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INTEGRATION USING PARTIAL FRACTIONS 429 Problem 9. Determine: 1 x 2 a 2 dx Let 1 x 2 a 2 Á A x a C B x C a Á Ax C a CBx a x C ax a Equating the numerators gives: 1 Á Ax Ca CBx a Let x D a,thenA D 1 2a ,andletx Da, then B D 1 2a Hence 1 x 2 a 2 dx Á 1 2a 1 xa 1 xCa dx D 1 2a [ lnx a lnx C a ] C c D 1 2a ln x − a x Y a Y c Problem 10. Evaluate: 4 3 3 x 2 4 dx, correct to 3 significant figures From Problem 9, 4 3 3 x 2 4 dx D 3 1 22 ln x 2 x C2 4 3 D 3 4 ln 2 6 ln 1 5 D 3 4 ln 5 3 D 0 .383, correct to 3 significant figures. Problem 11. Determine: 1 a 2 x 2 dx Using partial fractions, let 1 a 2 x 2 Á 1 a xa C x Á A a x C B a Cx Á Aa C x C Ba x a xa C x Then 1 Á Aa C x CBa x Let x D a then A D 1 2a .Letx Da then B D 1 2a Hence 1 a 2 x 2 dx D 1 2a 1 a x C 1 a Cx dx D 1 2a [lna x Clna Cx] Cc D 1 2a ln a Y x a − x Y c Problem 12. Evaluate: 2 0 5 9 x 2 dx, correct to 4 decimal places From Problem 11, 2 0 5 9 x 2 dx D 5 1 23 ln 3 Cx 3 x 2 0 D 5 6 ln 5 1 ln 1 D 1.3412, correct to 4 decimal places Now try the following exercise Exercise 175 Further problems on inte- gration using partial frac- tions with quadratic factors 1. Determine x 2 x 13 x 2 C 7x 2 dx lnx 2 C 7 C 3 p 7 tan 1 x p 7 lnx 2 C c In Problems 2 to 4, evaluate the definite inte- grals correct to 4 significant figures. 2. 6 5 6x 5 x 4x 2 C 3 dx [0.5880] 3. 2 1 4 16 x 2 dx [0.2939] 4. 5 4 2 x 2 9 dx [0.1865] 51 The t D tan  2 substitution 51.1 Introduction Integrals of the form 1 a cos  Cb sin  C c dÂ, where a, b and c are constants, may be determined by using the substitution t D tan  2 . The reason is explained below. If angle A in the right-angled triangle ABC shown in Fig. 51.1 is made equal to  2 then, since tangent D opposite adjacent ,ifBC D t and AB D 1, then tan  2 D t . By Pythagoras’ theorem, AC D p 1 Ct 2 C BA t 2 1 √ 1+ t 2 q Figure 51.1 Therefore sin  2 D t p 1 Ct 2 and cos  2 D 1 p 1 Ct 2 Since sin 2x D 2sinx cos x (from double angle formulae, Chapter 26), then sin  D 2sin  2 cos  2 D 2 t p 1 Ct 2 1 p 1 Ct 2 i.e. sin q = 2t .1 Y t 2 / 1 Since cos 2x D cos 2  2 sin 2  2 D 1 p 1 Ct 2 2 t p 1 Ct 2 2 i.e. cos q = 1 − t 2 1 Y t 2 2 Also, since t D tan  2 , dt d D 1 2 sec 2  2 D 1 2 1 Ctan 2  2 from trigonomet- ric identities, i.e. dt d D 1 2 1 Ct 2 from which, dq = 2dt 1 Y t 2 3 Equations (1), (2) and (3) are used to determine integrals of the form 1 a cos  Cb sin  C c d where a, b or c may be zero. 51.2 Worked problems on the t = tan q 2 substitution Problem 1. Determine: d sin  If t D tan  2 then sin  D 2t 1 Ct 2 and d  D 2 dt 1 Ct 2 from equations (1) and (3). Thus d sin  D 1 sin  d THE t D tanÂ/2 SUBSTITUTION 431 D 1 2t 1 Ct 2 2 dt 1 Ct 2 D 1 t dt D ln t Cc Hence d sin  D ln tan  2 C c Problem 