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1. 1. Basic Mathematics De
2. 2. nite Integration R Horan & M Lavelle The aim of this package is to provide a short self assessment programme for students who want to be able to calculate basic de
3. 3. nite integrals.Copyright 2004 rhoran@plymouth.ac.uk , mlavelle@plymouth.ac.uk cLast Revision Date: September 9, 2005 Version 1.0
5. 5. nite Integration3. The Area Under a Curve4. Final Quiz Solutions to Exercises Solutions to Quizzes The full range of these packages and some instructions, should they be required, can be obtained from our web page Mathematics Support Materials.
6. 6. Section 1: Introduction 31. IntroductionIt is possible to determine a function F (x) from its derivative f (x) bycalculating the anti-derivative or integral of f (x), i.e.,  dF if = f (x) ; then F (x) = f (x)dx + C dxwhere C is an integration constant (see the package on inde
7. 7. niteintegration). In this package we will see how to use integration tocalculate the area under a curve.As a revision exercise, try this quiz on inde
8. 8. nite integration. Quiz Select the inde
9. 9. nite integral of (3x2   1 x)dx with respect to x 2 1 3 3 (a) 6x   + C ; (b) x   x2 + C ; 2 2 1 2 1 (c) x + x + C ; 2 (d) x3   x2 + C : 4 4Hint: If n T=  1 ; the integral of x is x n n+1 =(n + 1).
10. 10. Section 2: De
11. 11. nite Integration 42. De
12. 12. nite IntegrationWe de
13. 13. ne the de
14. 14. nite integral of the function f (x) with respect tox from a to b to be  b b f (x)dx = F (x) = F (b)   F (a) ; a awhere F (x) is the anti-derivative of f (x). We call a and b the lowerand upper limits of integration respectively. The function being inte-grated, f (x), is called the integrand. Note the minus sign!Note integration constants are not written in de
15. 15. nite integrals sincethey always cancel in them:  b b f (x)dx = F (x) a a = (F (b) + C )   (F (a) + C ) = F (b) + C   F (a)   C = F (b)   F (a) :
16. 16. Section 2: De
17. 17. nite Integration 5 2Example 1 Calculate the de
18. 18. nite integral 1 x3 dx.  aFrom the rule axn dx = xn+1 we have n+1  2 1 3+1 2 3 x dx = x 1 3+1 1 2 1 1 1 = x4 = ¢ 24   ¢ 11 4 1 4 4 1 = 4 ¢ 16   1 = 4   1 = 15 : 4 4 4Exercise 1. Calculate the following de
19. 19. nite integrals: (click on thegreen letters for the solutions) 3 2 (a) 0 xdx ; (b)  1 xdx ; 2 2 (c) 1 (x2   x)dx ; (d)  1 (x 2   x)dx :
20. 20. Section 3: The Area Under a Curve 63. The Area Under a CurveThe de
21. 21. nite integral of a function f (x) which lies above the x axiscan be interpreted as the area under the curve of f (x).Thus the area shaded blue below y y = f (x) A x 0 a bis given by the de
22. 22. nite integral  b b f (x)dx = F (x) = F (b)   F (a) : a aThis is demonstrated on the next page.
23. 23. Section 3: The Area Under a Curve 7Consider the area, A, under the curve, y = f (x). If we increasethe value of x by x, then the increase in area, A, is approximately y A = y x A A = y : x Here we approximate the y area of the thin strip by a rectangle of width x and height y . In the limit A A as the strips become thin, x x+x x x 3 0, this means: 0 dA A = lim = y: dx x 30 xThe function (height of the curve) is the derivative of the area andthe area below the curve is an anti-derivative or integral ofthe function.N.B. so far we have assumed that y = f (x) lies above the x axis.
