

26 3g Find the mass of a 3 m bar with density (in g/m) of ρ(x) = 15e x/3 for x 3. m k = ρ(x k ) x = 15e x k/3 x m = 3 15e x/3 dx Practice Exam 1 Solutions 26 / 72

27 3h Find the total hydrostatic force on the face of a semi-circular dam with radius of 2 m, when its reservoir is full of water. Set an xy-coordinate system with the y-axis pointing downward and the origin at the center of the dam top. Area of k-th strip is A k = w k y w k = 2 4 yk 2 Depth is approximately y k Force on k-th strip P k area F k [ρ g y k ] 2 4 yk 2 y n Total Force [ρ g y k ] 2 4 yk 2 y F = 2 ρ g k=1 2 y 4 y 2 dy (g = 9.81, ρ = 1 for water) Practice Exam 1 Solutions 27 / 72

28 4 a, b Explain how ln x is defined rigorously. Explain how we can show from this that d dx ln x = 1 x d By the Fundamental Theorem, dx ln x = d x 1 dx 1 t dt = 1 x Practice Exam 1 Solutions 28 / 72

29 4 c Explain how e x is defined rigorously. d dx ln x = 1 x > = ln x is an increasing function. Thus y = ln x passes the horizontal line test (i.e. it is one-to-one) and has an inverse. e x is defined as the inverse of ln x Practice Exam 1 Solutions 29 / 72

30 5 a: True or False The half-life of a radioactive substance does not depend on its amount. Justify your answer. Half-life is the time it takes for the amount present to decay to one-half that amount. Q(t) = Ae kt Q() = A Find t such that Q(t ) = 1 2 Q( Ae kt = A 2 e kt = 1 2 kt = ln ( 1 2 ) t = ln ( 1 2 ) k Half-life Does not depend on A Practice Exam 1 Solutions 3 / 72

31 5 b A bacteria culture grows exponentially and starts with 2 bacteria. In 1 hour there are 4 bacteria. a) Write an equation for f(t), the number of bacteria t hours after the bacteria begins to grow. f(t) = Ae kt We are given f() = 2, f(1) = 4. f(t) = 2e kt Use f(1) = 4 to find k. 4 = f(1) = 2e k = e k = 2= k = ln 2 t ln (2) f(t) = 2e Practice Exam 1 Solutions 31 / 72

32 5 b A bacteria culture grows exponentially and starts with 2 bacteria. In 1 hour there are 4 bacteria. b) At what rate is the bacteria growing after 5 hours? (Please indicate units in your answer.) t ln (2) f(t) = 2e f t ln (2) (t) = 2 ln (2)e f (5) = 2 ln (2)e 5 ln (2) bacteria per hour (c) How long does it take for the number of bacteria to triple? 2e t ln (2) = 6 = e t ln (2) = 3 t ln (2) = ln 3 = t = ln 3 hours ln 2 Practice Exam 1 Solutions 32 / 72

33 6 a: Integration by parts 3x sec 2 x dx 1 Write down uv dx = uv u v dx 2 Choose values for u and v 3 4 u = 3x and v = sec 2 x u = 3 and v = sec 2 x dx = tan x sin x 3x sec 2 x dx = 3x tan x 3 cos x dx (use u-substitution) 3x sec 2 x dx = 3x tan x + 3 ln cos x + C Practice Exam 1 Solutions 33 / 72

34 6 b: Integration by parts x 2 e 5x dx 1 Write down uv dx = uv u v dx 2 Choose values for u and v u = x 2 and v = e 5x u = 2x e 5x dx = 1 5 e5x + C Choose v = 1 5 e5x 3 x 2 e 5x dx = 1 5 x2 e 5x 2 x e 5x dx 5 4 Repeat integration by parts on x e 5x dx Practice Exam 1 Solutions 34 / 72

35 x 2 e 5x dx = 1 5 x2 e 5x 2 x e 5x dx 5 1 x e 5x dx Integration by parts Write down uv dx = uv u v dx 2 Choose values for u and v 3 4 u = x and v = e 5x u = 1 v = 1 5 e5x xe 5x dx = 1 5 x e5x 1 e 5x dx = x e5x 1 25 e5x + C x 2 e 5x dx = 1 5 x2 e 5x 2 5 (1 5 x e5x 1 25 e5x ) + C Practice Exam 1 Solutions 35 / 72

