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- 1. INTEGRATION OF TRIGONOMETRIC INTEGRALSRecall the definitions of the trigonometric functions. The following indefinite integrals involve all of these well-knowntrigonometric functions. Some of the following trigonometryidentities may be needed. A.) B.) C.) so that D.) so that E.) F.) so that G.) so that
- 2. It is assumed that you are familiar with the following rules ofdifferentiation. These lead directly to the following indefinite integrals. o 1.) o 2.) o 3.) o 4.) o 5.) o 6.)The next four indefinite integrals result from trig identities and u-substitution. o 7.)
- 3. o 8.) o 9.) o 10.)We will assume knowledge of the following well-known, basicindefinite integral formulas : , where is a constant , where is a constant. Integrals with Inverse Trigonometric Functions1.2.
- 4. 3.4.5.6.7.8.9.10.
- 5. 11.12.13.14.15.16.17.
- 6. 18.19.20.21.22.23.24.
- 7. 25.26.27.28.29.
- 8. 30.31.32.33.34.35.
- 9. 36.37.38.
- 10. These formulas, if effectively, practiced would help you performwell in mathematics section especially calculus part of yourboards and entrances, thus helping you secure good marks inyour class XII exams and helping you secure a good rank inentrance exams.
- 11. How to score well Before you start the exam, utilize the first 15 minutes to scan the paper. Read the question paper thoroughly before jumping to write the answers. Among the questions with internal choices, select the ones that you plan to attempt, and frame skeletons of the answers you are going to write for these questions. Follow a pattern. For example, in case you start with long answer questions, complete that section and only then move to short or very short answer section. Highlight the important points and write your answer in points to enhance visibility. Points to remember Marks are deducted for missing steps. So remember to write down all the steps. Practice. Practice. Practice. This is the mantra for scoring good marks in CBSE Class 12 Mathematics Exam. Make NCERT book your bible. Revise and practise all the problems solved in the NCERT book.
- 12. Question: Integrate . Let u = x-1so that du = (1) dx = dx .In addition, we can "back substitute" with x = u+1 .Substitute into the original problem, replacing all forms of x,gettingQuestion: Integrate . Let u = 2x+3so that
- 13. du = 2 dx ,or (1/2) du = dx .In addition, we can "back substitute" with x = (1/2)(u-3) .Substitute into the original problem, replacing all forms of x,gettingQuestion: Integrate . Let u = x+2
- 14. so that du = (1) dx = dx .In addition, we can "back substitute" with x = u-2 .Substitute into the original problem, replacing all forms of x,gettingQuestion: Integrate . Let
- 15. so that .In addition, we can "back substitute" with .Substitute into the original problem, replacing all forms of x,gettingQuestion: Integrate . Use u-substitution. Let u = 1+3e-xso that (Dont forget to use the chain rule on e-x.)
- 16. du = 3e-x(-1) dx = -3e-x dx ,or (-1/3)du = e-x dx .However, how can we replace the term e-3x in the original problem? Note that .From the u-substitution u = 1+3e-x ,we can "back substitute" with e-x = (1/3)(u-1) .Substitute into the original problem, replacing all forms of x,getting(Recall that (AB)C = AC BC .)
- 17. Question: Integrate . Use u-substitution. Let u = e2x+6ex+ 1so that (Dont forget to use the chain rule on e2x.) du = (2e2x+6ex) dx = (2ex+x+6ex) dx = (2exex+6ex) dx = 2ex(ex+3) dx
- 18. = 2ex(3+ex) dxor (1/2) du = ex(3+ex) dx .Substitute into the original problem, replacing all forms of x,getting(Do not make the following very common mistake : . Why is this incorrect ?)
- 19. .Question: Integrate . First, factor out e9xfrom inside the parantheses. Then(Recall that (AB)C = AC BC .)(Recall that (AB)C = ABC .) .Now use u-substitution. Let u = 27+e3xso that (Dont forget to use the chain rule on e3x.) du = 3e3x dx ,
- 20. or (1/3) du = e3x dx .Substitute into the original problem, replacing all forms of x , andgetting . Question: Integrate . Use u-substitution. Let so that
- 21. ,or .Substitute into the original problem, replacing all forms of ,getting .Question: Integrate . First multiply by , getting . . .Now use u-substitution. Letso that .