24. 24. Section 3: The Area Under a Curve 8 3Example 2 Consider the integral 0 xdx. The integrand y = x (astraight line) is sketched below. The area underneath the line is theblue shaded triangle. The area of any triangle is half its base timesthe height. For the blue shaded triangle, this is 1 9 A= ¢3¢3= : 2 2 y y =x 3 A x 0 3As expected, the integral yields the same result:  3 3 = 3   0 = 9  0= 9: x2 2 2 xdx = 0 2 0 2 2 2 2
25. 25. Section 3: The Area Under a Curve 9Here is a quiz on this relation between de
26. 26. nite integrals and the areaunder a curve.Quiz Select the value of the de
27. 27. nite integral  3 2dx ; 1which is sketched in the following diagram: y y =2 2 A x 0 1 3 (a) 6 ; (b) 2 ; (c) 4 ; (d) 8 :Hint: 2 may be written as 2x , since x = 1. 0 0
28. 28. Section 3: The Area Under a Curve 10Example 3 Consider the two lines: y = 3 and y =  3. Let us integrate these functions in y 6 turn from x = 0 to x = 2. y = +3 3  a) For y = 3: -  2 2 (+3)dx = 3x = 3¢2 3¢0 = 6 : 0 j j 0 0 1 2 x and 6 is indeed the area of the rect-  3  angle of height 3 and length 2. y =  3 b) However, for y =  3:  2 2 ( 3)dx =  3x =  3¢2 ( 3¢0) =  6 : 0 0Although both rectangles have the same area, the sign of this resultis negative because the curve, y =  3, lies below the x axis. Thisindicates the sign convention: If a function lies below the x axis, its integral is negative. If a function lies above the x axis, its integral is positive.
29. 29. Section 3: The Area Under a Curve 11Exercise 2. y 6 y1 (x) A B 0 C D - x y2 (x)From the diagram above, what can you say about the signs of thefollowing de
30. 30. nite integrals? (Click on the green letters for the solu-tions) B D (a) A y1 (x)dx ; (b) B y1 (x)dx ; 0 D (c) A y2 (x)dx ; (d) C y2 (x)dx :
31. 31. Section 3: The Area Under a Curve 12    aExample 4 To calculate  42 6x2 dx, use axn dx = xn+1 . Thus n+1   2  2 6 6x2 dx = x2+1  4 2+1  4 6  2  2 = x3 = 2x3 3  4  4 = 2 ¢ ( 2)3   2 ¢ ( 4)3 =  16 + 128 = 112 :Note that even though the integration range is for negative x (from 4 to  2), the integrand, f (x) = 6x2 , is a positive function. Thede
32. 32. nite integral of a positive function is positive. (Similarly it isnegative for a negative function.)Quiz Select the de
33. 33. nite integral of y = 5x4 with respect to x if thelower limit of the integral is x =  2 and the upper limit is x =  1 (a)  31 ; (b) 31 ; (c) 29 ; (d)  27 :
34. 34. Section 3: The Area Under a Curve 13Exercise 3. Use the integrals listed below to calculate the followingde
35. 35. nite integrals. (Click on the green letters for the solutions) 1 f (x) xn for n T=  1 sin(ax) cos(ax) eax x  1 1 1 1 f (x)dx n+1 xn+1   a cos(ax) a sin(ax) a eax ln(x) 9 p 1 (a) 4 3 tdt ; (b)  1 (x 2   2x + 4)dx ;  3 (c) 0 sin(x)dx ; (d) 0 4e2x dx ; 2 3  (e) 1 dt ; (f) 2 2 cos(4w)dw : t 4