2. Determine: dx cos x If tan x 2 then cos x D 1 t 2 1 Ct 2 and d x D 2 dt 1 Ct 2 from equations (2) and (3). Thus dx cos x D 1 1 t 2 1 Ct 2 2 dt 1 Ct 2 D 2 1 t 2 dt 2 1 t 2 may be resolved into partial fractions (see Chapter 7). Let 2 1 t 2 D 2 1 t1 Ct D A 1 t C B 1 Ct D A1 Ct CB1 t 1 t1 Ct Hence 2 D A1 Ct CB1 t When t D 1, 2 D 2A, from which,A D 1 When t D1, 2 D 2B, from which,B D 1 Hence 2 dt 1 t 2 D 1 1 t C 1 1 C t dt Dln1 t Cln1 Ct Cc D ln 1 Ct 1 t C c Thus dx cos x D ln 1 Y tan x 2 1 − tan x 2 Y c Note that since tan 4 D 1, the above result may be written as: dx cos x D ln tan 4 C tan x 2 1 tan 4 tan x 2 C c D ln tan p 4 Y x 2 Y c from compound angles, Chapter 26, Problem 3. Determine: dx 1 Ccos x If tan x 2 then cos x D 1 t 2 1 Ct 2 and d x D 2 dt 1 Ct 2 from equations (2) and (3). Thus dx 1 Ccos x D 1 1 Ccos x dx D 1 1 C 1 t 2 1 Ct 2 2 dt 1 Ct 2 D 1 1 Ct 2 C1 t 2 1 Ct 2 2 dt 1 Ct 2 D dt Hence dx 1 Ccos x D t C c D tan x 2 Y c Problem 4. Determine: d 5 C4cos If t D tan  2 then cos  D 1 t 2 1 Ct 2 and d x D 2 dt 1 Ct 2 from equations (2) and (3). Thus d 5 C4cos D 2 dt 1 Ct 2 5 C4 1 t 2 1 Ct 2 D 2 dt 1 Ct 2 51 Ct 2 C41 t 2 1 Ct 2 D 2 dt t 2 C 9 D 2 dt t 2 C 3 2 D 2 1 3 tan 1 t 3 C c, 432 ENGINEERING MATHEMATICS from 12 of Table 49.1, page 418. Hence dq 5 Y 4cosq D 2 3 tan −1 1 3 tan q 2 Y c Now try the following exercise Exercise 176 Further problems on the t = tan q 2 substitution Integrate the following with respect to the variable: 1. d 1 Csin  2 1 Ctan  2 C c 2. dx 1 cos x C sin x ln tan x 2 1 Ctan x 2 C c 3. d˛ 3 C2cos˛ 2 p 5 tan 1 1 p 5 tan ˛ 2 C c 4. dx 3sinx 4cosx 1 5 ln 2tan x 2 1 tan x 2 C 2 D c 51.3 Further worked problems on the t = tan q 2 substitution Problem 5. Determine: dx sin x Ccos x If tan x 2 then sin x D 2t 1 Ct 2 ,cosx D 1 t 2 1 Ct 2 and dx D 2 dt 1 Ct 2 from equations (1), (2) and (3). Thus dx sin x Ccos x D 2 dt 1 Ct 2 2t 1 Ct 2 C 1 t 2 1 Ct 2 D 2 dt 1 Ct 2 2t C 1 t 2 1 Ct 2 D 2 dt 1 C2t t 2 D 2 dt t 2 2t 1 D 2 dt t 1 2 2 D 2 dt p 2 2 t 1 2 D 2 1 2 p 2 ln p 2 Ct 1 p 2 t 1 C c (see problem 11, Chapter 50, page 429), i.e. dx sin x Ccos x D 1 p 2 ln p 2 − 1 Y tan x 2 p 2 Y 1 − tan x 2 Y c Problem 6. Determine: dx 7 3sinx C6cosx From equations (1) and (3), dx 7 3sinx C 6cosx D 2 dt 1 Ct 2 7 3 2t 1 Ct 2 C 6 1 t 2 1 Ct 2 D 2 dt 1 Ct 2 71 Ct 2 32t C 61 t 2 1 Ct 2 D 2 dt 7 C7t 2 6t C 6 6t 2 D 2 dt t 2 6t C 13 D 2 dt t 3 2 C 2 2 D 2 1 2 tan 1 t 3 2 C c THE t D tanÂ/2 SUBSTITUTION 433 from 12, Table 49.1, page 418. Hence dx 7 3sinx C6cosx D tan −1 tan x 2 − 3 2 Y c Problem 7. Determine: d 4cos C 3sin From equations (1) to (3), d 4cos C3sin D 2 dt 1 Ct 2 4 1 t 2 1 Ct 2 C 3 2t 1 Ct 2 D 2 dt 4 4t 2 C 6t D dt 2 C3t 2t 2 D 1 2 dt t 2 3 2 t 1 D 1 2 dt t 3 4 2 25 16 D 1 2 dt 5 4 2 t 3 4 2 D 1 2 1 2 5 4 ln 5 4 C t 3 4 5 4 t 3 4 C c from problem 11, Chapter 50, page 429, D 1 5 ln 1 2 C t 2 t C c Hence d 4cos C3sin D 1 5 ln 1 2 Y tan q 2 2 − tan q 2 Y c or 1 5 ln 1 Y 2tan q 2 4 − 2tan q 2 Y c Now try the following exercise Exercise 177 Further problems on the t = tan q=2 substitution In Problems 1 to 4, integrate with respect to the variable. 