- 22. Substitute into the original problem, replacing all forms of ,getting .Question: Integrate . Let andso that and .Therefore, .SOLUTION 6 : Integrate . Let
- 23. andso that (Dont forget to use the chain rule when differentiating .) and .Therefore, .Now use u-substitution. Letso that ,or .Then
- 24. +C +C +C.Question: Integrate . Let andso that and .Therefore,(Add in the numerator. This will replicate thedenominator and allow us to split the function into two parts.) .Question: Integrate . Let andso that
- 25. and .Therefore, .Integrate by parts again. Let andso that and .Hence, .SOLUTION : Integrate . Use the powersubstitutionso that ,
- 26. ,and .Substitute into the original problem, replacing all forms of ,getting .SOLUTION 6 : Integrate . Use the powersubstitutionso that , ,and .Substitute into the original problem, replacing all forms of ,getting
- 27. (Use polynomial division.) .Question: Integrate . Because the degree of thenumerator is not less than the degree of the denominator, wemust first do polynomial division. Then factor and decomposeinto partial fractions, getting(After getting a common denominator, adding fractions, andequating numerators, it follows that ;let ;let .)
- 28. (Recall that .)Question: Integrate . Because the degree ofthe numerator is not less than the degree of the denominator,we must first do polynomial division. Then factor anddecompose into partial fractions, getting(After getting a common denominator, adding fractions, andequating numerators, it follows that ;let ;let .)
- 29. .SOLUTION : Integrate . Use the power substitutionso thatand .Substitute into the original problem, replacing all forms of ,getting(Use polynomial division.) .
- 30. SOLUTION 4 : Integrate . Use the powersubstitutionso that , ,and .Substitute into the original problem, replacing all forms of ,getting .SOLUTION : Integrate . Use the power substitutionso that
- 31. and .Substitute into the original problem, replacing all forms of ,getting(Use polynomial division.) .Use the method of partial fractions. Factor and decompose intopartial fractions, getting(After getting a common denominator, adding fractions, andequating numerators, it follows that ;let ;let .)(Recall that .) .
- 32. Question: Integrate . Decompose into partial fractions,getting(After getting a common denominator, adding fractions, andequating numerators, it follows that ;let ;let ;it follows that and .) . Question: Integrate . Use u-substitution. Let so that .
- 33. Now rewrite this rational function using rules of exponents. Then .Substitute into the original problem, replacing all forms of ,getting .Question: Integrate . First complete the squarein the denominator, getting .Now use u-substitution. Letso that .In addition, we can "back substitute" with
- 34. .Substitute into the original problem, replacing all forms of ,getting .In the first integral use substitution. Letso that ,or .Substitute into the first integral, replacing all forms of , and useformula 3 from the beginning of this section on the secondintegral, getting
- 35. Integrate . First, use polynomial division to divideby . The result is .In the second integral, use u-substitution. Letso that .Substitute into the original problem, replacing all forms of ,getting(Now use formula 1 from the introduction to this section.) .SOLUTION : Integrate . Let andso that and .Therefore,
- 36. .Use integration by parts again. let andso that and .Hence, .To both sides of this "equation" add , getting .Thus,(Combine constant with since is an arbitraryconstant.) .Question: Integrate . Use integration byparts. Let andso that
- 37. and .Therefore, .Use integration by parts again. let andso that and .Hence, .From both sides of this "equation" subtract ,getting .Thus,(Combine constant with since is an arbitrary constant.)