36. 36. Section 3: The Area Under a Curve 14Quiz Find the correct result for the de
37. 37. nite integral  2b x2 dx : a 8 3 1 3 (a) b   a ; (b) 4b   2a ; 3 3 8 3 1 3 1 3 1 3 (c) b + a ; (d) b   a : 3 3 3 3Quiz Select the correct result for the de
38. 38. nite integral  3 1 dx ; 2 x2from the answers oered below 1 1 1 (a)  1 ; (b) ; (c) ; (d) : 5 36 6
39. 39. Section 4: Final Quiz 154. Final QuizBegin Quiz Choose the solutions from the options given. 1. What is the area under the curve of the following positive function y = 10x4 + 3x2 between x =  1 and x = 2? (a) 75 ; (b) 53 ; (c) 69 ; (d) 57 : 2. What is the de
40. 40. nite integral of 3 sin(2x) from x = 0 to x = =2 ? 5 (a)  3 ; (b) 0 ; (c) 3 ; (d) : 2 3. Find the (non-zero) value of b for which the de
41. 41. nite integral b 0 (2s   3)ds vanishes (a) 1 ; (b) 5 ; (c) 3 ; (d) 2 :  4. Select  below the de
42. 42. nite   2 integral ¡ 2 e2x dx with respect to x. ¡ ¡ (a) 2 e4   e 4 ; (b) 1 e4   4 pe ; (c) 0 ; (d) 1  e4   e 4 : 2 2End Quiz
43. 43. Solutions to Exercises 16Solutions to Exercises Exercise 1(a) To calculate 3 0 xdx, use the formula  1 n+1 xn dx = x n+1with n = 1. This yields  3 1 1+1 3 1 2 3 = x xdx = x 0 1+1 0 2 0 1 = 2 ¢ (3)2   1 ¢ (0)2 2 1 = 2 ¢9 0= 9: 2Click on the green square to return £
44. 44. Solutions to Exercises 17 Exercise 1(b) To calculate  1 xdx, use the formula for the inde
45. 45. nite 2integral  1 xn dx = xn+1 n+1with n = 1. This yields  2 1 1+1 2 2 = 1 x2 xdx = x  1 1+1  1 2  1 1 = 2 ¢ (2)2   1 ¢ ( 1)2 2 1 1 = 2 ¢ 4   2 ¢ (+1) 1 3 = 2  = : 2 2Click on the green square to return £
46. 46. Solutions to Exercises 18 2Exercise 1(c) To evaluate the de
47. 47. nite integral  (x2   x)dx we 1rewrite it as the sum of two integrals and use = n+1 xn+1 with 1 xn dxn = 2 in the
48. 48. rst integral and with n = 1 in the second one   2 2 1 2+1 2   1 x1+1 2 2 x dx   x dx = 2+1 x 1+1 1 1 1 1 2 2 1 3   1 x2 = x 3 1 2 1 1 1 1 1 = 3 ¢2   3 ¢1   2 ¢2   2 ¢1 3 3 2 2 1 = 3 ¢8  1 ¢1  1 ¢4  2 ¢1 3 2 1 7 3 14 9 5 =   = 6  6 = 6: 3 2Click on the green square to return £
49. 49. Solutions to Exercises 19 Exercise 1(d) To
50. 50. nd the integral  1 (x2   x)dx we rewrite it as the 2sum of two integrals and use the result of the previous part to writeit as 2  2 2 1 3 2   1 x2 x dx   2 x dx = x  1  1 3  1 2  1 1 = 3 ¢ 23   1 ¢ ( 1)3   1 ¢ 22   1 ¢ ( 1)2 3 2 2 1 = 3 ¢8+ 1 ¢1  1 ¢4  2 ¢1 = 9   2 3 2 1 3 3 3 3 = 3  = : 2 2Click on the green square to return £
51. 51. Solutions to Exercises 20Exercise 2(a) y 6 y1 (x) A B 0 C D - x  BThe sign of the de
52. 52. nite integral, A y1 (x)dx, must be negative. Thisis because the function y1 (x) is negative for all values of x between Aand B . The area is all below the x axis.Click on the green square to return £