1. d 5 C4sin 2 3 tan 1 5tan x 2 C 4 3 C c 2. dx 1 C2sinx 1 p 3 ln tan x 2 C 2 p 3 tan x 2 C 2 C p 3 C c 3. dp 3 4sinp C 2cosp 1 p 11 ln tan p 2 4 p 11 tan p 2 4 C p 11 C c 4. d 3 4sin 1 p 7 ln 3tan  2 4 p 7 3tan  2 4 C p 7 C c 5. Show that dt 1 C3cost D 1 2 p 2 ln p 2 Ctan t 2 p 2 tan t 2 Cc 6. Show that /3 0 3 d cos  D 3.95, correct to 3 significant figures. 7. Show that /2 0 d 2 Ccos  D 3 p 3 52 Integration by parts 52.1 Introduction From the product rule of differentiation: d dx u v D v du dx C u d v dx , where u and v are both functions of x. Rearranging gives: u d v dx D d dx u v v du dx Integrating both sides with respect to x gives: u d v dx dx D d dx u v dx v du dx dx i.e. u d v dx dx D u v v du dx dx or udv = uv − v du This is known as the integration by parts for- mula and provides a method of integrating such products of simple functions as xe x dx, t sin t dt, e  cos  d and x ln x dx. Given a product of two terms to integrate the initial choice is: ‘which part to make equal to u’ and ‘which part to make equal to d v’. The choice must be such that the ‘u part’ becomes a constant after successive differentiation and the ‘d v part’ can be integrated from standard integrals. Invariable, the following rule holds: ‘If a product to be integrated contains an algebraic term (such as x, t 2 or 3Â)then this term is chosen as the u part. The one exception to this rule is when a ‘ln x ’ term is involved; in this case ln x is chosen as the ‘u part’. 52.2 Worked problems on integration by parts Problem 1. Determine x cos x dx From the integration by parts formula, u dv D uv v du Let u D x, from which du dx D 1, i.e. d u D dx and let d v D cos x dx, from which v D cos x dx D sin x. Expressions for u, d u and v are now substituted into the ‘by parts’ formula as shown below. u x dv cos x dx = = u (x) v (sin x) v (sin x) − − ∫ ∫ ∫ ∫ du (dx) i.e. x cos x dx D x sin x cos x C c D x sin x Y cos x Y c [This result may be checked by differentiating the right hand side, i.e. d dx x sin x C cos x C c D [xcos x Csin x 1] sin x C 0 using the product rule D x cos x, which is the function being integrated] Problem 2. Find: 3te 2t dt Let u D 3t, from which, du dt D 3, i.e. d u D 3 dt and let d v D e 2t dt, from which, v D e 2t dt D 1 2 e 2t Substituting into u dv D uv v du gives: 3te 2t dt D 3t 1 2 e 2t 1 2 e 2t 3 d t D 3 2 te 2t 3 2 e 2t dt D 3 2 te 2t 3 2 e 2t 2 C c INTEGRATION BY PARTS 435 