- 38. INTEGRATION AS LIMIT OF A SUM AND ITS EVALUATIONQuestion: Use the limit definition of definite integral to evaluateDivide the interval into equal parts each of lengthfor . Choose the sampling points to be the right-hand endpoints of the subintervals and given byfor . The function is .Then the definite integral is
- 39. (Use summation rule 6 from the beginning of this section.)(Use summation rules 5 and 1 from the beginning of this section.)(Use summation rule 2 from the beginning of this section.) . Application of Integrals
- 40. Q. 1. Find the area of the region in the first quadrantenclosed by the x-axis, the line y = x and the circleQ. 2. Find the area of the region bounded by the ellipse .Q. 3. Find the area of the region bounded by the parabolay = x2 and y = .Q. 4. Find the area of the smaller part of the circle x2 + y2 =a2 cut off by the linex= .Q. 5. Using integration, find the area of the region boundedby the triangle whose vertices are (1, 0), (2,2) and (3, 1).Q. 6. Prove that the curves y2 = 4x and x2 = 4y divide thearea of the square bounded by x=0, x=4, y=4 and y=0 intothree equal parts.Q. 7. Sketch the graph of y=Q. 8. Using the method of integration, find the areabounded by the curve .Q. 9. Find the area of the smaller region bounded by theellipse .Q. 10. Using integration, find the area of the triangularregion, the equations of whose sides are y=2x + 1, y=3x +1and x = 4.Q. 11. Find the area of the region
- 41. Q. 12. Find the area of the region between the circles x2 +y2 = 4 and (x – 2)2 + y2 = 4.Q. 13. Find the area bounded by the ellipse and 2 2 2the co-ordinates x = ae and x = 0, where b =a (1 – e ) ande<1.Q. 14. Find the area bounded by the curve y2 = 4a2(x – 1)and the lines x = 1and y = 4a.Q. 15. Using integration, find the area of the regionbounded by the following curves, after making a roughsketch: y = 1 +Q. 16. Draw a rough sketch of the curves y = sinx and y =cosx as x varies from o to and find the area of theregion enclosed by them and x-axis.Q. 17. Find the area lying above x-axis and includedbetween the circle x2+ y2 = 8x and the parabola y2 = 4x.Q. 18. Using integration find the area of the triangularregion whose sides have the equations y = 2x + 1, y = 3x +1 and x = 4.Q. 19. Find the area enclosed between the parabola y2 =4ax and the line y = mx.Q. 20. Find the area of the region bounded by theparabolas y2 = 4 ax and x2 = 4 by(NCERT)Question 12: [ use: sin2x = 1- cos2x, ans. is x – sinx+c]
- 42. Question 14: [ Use: 1+sin2x= (cosx+sinx)2 , putcosx+sinx = t , ans.is -1/(cosx+sinx)+c ]Question 18: dx [use: cos2x = cos²x-sin²x , ans.tanx+c]Question 22: [multiply & divide by sin(a-b), writesin(a-b) = sin{(x-b)-(x-a)} in Nr., use formula of sin(A-B).ans. is +c ]Question 23: is equal to [(A) ]A. tan x + cot x + C B. tan x + cosec x + C C. − tan x + cot x + CD. tan x + sec x + CQuestion 24: equals [(B) ]A. − cot (exx) + C B. tan (xex) + C C. tan (ex) + C D. cot (ex) + CQuestion 5: * Put x² = t , ans. is (3/2√2)tan-1(√2X²) +C +Question 9: [Put tanx = t, ans. is log|tanx+ |+c]Question 14: [ Dr. Can be written as = = , ans. is sin-1( )+c ]Question 17: [ + dx, put x²-1=t in 1stintegral, ans. is +2 log|x+ | +c ]Question 18: [ let 5x-2 = P.d/dx(1+2x+3x²)+Q, P=5/6 &Q=-11/3 , Ans. is 5/6 log (1+2x+3x²) – (11/3√2) tan-1( ) +c]
- 43. Question 25: equals *Dr. √–(4x²-9x)⇨ )² ,(B)]A B C DQuestion 3: [ by partial fraction, A/(x-1)+B/(x-2)+C/(x-3) ⇨A=1,B=-5 & C=4, ans. is log|x-1|-5log|x-2+4log|x-3|+c]Question 8: [ A/(x-1)+B/(x-1)2 +C/(x+2)⇨ A=-C=2/9,B=1/3, Ans. is 2/9log| |-1/3(x-1) +c]Question 10: [same as Q.3, A=-1/10,B=5/2 & C=-24/5,Ans. is 5/2 log|x+1|-1/10log|x-1|-12/5log|2x+3|+c ]Question 12: [after division, we get x+ ⇨A=1/2,B=3/2 , ans. is x²/2+1/2log|x+1|+3/2log|x-1|+c ]Question 15: [ A/(x+1)+B/(x-1)+Cx+D/(x²+1)⇨ A=-1/4,B=1/4,C=0 & D=-1/2 , ans. is ¼ log| |-1/2 tan-1x+c ]Question 17: [Hint: Put sin x = t, ans.is log| |+c]Question 18: [ put x²=y , , after dividing , weget , 1- , by partial fraction A/(y+3) +B/(y+4) ⇨ A=-1, B=3,ans. is x+(2/√3)tan-1(x/√3)-3tan-1(x/2) +c ]Question 23: A. [(A) , multiply & 2divide by x, put x = t, by partial fraction.]