53. 53. Solutions to Exercises 21Exercise 2(b) y 6 y1 (x) A B 0 C D - x DThe sign of the de
54. 54. nite integral, B y1 (x)dx, must be positive. Thisis because, between the integration limits B and D, there is more areaabove the x axis than below the x axis.Click on the green square to return £
55. 55. Solutions to Exercises 22Exercise 2(c) y 6 A B 0 C D - x y2 (x) 0The sign of the de
56. 56. nite integral, A y2 (x)dx, must be positive. Thisis because, between the integration limits A and 0, there is more areaabove the x axis than below it.Click on the green square to return £
57. 57. Solutions to Exercises 23Exercise 2(d) y 6 A B 0 C D - x y2 (x) D The sign of the de
58. 58. nite integral, C y2 (x)dx, must be negative. Thisis because, between the integration limits C and D, the integrandy2 (x) is always negative.Click on the green square to return £
59. 59. Solutions to Exercises 24 9 pExercise 3(a) To calculate the de
60. 60. nite integral 4 3 tdt we rewriteit as  9 p  9 3 tdt =3¢ t1=2 dt 4 4  1 n+1and use xn dx = x for n = 1 n+1 2  9 9 9 1 1 +1 1 3 2 3 93¢ = 3¢ 1 t2 = 3 ¢ 3 ¢ = 3 ¢ t2 1 t 2 dt t2 4 2 +1 4 2 4 3 4 9 3 = 2 ¢ (9) 2   2 ¢ (4) 3 = 2 ¢ (9 1 )3   2 ¢ (4 2 )3 3 1 = 2 t2 2 2 4 = 2 ¢ (3)3   2 ¢ (2)3 = 2 ¢ 27   2 ¢ 8 = 54   16 = 38 :N.B. dividing by a fraction is equivalent to multiplying by its inverse(see the package on fractions).Click on the green square to return £
61. 61. Solutions to Exercises 25 Exercise 3(b) To calculate the de
62. 62. nite integral  1 (x2   2x +4)dx we 1 1 2 1 1rewrite it as a sum of integrals  1 x dx   2 ¢  1 xdx + 4 ¢  1 1dx  1 n+1and use xn dx x= with n = 2 in the
63. 63. rst integral, n+1  1 1 1 2+1 1 = 1 x3 = 1 13   ( 1)3 = 2 ;   ¡ x2 dx = x  1 2+1  1 3  1 3 3with n = 1 in the second integral  1 1 1+1 1 1   ¡ 2¢ xdx = 2 ¢ x = x2 =   12   ( 1)2 = 0 ;  1 1+1  1  1and with n = 0 in the last integral  1 1 0+1 1 1 4¢ 1dx = 4 ¢ x = 4x = 4 (1   ( 1)) = 8 :  1 0+1  1  1Summing up these numbers we obtain 2=3 + 0 + 8 = 26=3.Click on the green square to return £
64. 64. Solutions to Exercises 26 Exercise 3(c) To calculate the de
65. 65. nite integral 0 sin(x)dx we notefrom the table that  1 sin(ax)dx =   cos(x): aThis yields (with a = 1)  sin(x)dx =   cos(x) 0 0 =   (cos()   cos(0)) =   (( 1)   1) = 2 :N.B. It is worth emphasizing that the angles in calculus formulae fortrigonometric functions are measured in radians.Click on the green square to return £
66. 66. Solutions to Exercises 27 3Exercise 3(d) To calculate the de
67. 67. nite integral 0 4e2x dx , write  3  3 4e2x dx = 4 ¢ e2x dx 0 0and use from the table  1 eax dx = eax : aThis gives for a = 2  3 1 3 3 4¢ e 2x dx = 4 ¢ e2x = 2e2x 0 2 0 0 = 2e(2¢3)   2e(2¢0) = 2e6   2e0 = 2e6   2 ¢ 1 = 2e6   2 :Click on the green square to return £