36 6 c: Integration by parts e 2x cos 4x dx uv dx = uv u v dx u = e 2x and v = cos 4x = u = 2e 2x and v = 1 sin 4x 4 e 2x cos 4x dx = 1 4 e2x sin 4x 1 e 2x sin 4x dx 2 Repeat Integration by Parts on e 2x sin 4x dx Practice Exam 1 Solutions 36 / 72

37 e 2x cos 4x dx = 1 4 e2x sin 4x 1 2 e 2x sin 4x dx u = e 2x and v = sin 4x = u = 2e 2x and v = 1 cos 4x 4 e 2x sin 4x dx = 1 4 e2x cos 4x + 1 e 2x cos 4x dx e 2x cos 4x dx = 1 4 e2x sin 4x 1 2 ( e 2x sin 4x dx) e 2x cos 4x dx = 1 4 e2x cos 4x 1 2 ( 1 4 e2x cos 4x e 2x cos 4x dx = e 2x sin 4x ( + 4 cos 4x ) + C 8 e 2x cos 4x dx) Practice Exam 1 Solutions 37 / 72

38 6 d: Integration by parts x 3 ln x dx 1 Write down uv dx = uv u v dx 2 Choose values for u and v u = ln x and v = x 3 = u = 1/x x 3 dx = 1 4 x4 + C Choose v = 1 4 x4 3 x 3 ln x dx = x4 ln x x 3 dx = x4 ln x 4 x C Practice Exam 1 Solutions 38 / 72

39 6 e: Integration by parts uv dx = uv u v dx π/2 (x 4) sin x dx u = x 4 and v = sin x = u = 1 and v = cos x (x 4) sin x dx = (x 4) cos x + cos x dx (x 4) sin x dx = (4 x) cos x + sin x + C π/2 π/2 (x 4) sin x dx = ((4 π 2 ) cos π 2 + sin π ) (4 cos + sin ) 2 (x 4) sin x dx = 1 4 = 3 Practice Exam 1 Solutions 39 / 72

40 Integrals involving trig functions sin 2 x + cos 2 x = 1 tan 2 x + 1 = sec 2 x Double-angle formulas sin (2t) = 2 sin t cos t cos (2t) = 2 cos 2 t 1 cos (2t) = 2 cos 2 t 1 = 2 cos 2 t = 1 + cos (2t) = (cos x) 2 = 1 (1 + cos (2x)) 2 Half-angle formulas (sin x) 2 = 1 (1 cos (2x)) 2 (cos x) 2 = 1 (1 + cos (2x)) 2 Practice Exam 1 Solutions 4 / 72

41 7a : 3π/4 sin 5 x cos 3 x dx π/2 using sin 2 x = 1 cos 2 x sin 5 x cos 3 x dx = (sin 2 x) 2 sin x cos 3 x dx = (1 2 cos 2 x + cos 4 x) sin x cos 2 x dx = (cos 2 x 2 cos 4 x + cos 6 x) sin x dx Use substitution u = cos x, du = sin x dx = (u 2 2u 4 + u 6 ) du = 1 3 u u5 1 7 u7 + C = 1 3 cos3 x cos5 x 1 7 cos7 x + C Practice Exam 1 Solutions 41 / 72

42 7c : cos 3 (9x) sin 2 (9x) dx using cos 2 (9x) = 1 sin 2 (9x) cos 3 (9x) sin 2 (9x) dx = cos 2 (9x) cos (9x) sin 2 (9x) dx = (1 sin 2 (9x)) cos (9x) sin 2 (9x) dx = (sin 2 (9x) 1) cos (9x) dx Use substitution u = sin (9x), du = 9 cos (9x) dx = 1 (u 2 1) du = ( 1 u + u) + C = 1 9 ( 1 + sin (9x)) + C sin (9x) Practice Exam 1 Solutions 42 / 72

43 7 b: tan 5 x sec 4 x dx 1 tan 2 x + 1 = sec 2 x (tan x) = sec 2 x tan 5 x sec 4 x dx = tan 5 x sec 2 x sec 2 x dx tan 5 x (tan 2 x + 1) sec 2 x dx 2 = 3 Using the substitution u = tan x, du = sec 2 x dx tan 5 x sec 4 x dx = u 5 (u 2 + 1) du = u 7 + u 5 du 4 = 1 8 u u6 + C = 1 8 (tan x) (tan x)6 + C Practice Exam 1 Solutions 43 / 72

44 The three basic trig substitutions a2 + x 2 a2 x 2 x2 a 2 x = a tan θ x = a sin θ x = a sec θ Practice Exam 1 Solutions 44 / 72