- 44. B. C. D.Question 5: x log 2x [integral by parts, (log2x).x²/2- ) dx⇨ (log2x).x²/2 – x²/4+c ]Question 14: [ integral by parts, (logx)².x²/2- 2 .1/x](x /2)dx , again by parts ⇨ (logx)².x²/2- [log x.(x²/2)- ] ⇨ (logx)².x²/2- x2/2(logx)+(1/4)x2 +c]Question 6: [ ⇨ –(9/2)log|(x+2)+ +c ]Question 7: [ )2 ] ⇨(2x-3)/4+(13/8)sin-1 (2x-3)/√3 +c +Question 20: [ , by parts⇨xex -ex – 4/п[cos(пx/4)] at x=0 to 1⇨ 1+4/п - 2√2/п ]Question 4: [ ans. is 16/15(2+√2) +Question 6: [ Dr. (17/4)- (x-1/2)² ⇨ (1/√17)log| |Put x=0 to 2 ⇨ (1/√17) log( )]Question 8: [by parts , ans is (e2/4)(e²-2) ]Question 9:The value of the integral is A. 6 B. 0 C. 3 D. 4 -1[ put x=sinѲ , limit will change from Ѳ=sin 1/3 when x=1/3 &Ѳ=п/2 when x=1, ⇨ dѲ , put cotѲ=t, againlimit will change from √8 to 0 ans. is (A) =6]
- 45. IMPORTANT PROPERTIES OF DEFINITE INTEGRALS:1. =2. = a<c<b3. =4. =5. = +6. =2 , if f(2a-x) = f(x) = 0 , if f(2a-x) = - f(x)7. = 2 , if f is an even function i.e., f-x) = f(x) = 0, if f is an odd function i.e., f(-x) = -f(x).Question 6: [ ⇨ 9]Question 10: [ = ⇨ - =- ]Question 12: [use property 4.⇨2I = . =2п ]Question 15: [ use property 4. ⇨ 2I = =0 ]Question 16: [ use property 4. ⇨ 2I =⇨ I= =2 (BY PROP.6)= 2I1 …..(i) , where I1= (by using prop. 4) ⇨ 2I1 =⇨ - = I2 - п/2.log2 …..(ii) , where I2= = (put 2x=y) = (byprop. 6), from (i) & (ii) we get I = - п log2. ]Question 19: Show that if f and g are definedas and
- 46. [ by prop. 4 ⇨ I = = = (according to given part) ⇨ I = 2 ]Question 21:The value of is A. 2 B. C. 0 D.[ use prop. 4 ⇨ 2I = =0 ]Misc. Question 3: [Hint:Put ,ans. Is -2/a +c]Question 4: [ put x= , ans. Is – ]Question 5: * ans. Is 2√x- 3x1/3 +6x1/6 -6log|x1/6 +1| +c ]Question 10: [ Nr. can be written as 2 2(1-2sin xcos x)(-cos2x), ans. Is -1/2 sin2x +c ]Question 11: [ same as Ques. 22]Question 15: [ put cosx =y , ans. Is -1/4 cos4x +c ]Question 16: [ put t = x4+1 , ans. Is ¼ log| x4+1| +c ]Question 18: [ Dr. = sin4x cos + sin3xcosxsin =sin4x(cos + cotxsin ) ( by using formula of sin(A+B)) , put t= cos+ cotxsin , ans. Is -2/sin . +c ]Question 19: [ use sin-1√x + cos-1√x =п/2 ⇨
- 47. 2/ dx – x , put √x= t, integrate by parts & useformula of dt , ans. Is 4/п{ sin-1√x+ } –x +c]Question 20: [ put x2 = cosy , use cos2y= (1+cos2y)/2, ans. Is-2 √(1-x) + cos-1√x + √x . √(1-x) +c ]Question 21: [use (f(x)+f’(x))dx , sin2x = 2sinx.cosx & 2 x1+cos2x = 2cos x , ans. Is e tanx +c ]Question 22: [ by partial fraction , we get A/(x+2) +B/(x+1) +C/(x+1)2 ⇨ A=3, B=-2 & C=1, ans. Is log - (X+1)-1 +c ]*Question 24: [ after simplification,weintegrate . log(1+ ) dx , put x = tanѲ , then put sinѲ=t (byparts) , ans. Is -1/3 (1+ )3/2 { log(1+ )-2/3 } +c ]Question 25: [ same as Ques. 21, ans. Is eп/2 ]Question 26: [ divide Nr. & Dr. by cos4x , put tan2x =y & limit will change from 0 to 1, ans. Is п/8 ]Question 27: [ use sin2x = 1 – cos2x , Nr. Can bewritten as 4-3cos2x-4 ⇨ -п/6+ 4/3 dx , put tanx = t,limit will change from 0 to ∞, ans. Is п/6 or we can do it by anothermethod ( by partial fraction) divide Nr. & Dr. by cos2x , put tanx = t]Question 28: [ put sinx-cosx=t ∵ sin2x=1-(sinx-cosx)2Limit will change from –(√3-1)/2 to (√3-1)/2 ⇨ ∵ evenfn. , ans. Is 2 ]