68. 68. Solutions to Exercises 28 2 3Exercise 3(e) To evaluate the de
69. 69. nite integral 1 dt we write t   2 3 2 1 dt =3¢ dt 1 t 1 tand use  1 dt = ln(t) : tThis yields  2 2 1 3¢ dt = 3 ¢ ln(t) 1 t 1 = 3 ¢ ln(2)   3 ¢ ln(1) = 3 ¢ ln(2)   3 ¢ 0 = 3 ln(2) :N.B. ln(0) = 1 , since e0 = 1 .Click on the green square to return £
70. 70. Solutions to Exercises 29 Exercise 3(f) To
71. 71. nd the de
72. 72. nite integral 4 2 2 cos(4w)dw use   2 2 2 cos(4w)dw = 2 ¢ cos(4w)dw : 4 4  1and cos(ax)dx = sin(x) . This gives for a = 4 a  2 1 2 1 2 2¢ cos(4w)dw = 2 ¢ sin(4w) = sin(4w) 4 4 4 2 4 1 1 = sin(4 ¢ )   sin(4 ¢ ) 2 2 2 4 1 1 = sin(2 )   sin( ) 2 2 1 1 = 2 ¢ 0   2 ¢ 0 = 0:Click on the green square to return £
73. 73. Solutions to Quizzes 30Solutions to Quizzes Solution to Quiz: To
74. 74. nd the inde
75. 75. nite integral (3x2   1 x)dx we 2use the sum rule for integrals, rewriting it as the sum of two integrals    1 1 (3x   x) dx = 2 3x dx + (  x) dx 2 2   2 1 = 3 x dx  2 x dx : 2 Using xn dx = n+1 xn+1 ; n T=  1 with n = 2 in the
76. 76. rst integral and 1with n = 1 in the second one gives   1 1 1+2 1 1 3 x dx   2 x dx = 3 ¢ x   2 ¢ 1 + 1 x1+1 + C 2 1+2 3 1 1 = x3   x2 + C = x3   x2 + C : 3 2(1 + 1) 4Check that dierentiation of this result gives 3x   2 x. 2 1 End Quiz
77. 77. Solutions to Quizzes 31Solution to Quiz: Using 2 = 2x0 , the integral   3 3 3 2dx = 2 x0 dx = 2x 1 1 1 = 2¢3 2¢1=6 2 = 4:Indeed from the diagram y y =2 2 A x 0 1 3the area under the curve between the integration limits is the area ofa square of side 2. This has area 2 ¢ 2 = 4. End Quiz
78. 78. Solutions to Quizzes 32Solution to Quiz: The de
79. 79. nite integral of y = 5x4 with respect to xif the lower limit of the integral is x =  2 and the upper limit x =  1can be written as   1 5x4 dx :  2 B B a From the basic result n A ax dx = xn+1 we obtain n+1 A 5 5  1   1  1 5x4 dx = x = x5  2 5  2  2 = ( 1)5   ( 2)5 =  1   ( 32) =  1 + 32 = 31 :Note that since the integrand 5x4 is positive for all x, the negativesuggested solutions could not be correct. End Quiz
80. 80. Solutions to Quizzes 33  2bSolution to Quiz: To calculate the de
81. 81. nite integral a x2 dx usethe basic inde
82. 82. nite integral  1 n+1 xn dx = x : n+1with n = 2. This gives  2b 1 (2+1) 2b 1 3 2b = x x2 dx = x a 2+1 a 3 a 1 = 3 ¢ (2b)3   1 ¢ (a)3 3 1 = 3 ¢ (2) ¢ b   1 ¢ a3 3 3 3 1 1 = 3 ¢ 8 ¢ b   3 ¢ a3 3 8 3 1 3 = b   a : 3 3 End Quiz
83. 83. Solutions to Quizzes 34Solution to Quiz: To evaluate the de
84. 84. nite integral   3 1 3 dx = x 2 dx 2 x2 2 B 1 n+1 Buse xn dx = x with n =  2 A n+1 A  3 1 3 1  1 3 x 2 dx = x 2+1 = x 2 ( 2 + 1) 2 ( 1) 2 3 3 1 1 1 1 = ( 1) ¢ =   =     (  ) x 2 x 2 3 2 1 1 2 3 =   + =  + 3 2 6 6 =  2 + 3 = 1 : 6 6 End Quiz