45 8 a: Trig Substitution Take a = 5 and choose x 5 = sin θ 1 x 2 25 x 2 dx x = 5 sin θ 1 x 2 25 x dx = 2 dx = 5 cos θ dθ 25 x 2 = cos θ 5 5 cos θ (5 sin θ) 2 (5 cos θ) dθ = 1 csc 2 θ dθ 25 = 1 25 cot θ + C = x 2 x + C Practice Exam 1 Solutions 45 / 72

46 8 b: Trig Substitution x x2 + 9 dx Take a = 3 θ 2 2 x + a a x= a tan θ x 3 tan θ x2 + 9 dx = (Take u = cos θ) = 3 x Choose x 3 dx = 3 sec 2 θ dθ x2 + 9 = sec θ 3 3 sec θ 3 sec2 θ dθ = 3 = 3 = tan θ x = 3 tan θ 1 u 2 du = 3 1 u + C 1 cos θ + C = 3 sec θ + C = x C sin θ cos 2 θ dθ Practice Exam 1 Solutions 46 / 72

47 8 c: Trig Substitution x 2 x2 16 dx Let a = 4 and take x = 4 sec θ, dx = 4 sec θ tan θ x2 16 = 4 tan θ Practice Exam 1 Solutions 47 / 72

48 x 2 x2 16 dx Take a = 4 x 2 2 x a x 4 = sec θ x = 4 sec θ θ a x= a sec θ dx = 4 sec θ tan θ dθ x2 16 = 4 tan θ x 2 x2 16 dx dx = (4 sec θ) 2 We evaluate sec 3 θ dθ 4 tan θ 4 tan θ sec θ dθ = 16 sec 3 θ dθ Practice Exam 1 Solutions 48 / 72

49 1 sec 3 x dx Use uv dx = uv u v dx u = sec x and v = sec 2 x u = sec x tan x and v = tan x sec 3 x dx = sec x tan x (sec x tan 2 x dx 2 = sec x tan x 3 = sec x tan x sec x (sec 2 x 1) dx sec 3 x dx + sec 3 x dx = sec x tan x + sec x dx sec x dx sec 3 x dx = 1 2 sec x tan x + 1 ln sec x + tan x + C 2 Practice Exam 1 Solutions 49 / 72

50 Completion of 8 c Step 1. Take x = 4 sec θ, dx = 4 sec θ tan θ, x2 16 = 4 tan θ Step 2. x 2 (4 sec θ) 2 x2 16 dx dx = 4 tan θ sec θ dθ = 16 4 tan θ Step 3. = 16( 1 2 sec θ tan θ + 1 ln sec θ + tan θ + C 2 = 8( x 4 x x 2 x2 16 dx = x x ln x 4 + x C ln x 4 + x C 4 sec 3 θ dθ Practice Exam 1 Solutions 5 / 72

51 9 a: (Partial Fractions) True or False: x + 2 x 2 (x 2 1) can be expressed in the form A x 2 + x + 2 x 2 (x 2 1) = A x + B x 2 + Adding up the right side x + 2 x 2 (x 2 1) C x 1 + = Ax(x 1)(x + 1) x 2 (x 1)(x + 1) + D x + 1 B x 1 + C x + 1 B(x 1)(x + 1) x 2 (x 1)(x + 1) + Cx2 (x + 1) x 2 (x 1)(x + 1) + Dx2 (x 1) x 2 (x 1)(x + 1) Practice Exam 1 Solutions 51 / 72

52 x + 2 x 2 (x 2 1) = = Ax(x 1)(x + 1) + B(x 1)(x + 1) + Cx2 (x + 1) + Dx 2 (x 1) x 2 (x 1)(x + 1) Choose A, B, C so x + 2 = Ax(x 1)(x + 1) + B(x 1)(x + 1) + Cx 2 (x + 1) + Dx 2 (x 1) (comparing numerators) Set x = = 2 = B = B = 2 Set x = 1 = 1 = 2D = D = 1/2 Set x = 1 = 3 = 2C = C = 3/2 Set x = 2 = 4 = 6A + 3B + 12C + 4D = A = 1 Practice Exam 1 Solutions 52 / 72