- 48. Question 30: [ put sinx-cosx = t, same as Ques. 28 ,limit will change from -1 to 0, ans. Is 1/40 log9 ]Question 31: [ use sin2x formula , put sinx=t ,integrate by parts , ans. Is п/2 -1 ]Question 32: [ use prop. 4 ⇨ 2I = dx= - , ans. Is ]Question 33: [ + + dx+ dx + dx = 19/2 ]Question 34: [ by partial fraction A/x +B/x2+C/(x+1) ⇨ A= -1, B=1 & C=1 ]Question 39: [ by parts ∫1. Sin-1xdx ]Question 40: Evaluate dx as a limit of sum.[nh =1, = +………+f((n-1)h)] +………….. ] . ( ∵ nh=1) =( ( =1 ]Question 43:If then is equal toA. B. C. D.Question 44:The value of is A. 1 B. 0 C. – 1 D.[Nr.=x+(x-1) & Dr.=1-x(1-x), use prop. 4 , it gives tan-1x+tan-1(x-1),B]Definite IntegralQuestion 2: Find the area of the region bounded by y2 = 9x, x = 2, x= 4 and the x-axis in the first quadrant.
- 49. Question 5:Find the area of the region bounded by the ellipseQuestion 6:Find the area of the region in the first quadrantenclosed by x-axis, line and the circleQuestion 7:Find the area of the smaller part of the circle x2 + y2 = a2cut off by the lineQuestion 9:Find the area of the region bounded by the parabola y =x2 andQuestion 10:Find the area bounded by the curve x2 = 4y and the linex = 4y – 2Question 1:Find the area of the circle 4x2 + 4y2 = 9 which is interiorto the parabola x2 = 4yQuestion 2:Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 +y2 = 1Question 5:Using integration find the area of the triangular regionwhose sides have the equations y = 2x +1, y = 3x + 1 and x = 4.Question 4:Using integration finds the area of the region boundedby the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).Question 6:Smaller area enclosed by the circle x2 + y2 = 4 and theline x + y = 2 isA. 2 (π – 2) B. π – 2 C. 2π – 1 D. 2 (π + 2)Question 7:Area lying between the curve y2 = 4x and y = 2x isA. B. C. D.Question 4:Sketch the graph of and evaluateQuestion 8:Find the area of the smaller region bounded by theellipse and the lineQuestion 11:Using the method of integration find the area boundedby the curve
- 50. [Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x+ y = 1 and – x – y = 11]Question 12:Find the area bounded by curvesQuestion 14:Using the method of integration find the area of theregion bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0Question 15:Find the area of the regionQuestion 17:The area bounded by the curve , x-axis and theordinates x = –1 and x = 1 is given by [Hint: y = x if x > 0 and y = –x2 2if x < 0]A. 0 B. C. D.Question 18:The area of the circle x2 + y2 = 16 exterior to theparabola y2 = 6x isA. B. C. D.Question 19:The area bounded by the y-axis, y = cos x and y = sin xwhenIntegration Problems1. ∫(2x3 + 5x + 1)e2x dx [ans. e2x(x3 – 3/2x2 + 4x – 3/2) + C ]2. ∫cos2 x tan2 x dx [ x/2− 1/4 sin 2x + C]3. ∫e cos 2x sin x cos x dx [ (−1/4) ecos 2x + C]4. dx [ ln |2 + tan x| + C]5. [ x2/2 − 3x + 8 ln |x + 3| + C]6. dx [ 1/3(x2 + 4)3/2 − 4(x2 + 4)1/2 + C] 7. dx [ 2/3 ln |1 + 3√x| + C]
- 51. 8. dx [x +ln|x|+1/2ln|x2+4|−1/2arctan(x/2)+c]9. [2/3 sin3x – cosx+c]10. [ (2 - √2)/3 ]**Question: Integrate . Let .In addition, we can "back substitute" with ,or x = (4-u)2 = u2-8u+16 .Then dx = (2u-8) du .In addition, the range of x-values is ,so that the range of u-values is ,or