Hence 3te 2t dt = 3 2 e 2t t − 1 2 Y c, which may be checked by differentiating. Problem 3. Evaluate 2 0 2 sin  d  Let u D 2Â, from which, du d D 2, i.e. du D 2 d  and let d v D sin  d , from which, v D sin  d Dcos  Substituting into u d v D uv v du gives: 2 sin  d  D 2Âcos  cos Â2 d  D2 cos  C 2 cos  d D2 cos  C 2sin C c Hence 2 0 2 sin  d  D [ 2 cos  C 2sin ] 2 0 D 2 2 cos 2 C 2sin 2 [0 C2sin0] D 0 C 2 0 C0 D 2 since cos 2 D 0and sin 2 D 1 Problem 4. Evaluate: 1 0 5xe 4x dx, correct to3significant figures Let u D 5x, from which du dx D 5, i.e. du D 5 dx and let d v D e 4x dx, from which, v D e 4x dx D 1 4 e 4x Substituting into u d v D uv v du gives: 5xe 4x dx D 5x e 4x 4 e 4x 4 5 d x D 5 4 xe 4x 5 4 e 4x dx D 5 4 xe 4x 5 4 e 4x 4 C c D 5 4 e 4x x 1 4 C c Hence 1 0 5xe 4x dx D 5 4 e 4x x 1 4 1 0 D 5 4 e 4 1 1 4 5 4 e 0 0 1 4 D 15 16 e 4 5 16 D 51.186 C0.313 D 51.499 D 51.5, correct to 3 significant figures. Problem 5. Determine: x 2 sin x dx Let u D x 2 , from which, du dx D 2x,i.e.d u D 2x dx, and let d v D sin x dx, from which, v D sin x dx Dcos x Substituting into u dv D uv v du gives: x 2 sin x dx D x 2 cos x cos x2x dx Dx 2 cos x C 2 x cos x dx The integral, x cos x dx, is not a ‘standard inte- gral’ and it can only be determined by using the integration by parts formula again. From Problem 1, x cos x dx D x sin x C cos x Hence x 2 sin x dx Dx 2 cos x C2fx sin x C cos xgCc Dx 2 cos x C2x sin x C 2cosx C c D .2 − x 2 / cos x Y 2x sin x Y c In general, if the algebraic term of a product is of power n, then the integration by parts formula is applied n times. Now try the following exercise Exercise 178 Further problems on inte- gration by parts Determine the integrals in Problems 1 to 5 using integration by parts. 436 ENGINEERING MATHEMATICS 1. xe 2x dx e 2x 2 x 1 2 C c 2. 4x e 3x dx 4 3 e 3x x C 1 3 C c 3. x sin x dx [x cos x Csin x C c] 4. 5 cos 2 d 5 2  sin 2 C 1 2 cos 2 C c 5. 3t 2 e 2t dt 3 2 e 2t t 2 t C 1 2 C c Evaluate the integrals in Problems 6 to 9, correct to 4 significant figures. 6. 2 0 2xe x dx [16.78] 7. 4 0 x sin 2x d x [0.2500] 8. 2 0 t 2 cos t d t [0.4674] 9. 2 1 3x 2 e x 2 dx [15.78] 52.3 Further worked problems on integration by parts Problem 6. Find: x ln x dx The logarithmic function is chosen as the ‘u part’ Thus when u D ln x,then du dx D 1 x ,i.e.du D dx x Letting d v D x dx gives v D x dx D x 2 2 Substituting into u d v D uv v du gives: x ln x dx D ln x x 2 2 x 2 2 dx x D x 2 2 ln x 1 2 x dx D x 2 2 ln x 1 2 x 2 2 C c Hence x ln x dx = x 2 2 ln x − 1 2 Y c or x 2 4 .2lnx − 1/ Y c Problem 7. Determine: ln x dx ln x dx is the same as 1 ln x dx Let u D ln x, from which, du dx D 1 x ,i.e.du D dx x and let d v D 1 d x, from which, v D 1 dx D x Substituting into u dv D uv v du gives: ln x