53 9 b: x 2 + 2x 1 2x 3 + 3x 2 2x dx Factor 2x 3 + 3x 2 2x = x(2x 2 + 3x 2) = x(2x 1)(x + 2) x 2 + 2x 1 x(2x 1)(x + 2) = A x + B x C 2x 1 Place right side over a common denominator and add = A(2x 1)(x + 2) x(2x 1)(x + 2) = + Bx(2x 1) x(2x 1)(x + 2) + Cx(x + 2) x(2x 1)(x + 2) A(2x 1)(x + 2) + Bx(2x 1) + Cx(x + 2) x(2x 1)(x + 2) Comparing numerators, choose A, B, C so that x 2 + 2x 1 = A(2x 1)(x + 2) + Bx(2x 1) + Cx(x + 2) Practice Exam 1 Solutions 53 / 72

54 Solving for A, B, C x 2 + 2x 1 = A(2x 1)(x + 2) + Bx(2x 1) + Cx(x + 2) for all x Set x = = 1 = 2A = A = 1/2 Set x = 2 = 1 = 1B = B = 1/1 Set x = 1/2 = 4/9 = C(2/3)(8/3) = C = 1/5 x 2 + 2x 1 2x 3 + 3x 2 2x dx = 1 x + 1/2 x /2 2x 1 dx x 2 + 2x 1 2x 3 + 3x 2 2x dx = ln x ln x ln 2x 1 + C 4 Practice Exam 1 Solutions 54 / 72

55 9 c: x 2 + 2x 1 dx (Partial Fractions) x 3 x x 2 + 2x 1 x 3 x = x2 + 2x 1 x(x 1)(x + 1) = A x + B x 1 + C x + 1 Add up the right side and compare numerators A(x 1)(x + 1) x(x 1)(x + 1) x 2 + 2x 1 x(x 1)(x + 1) Bx(x + 1) Cx(x 1) + + x(x 1)(x + 1) x(x 1)(x + 1) = A(x 1)(x + 1) + Bx(x + 1) + Cx(x 1) x(x 1)(x + 1) x 2 + 2x 1 = A(x 1)(x + 1) + Bx(x + 1) + Cx(x 1) Practice Exam 1 Solutions 55 / 72

56 x 2 + 2x 1 = A(x 1)(x + 1) + Bx(x + 1) + Cx(x 1) Solve for A, B and C When x = 1 = A = A = 1 When x = 1 2 = 2B = B = 1 When x = 1 2 = 2C = C = 1 x 2 + 2x 1 x 3 x = x2 + 2x 1 x(x 1)(x + 1) = A x + with these values of A, B, C B x 1 + C x + 1 Practice Exam 1 Solutions 56 / 72

57 x 2 + 2x 1 x 3 x = x2 + 2x 1 x(x 1)(x + 1) = 1 x + 1 x 1 1 x + 1 x 2 + 2x 1 x 3 x dx = 1 x + 1 x 1 1 x + 1 dx x 2 + 2x 1 x 3 x dx = ln x + ln x 1 ln x C x 2 + 2x 1 x 3 x x(x 1) dx = ln (x + 1) + C Practice Exam 1 Solutions 57 / 72

58 p(x) We use the following guidelines to simplify q(x) dx degree p(x) < degree q(x) (if not, use long division) q(x) is completely factored 1 Single linear factor A factor (x r) in the denominator A requires the partial fraction x r. 2 Repeated linear factor A factor (x r) k in the denominator requires the partial fractions A 1 x r + A 2 (x r) A k 2 (x r). k 3 Single irreducible quadratic factor A factor x 2 + x + 1 in the denominator requires the partial fraction Ax + B x 2 + x + 1 Practice Exam 1 Solutions 58 / 72

59 1 a: Improper Integrals of type II What does it mean when you say 1 1 x 1 dx diverges? Answer: It means the limit 1 1 x 1 does not exists. This is the definition. b 1 dx = lim b 1 x 1 dx 1 1 x 1 b 1 dx = lim b 1 x 1 dx = lim b 1 ln x 1 b = lim ln b 1 ln 1 = divergent b 1 Practice Exam 1 Solutions 59 / 72

60 1 b: Improper Integrals of type I What does it mean for It means the limit This is the definition. 6 1 dx = lim (x 5) 1/3 b a a f(x) dx to converge? f(x) dx = lim b 6 b b a (x 5) 1/3 dx f(x); dx exists. 3 = lim b 2 (x 5)2/3 b = ( lim (b b 5)2/3 1) = Diverges Practice Exam 1 Solutions 6 / 72

61 1 c If 2 1 f(x) dx is an improper integral, then be an improper integral. Justify. 1 f(x) dx must also Here is an example showing that the conclusion of this statement is not always True. It a False statement (2 x) 1 (2 x) dx is improper dx is not improper Practice Exam 1 Solutions 61 / 72