- 52. .Substitute into the original problem, replacing all forms of x,getting .INTEGRATION AS LIMIT OF A SUM AND ITS EVALUATION
- 53. Question: Divide the interval into equal parts each oflengthfor . Choose the sampling points to be the right-hand endpoints of the subintervals and given byfor . The function is .Then the definite integral is(Use summation rule 6 from the beginning of this section.)
- 54. (Use summation rules 5 and 1 from the beginning of this section.)(Use summation rule 2 from the beginning of this section.) . .SOLUTION : Divide the interval into equal parts each oflength
- 55. for . Choose the sampling points to be the right-hand endpoints of the subintervals and given byfor . The function is .Then the definite integral is(Use summation rule 6 from the beginning of this section.)(Use summation rules 5 and 1 from the beginning of this section.)
- 56. (Use summation rule 2 from the beginning of this section.) .Question: Divide the interval into equal parts each of lengthfor . Choose the sampling points to be the right-hand endpoints of the subintervals and given byfor . The function is .
- 57. Then the definite integral is (Recall that .)
- 58. (Use LHopitals rule since the limit is in the indeterminate form of .) .**Question: Integrate . First, split the function intotwo parts, getting(Recall that .)(Use formula 2 from the introduction to this section on integratingexponential functions.)
- 59. .**Question: Integrate . Note that .Let andso that and .Therefore,
- 60. .**SOLUTION : Integrate . First, use polynomialdivision to divide by . The result is .In the third integral, use u-substitution. Letso that ,or .
- 61. For the second integral, use formula 2 from the introduction tothis section. In the third integral substitute into the originalproblem, replacing all forms of , getting(Now use formula 1 from the introduction to this section.)**Question: Integrate . First rewrite thisrational function as .Now use u-substitution. Let .so that ,or .In addition, we can "back substitute" with .
- 62. Substitute into the original problem, replacing all forms of ,getting =**Question: Integrate . Because the degree of thenumerator is not less than the degree of the denominator, wemust first do polynomial division. Then factor and decomposeinto partial fractions (There is a repeated linear factor !), getting(After getting a common denominator, adding fractions, andequating numerators, it follows that
- 63. ; let ; let ; let ; let ; it follows that and .)**Question: Integrate . Factor anddecompose into partial fractions (There is a repeated linear factor!), getting
- 64. (After getting a common denominator, adding fractions, andequating numerators, it follows that ;let ;let ;let ;let .) .
- 65. **SOLUTION : Integrate . Factor and decomposeinto partial fractions (There are two repeated linear factors !),getting(After getting a common denominator, adding fractions, andequating numerators, it follows that ;let ;let ;let ;let ;it follows that and .)
- 66. .**Question: Integrate . Begin by rewriting thedenominator by adding , getting(The factors in the denominator are irreducible quadratic factorssince they have no real roots.)