dx D ln xx x dx x D x ln x dx D x ln x x C c Hence ln xdx= x .ln x − 1/ Y c Problem 8. Evaluate: 9 1 p x ln x dx, correct to 3 significant figures Let u D ln x, from which d u D dx x and let d v D p x dx D x 1 2 dx, from which, v D x 1 2 dx D 2 3 x 3 2 Substituting into u dv D uv v du gives: p x ln x dx D ln x 2 3 x 3 2 2 3 x 3 2 dx x D 2 3 p x 3 ln x 2 3 x 1 2 dx D 2 3 p x 3 ln x 2 3 2 3 x 3 2 C c D 2 3 p x 3 ln x 2 3 C c Hence 9 1 p x ln x dx D 2 3 p x 3 ln x 2 3 9 1 INTEGRATION BY PARTS 437 D 2 3 p 9 3 ln 9 2 3 2 3 p 1 3 ln 1 2 3 D 18 ln 9 2 3 2 3 0 2 3 D 27.550 C0.444 D 27.994 D 28.0, correct to 3 significant figures. Problem 9. Find: e ax cos bx dx When integrating a product of an exponential and a sine or cosine function it is immaterial which part is made equal to ‘u’. Let u D e ax , from which du dx D ae ax , i.e. du D ae ax dx and let dv D cos bx d x, from which, v D cos bx d x D 1 b sin bx Substituting into u d v D uv v du gives: e ax cos bx dx D e ax 1 b sin bx 1 b sin bx ae ax dx D 1 b e ax sin bx a b e ax sin bx dx 1 e ax sin bx dx is now determined separately using integration by parts again: Let u D e ax then du D ae ax dx,andlet d v D sin bx dx, from which v D sin bx dx D 1 b cos bx Substituting into the integration by parts formula gives: e ax sin bx dx D e ax 1 b cos bx 1 b cos bx ae ax dx D 1 b e ax cos bx C a b e ax cos bx dx Substituting this result into equation (1) gives: e ax cos bx dx D 1 b e ax sin bx a b 1 b e ax cos bx C a b e ax cos bx dx D 1 b e ax sin bx C a b 2 e ax cos bx a 2 b 2 e ax cos bx dx The integral on the far right of this equation is the same as the integral on the left hand side and thus they may be combined. e ax cos bx dx C a 2 b 2 e ax cos bx d x D 1 b e ax sin bx C a b 2 e ax cos bx i.e. 1 C a 2 b 2 e ax cos bx d x D 1 b e ax sin bx C a b 2 e ax cos bx i.e. b 2 C a 2 b 2 e ax cos bx d x D e ax b 2 b sin bx C a cos bx Hence e ax cos bx dx D b 2 b 2 C a 2 e ax b 2 b sin bx C a cos bx D e ax a 2 Y b 2 .b sin bx Y a cos bx/ Y c Using a similar method to above, that is, integrating by parts twice, the following result may be proved: e ax sin bx dx = e ax a 2 Y b 2 .a sin bx − b cos bx/ Y c 2 Problem 10. Evaluate 4 0 e t sin 2t d t, correct to 4 decimal places 438 ENGINEERING MATHEMATICS Comparing e t sin 2t d t with e ax sin bx dx shows that x D t, a D 1andb D 2. Hence, substituting into equation (2) gives: 4 0 e t sin 2t d t D e t 1 2 C 2 2 1sin2t 2cos2t 4 0 D e 4 5 sin 2 4 2cos2 4 e 0 5 sin 0 2cos0 D e 4 5 1 0 1 5 0 2 D e 4 5 C 2 5 D 0 .8387, correct to 4 decimal places Now try the following exercise Exercise 179 Further problems on inte- gration by parts Determine the integrals in Problems 1 to 5 using integration by parts. 1. 2x 2 ln x dx 2 3 x 3 ln x 1 3 C c 2. 