62 1 d: 1 ln x x dx Evaluate if it converges or show that it diverges. 1 ln x 1 ln x dx = lim x a + a x 1 Let u = ln x, du = 1 1 ln x x ln x x ln x x dx = x dx u du = u2 2 (ln x) 2 dx = lim a a 1)2 dx = lim ((ln a + 2 dx + C = (ln x)2 2 + C (ln a)2 ) diverges 2 Practice Exam 1 Solutions 62 / 72

63 1 e: (a) 1 1 (x 1) 2 dx Evaluate if it converges or show that it diverges b 1 dx = lim (x 1) 2 b 1 (x 1) dx 2 Let u = x 1, du = dx 1 (x 1) 2 dx = u 2 du = 1 u + C = 1 x 1 + C 1 dx = lim (x 1) 2 b 1 1 x 1 1 dx diverges (x 1) 2 b = lim b 1 [ b 1 ] Practice Exam 1 Solutions 63 / 72

64 1 e: (b) Evaluate if it converges or show that it diverges. x e x2 dx x e x2 dx = lim b b ( 1 2 ) 2x e x2 dx = lim b ( 1 2 e x2 ) b = lim b ( e b2 ) = 1/2 convergent Practice Exam 1 Solutions 64 / 72

65 1 f: True or False The improper integral e x x dx converges. Plan: The graph of y = 2 + e x lies above the graph of y = 2 x x. We will show that the area under the curve y = 2 x, 1 x < is infinite. This will show that the area under the curve y = 2 + e x, 1 x < is also infinite. x 2 b 2 dx = lim dx = lim 1 x b 1 x 2 ln x b = lim 2 ln b = b 1 b 2 + e x x > 2 x = 2 + e x 1 x dx diverges Practice Exam 1 Solutions 65 / 72

66 1 g: True or False The improper integral 1 1 dx converges. x dx = lim x1.1 a + 1 a x 1.1 dx 2 = lim a + 1x.1 1 a 1 3 = lim 1 = a + a divergent.1 Practice Exam 1 Solutions 66 / 72

67 Additional Examples 1 x 3 9 x 2 dx 2 tan 3 x dx 3 sec 4 x dx Practice Exam 1 Solutions 67 / 72

68 Example 1: x 3 9 x 2 dx Step 1. Find the appropriate change of variable for the integral by drawing a triangle. Choose x 3 = sin θ Then x = 3 sin θ dx = 3 cos θ dθ and 9 x 2 3 = cos θ Practice Exam 1 Solutions 68 / 72

69 Step 2. x 3 9 x 2 dx Choose x = 3 sin θ dx = 3 cos θ dθ 9 x2 = 3 cos θ x 3 (3 sin θ) 3 dx = 3 cos θ dθ = 27 sin 3 θ dθ 9 x 2 3 cos θ = 27 sin 2 θ sin θ dθ = 27 (1 cos 2 θ) sin θ dθ Practice Exam 1 Solutions 69 / 72

70 27 (1 cos 2 θ) sin θ dθ Substitution: u = cos θ, du = sin θ dθ = 27 (1 u 2 ) du = 27u + 9u 3 + C = 27 cos θ + 9 cos 3 θ + C Step 3. x 3 9 x 2 dx = 27 9 x 2 3 x = 3 sin θ 9 x 2 + 9( ) 3 + C 3 9 x2 = 3 cos θ Practice Exam 1 Solutions 7 / 72

71 Example 2: tan 3 x dx 1 tan 3 x dx = 2 = 3 tan 2 x + 1 = sec 2 x tan x tan 2 x dx = tan x sec 2 x dx (tan x) = sec 2 x sin x cos x dx tan x (sec 2 x 1) dx Using the substitutions u = tan x, du = sec 2 x dx and w = cos x 1 tan 3 x dx = u du ( w dw) + C = 1 2 u2 + ln w + C = 1 2 (tan x)2 + ln cos x + C Practice Exam 1 Solutions 71 / 72

72 Example 3: Even powers of sec x sec 4 x dx tan 2 x + 1 = sec 2 x (sec x) = sec x tan x (tan x) = sec 2 x 1 sec 4 x dx = sec 2 x sec 2 x dx = (tan 2 x + 1) sec 2 x dx 2 Using the substitution u = tan x, du = sec 2 xdx 3 sec 4 x dx = (u 2 + 1) du = 1 3 u3 + u + C 4 sec 4 x dx = 1 3 (tan x)3 + tan x + C Practice Exam 1 Solutions 72 / 72