- 67. (After getting a common denominator, adding fractions, andequating numerators, it follows that ;let ;it follows that and and ;letit follows that and and .) .Now use the method of substitution. In the first integral, letso that .In the second integral, let
- 68. so that .In addition, we can ``back substitute", usingin the first integral andin the second integral. Now substitute into the original problems,replacing all forms of , getting
- 69. (Recall that .) .**Solution: Integrate. U se the power substitutionso thatand .
- 70. Substitute into the original problem, replacing all forms of ,getting(Use polynomial division.) .**Question: Integrate . Because we want tosimultaneously eliminate a square root and a cube root, use thepower substitution
- 71. so that , , ,and .Substitute into the original problem, replacing all forms of ,getting(Use polynomial division. PLEASE INSERT A FACTOR OF 6 WHICHWAS ACCIDENTLY LEFT OUT.)
- 72. .**SOLUTION : Integrate . Remove the ``outside"square root first. Use the power substitutionso that , , ,and (Use the chain rule.)
- 73. .Substitute into the original problem, replacing all forms of ,getting .**SOLUTION : Integrate . Remove the cube rootfirst. Use the power substitution
- 74. so that , , ,and (Use the chain rule.) .Substitute into the original problem, replacing all forms of ,getting .
- 75. **Question: Integrate . Remove the``outside" square root first. Use the power substitutionso that ,and (Use the chain rule.) ,or .Substitute into the original problem, replacing all forms of ,getting
- 76. .**SOLUTION : Integrate . Use the power substitutionso that ,
- 77. ,and .Substitute into the original problem, replacing all forms of ,getting .Use the method of partial fractions. Factor and decompose intopartial fractions, getting (There are repeated linear factors!)(After getting a common denominator, adding fractions, andequating numerators, it follows that ;let ;
- 78. let ;let ;let ;it follows that and .)(Recall that .)
- 79. .**SOLUTION : Integrate . Use the powersubstitutionso thatand .Substitute into the original problem, replacing all forms of ,getting(Use polynomial division.)
- 80. .Use the method of partial fractions. Factor and decompose intopartial fractions, getting(After getting a common denominator, adding fractions, andequating numerators, it follows that ;let ;let ;let ;it follows that and and .)
- 81. .DEFINITE INTEGRALTheory:To find the area between two intersecting curves that onlyintersect at two points, we first find the ‘x’ coordinates of the twointersection points: x = a and x = b. Definite integrals give us thearea under each curve from x = a to b, then we subtract the twoareas to obtain the area between the curves. In the diagram below,the area between the two graphs is shaded:
- 82. Area under a CurveThe area between the graph of y = f(x) and the x-axis is given by thedefinite integral below. This formula gives a positive result for agraph above the x-axis, and a negative result for a graph below thex-axis.Note: If the graph of y = f(x) is partly above and partly below the x-axis, the formula given below generates the net area. That is, thearea above the axis minus the area below the axis. Formula:
- 83. Example Find the area between y = 7 – x2 and the x- 1: axis between the values x = –1 and x = 2. Example Find the net area between y = sin x and the 2: x-axis between the values x = 0 and x = 2π.Area between CurvesThe area between curves is given by the formulas below.Formula 1:
- 84. for a region bounded above and below by y = f(x) and y = g(x), and on the left and right by x = a and x = b.Formula 2: for a region bounded left and right by x = f(y) and x = g(y), and above and below by y = c and y = d.Example 1:1 Find the area between y = x and y = x2 from x = 1 to x = 2.Example 2:1 Find the area between x = y + 3 and x = y2 from y = –1 to y = 1.
- 85. Area Under a CurveDefinite IntegralsSo far when integrating, there has always been a constant termleft. For this reason, such integrals are known as indefiniteintegrals. With definite integrals, we integrate a function between2 points, and so we can find the precise value of the integral andthere is no need for any unknown constant terms [the constantcancels out].
- 86. The Area Under a CurveThe area under a curve between two points can be found by doinga definite integral between the two points.To find the area under the curve y = f(x) between x = a and x = b,integrate y = f(x) between the limits of a and b.
- 87. Areas under the x-axis will come out negative and areas above thex-axis will be positive. This means that you have to be careful whenfinding an area which is partly above and partly below the x-axis.You may also be asked to find the area between the curve and they-axis. To do this, integrate with respect to y.ExampleFind the area bounded by the lines y = 0, y = 1 and y = x2.