2ln3x dx [2xln 3x 1 Cc] 3. x 2 sin 3x dx cos 3x 27 2 9x 2 C 2 9 x sin 3x C c 4. 2e 5x cos 2x dx 2 29 e 5x 2sin2x C 5cos2x Cc 5. 2 sec 2  d [2[ tan  lnsec Â] Cc] Evaluate the integrals in Problems 6 to 9, correct to 4 significant figures. 6. 2 1 x ln x dx [0.6363] 7. 1 0 2e 3x sin 2x dx [11.31] 8. 2 0 e t cos 3t d t [1.543] 9. 4 1 p x 3 ln x dx [12.78] 10. In determining a Fourier series to repre- sent fx D x in the range to , Fourier coefficients are given by: a n D 1 x cos nx dx and b n D 1 x sin nx dx where n is a positive integer. Show by using integration by parts that a n D 0 and b n D 2 n cos n 11. The equations: C D 1 0 e 0.4 cos 1.2 d and S D 1 0 e 0.4 sin 1.2 d are involved in the study of damped oscillations. Determine the values of C and S.[C D 0.66, S D 0.41] [...]... to C4 1 2 R.m.s deviation 4 1 D 4 1 1 5 D 4 1 5 D 4 2x 2 1 5 D 2 1 dx 3 1 4x 4 4x 2 C 1 dx 1 1 4x 5 5 5 D y 2 dx 1 4 4 5 4x 3 Cx 3 5 4 5 D 1 [ 737.87 5 D 4 1 4 3 4 C4 3 4 15 1 3C 3 1 [738. 34] 5 147 .67 D 12. 152 D 12. 2, correct to 3 significant figures Determine the r.m.s values of: (a) y D 3x from x D 0 to x D 4 (b) y D t2 from t D 1 to t D 3 (c) y D 25 sin  from  D 0 to  D 2 25 (a) 6.928 (b) 4. 919... 54. 9 t y D 3et /4 1 2. 34 0 3.0 1 3.85 2 4. 95 3 6.35 4 8.15 Since all the values of y are positive the area required is wholly above the t-axis Since y D x 2 C 5 then x 2 D y 5 and p xD y 5 The area enclosed by the curve y D x 2 C 5 (i.e p x D y 5 , the y-axis and the ordinates y D 5 and y D 14 (i.e area ABC of Fig 54. 9) is given by: 4 4 3 3et /4 d t D 1 4 1 1 D 12[ et /4 ]4 1 D 12 e D 12 2.7183 14 et /4. .. ordinates y3 , y4 and y5 , and so on Then /3 p 4 cos3 x d x (Use 6 intervals) 0 [0.799] y y1 y = a + bx + cx 2 y2 y3 b −d Figure 53.3 0 d x y dx ³ a d, y1 D a bd C cd2 4 1 1 d y1 C 4y2 C y3 C d y3 C 4y4 C y5 3 3 C 1 d y2n 3 1 C 4y2n C y2nC1 44 4 ENGINEERING MATHEMATICS Thus, from equation (5): 3 2 1 p d x ³ 0.5 [ 2.0000 C 1.1 547 x 3 1 y y = f (x ) C 4 1.6330 C 1.2 649 C 2 1 .41 42 D y1 0 y2 y3 y4 y2n + 1 a... 4 s Since 2t2 C 5 is a quadratic expression, the curve v D 2t2 C 5 is a parabola cutting the v-axis at v D 5, as shown in Fig 54. 4 The distance travelled is given by the area under the v/t curve (shown shaded in Fig 54. 4) By integration, shaded area y = 2x + 3 12 4 v dt D 10 0 8 4 D 6 2t2 C 5 d t 0 4 D 2 2t3 C 5t 3 4 0 3 0 1 Figure 54. 3 2 3 4 5 D x i.e 24 3 C5 4 0 distance travelled = 62.67 m 45 0 ENGINEERING. .. 5 dy y 5 yD5 4 D 12 1.9395 D 23.27 square units x y x dy D Area D 1 D 14 yD 14 y dt Hence area D 0 3 x3 C 5x 3 0 D 24 square units D When x D 3, y D 32 C 5 D 14, and when x D 0, y D 5 y D x3 2x 2 8x D x x 2 2x Dx xC2 x 8 4 When y D 0, then x D 0 or x C 2 D 0 or x 4 D 0, i.e when y D 0, x D 0 or 2 or 4, which means that the curve crosses the x-axis at 0, 2 and 45 4 ENGINEERING MATHEMATICS 4 Since the curve... interval from t D 1 s to t D 5 s [ 140 m] 4 x3 2x 2 8x d x 0 D x4 4 2x 3 3 0 8x 2 2 x4 4 54. 4 The area between curves 2 2x 3 3 8x 2 2 4 0 The area enclosed between curves y D f1 x and y D f2 x (shown shaded in Fig 54. 11) is given by: 2 2 D 6 42 3 3 1 D 49 square units 3 b shaded area D b f2 x d x f1 x d x a a b D f2 x / − f1 x / dx a Now try the following exercise y Exercise 1 84 Further problems on areas under... answer correct to 4 significant figures /2 With 6 intervals, each will have a width of 2 0 , 6 rad (or 15° ) and the ordinates occur at 0, i.e 12 5 , , , , and Corresponding values of 12 6 4 3 12 2 NUMERICAL INTEGRATION 1 are shown in the table below: 1 C sin x 44 1 /3p 3 sin  d  (Use 6 intervals) 0 [0.672] 1 1 C sin x x 1 .4 4 e x2 dx (Use 7 intervals) 0 0 12 [0. 843 ] 1.0000 (or 15° ) 0.7 944 0 (or 30° ) 0.66667... 0 and x D 5 (see Fig 56 .4) then: y = x2 + 4 A 0 D C 1 2 3 4 5 x Figure 56.5 y y = 2x 10 10 5 Revolving the shaded area shown in Fig 56.5 about the x-axis 360° produces a solid of revolution given by: 4 0 1 2 3 4 5 x Volume D 1 −5 x2 C 4 2 d x 1 4 x 4 C 8x 2 C 16 d x D −10 Figure 56 .4 4 y2 d x D 1 D x5 8x 3 C C 16x 5 3 4 1 VOLUMES OF SOLIDS OF REVOLUTION D [ 2 04. 8 C 170.67 C 64 6 7 y D 3x 2 8 yD 9 0.2... 81 4 D 2 5x 2 2 3 1 12 1 or 21.08 square units 12 15 3 4 Problem 4 Determine the area enclosed by the curve y D 3x 2 C 4, the x-axis and ordinates x D 1 and x D 4 by (a) the trapezoidal rule, (b) the mid-ordinate rule, (c) Simpson’s rule, and (d) integration AREAS UNDER AND BETWEEN CURVES 45 1 (c) By Simpson’s rule, x 0 1.0 1.5 2.0 2.5 3.0 3.5 4. 0 y 4 7 10.75 16 22.75 31 40 .75 52 y y= 3x 2 area D +4. .. +4 1 3 width of interval first + last ordinates C4 40 sum of even ordinates C2 50 sum of remaining odd ordinates Selecting 6 intervals, each of width 0.5, gives: 30 area D 20 1 0.5 [ 7 C 52 C 4 10.75 C 22.75 3 C 40 .75 C 2 16 C 31 ] D 75 square units 10 (d) By integration, shaded area 4 4 y dx D 0 1 2 3 4 1 x 4 3x 2 C 4 d x D 1 Figure 54. 6 D [x 3 C 4x ]4 1 D 75 square units Integration gives the precise . gives: 5xe 4x dx D 5x e 4x 4 e 4x 4 5 d x D 5 4 xe 4x 5 4 e 4x dx D 5 4 xe 4x 5 4 e 4x 4 C c D 5 4 e 4x x 1 4 C c Hence 1 0 5xe 4x dx D 5 4 e 4x x 1 4 1 0 D 5 4 e 4 1. ordinates y 3 , y 4 and y 5 , and so on. Then b a y dx ³ 1 3 dy 1 C 4y 2 C y 3 C 1 3 dy 3 C 4y 4 C y 5 C 1 3 dy 2n1 C 4y 2n C y 2nC1 44 4 ENGINEERING MATHEMATICS y y 1 y 2 y 3 y 4 y 2 n +. 2, page 44 0. Thus, from equation (5): 3 1 2 p x dx ³ 1 3 0.25 [ 2.0000 C1.1 547 C 4 1.7889 C1.5119 C1.3333 C 1.2060 C21.6330 C 1 .41 42 C1.2 649 D 1 3 0.25[3.1 547 C23.36 04 C 8.6 242 ] D
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