- 88. EXAMPLE 4: Find the area between the curve f (x) = cos п x on the interval [0, 2].SOLUTION:STEP 1: Graph the function.(See figure 3)STEP 2: Set up the integralsand evaluate.Notice that the area wehave to find is in three figure 3pieces. The intervals [0, .5]and [1.5, 2] are above the x-axis, and the interval [.5,1.5] is below. Therefore, wewill need to have threeintegrals. Also notice thatsymmetry cannot be used in
- 89. this problem.EXAMPLE 5: Find the area between the curves f (x) = 4 - x 2 and g (x) = x 2 - 4.SOLUTION:STEP 1: Graph the functions.(See figure 4)The reason for graphing thetwo equations is to be ableto determine which functionis on top and which one ison the bottom. Sometimes, figure 4you can also determine thepoints of intersection. Fromthis graph, it is cleat that f(x) is the upper function, g(x) is the lower function, andthat the points ofintersection are x = -2 and x= 2.STEP 2: Determine the points of intersection.If you did not determine the points of intersection from the graph,solve for them algebraically or with your calculator. To find themalgebraically, set each equation equal to each other.
- 90. 4 - x 2 = x 2 – 4 ⇨ -2x 2 = -8 ⇨ x 2 = 4 ⇨ x = -2 or x = 2STEP 3: Set up and evaluate the integral.Recall from early in the notes, when we were finding the areabetween the curve and the x-axis, we had to determine the upperand the lower curve. Then the area was defined to be the followingintegral.So the definite integral would be the following.Now, let us evaluate the integral.If you look at the graph of the two functions carefully, you shouldhave noticed that we could have used some symmetry whensetting up the integral. The region is symmetric with respect toboth the x- and the y-axis. If we had used the y-axis symmetry, theresulting integral would have had bounds of 0 and 2, and we wouldhave had to take 2 times the area to find the total area. Here is thisintegral.
- 91. If we had used both symmetries, the resulting integral would stillhave bounds of 0 and 2, but the upper function would have been f(x) and the lower function would be y = 0 (the x-axis). To find thetotal area, we would have to take this area times 4. Here is thisintegral.EXAMPLE 7: Find the area between the curves x = y 3 and x = y 2 that is contained in the first quadrant.SOLUTION: STEP 1: Graph the functions. (See figure 6)Since both equations are x in terms of y, we will integrate withrespect to y. When integrate with respect to x, we have todetermine the upper function and the lower function. Now that weare integrating with respect to y, we must determine what functionis the farthest from the y-axis. The function that is the farthestfrom the y-axis is x = y 2. So that will be our upper curve. The lowercurve will be the curve that is nearest to the y-axis. In this case, it isthe function x = y 3.
- 92. figure 6STEP 2: Find the points of intersection.Set the two equations equal to each other.y 2 = y 3 ⇨ y 2 - y 3 = 0 ⇨ y 2 (1 - y) = 0 ⇨ y = 0 or y = 1STEP 3: Set up and evaluate the integral. using definite integrals to find the area between two curves From the figure we can easily get that the area of the shaded portion spqr = area tpqu - area tsru. This is equivalent to the area enclosed between the curve y = f(x),
- 93. the x-axis and the ordinates x=a and x = b Minus the areaenclosed between the curve y = g(x), the x-axis and the ordinatesx = a and x = b. this is expressed mathematically as follows: aTherefore, the area between the two curves can be expressed as aExample - 3 Find the area bounded by the curves y = x2 and y = 2x. Solution: Step 1: To find the region we need to sketch the graph and find where the two curves intersect. To find where the curves intersect, we will set them equal to each other and solve for x. 2x = x2 X2 - 2x = 0 X(x - 2) = 0 X = 0 or x - 2 = 0 X = 0 or x = 2 Plugging x = 0 into y = 2x gives us y = 2(0) = 0 Plugging x = 2 into y = 2x gives us y = 2(2) = 4 Therefore, the curves intersect at the points (0, 0) and (2, 4)
- 94. Step 2: as we can see in the figure, we are to find the area of the shaded portion oabdo. Area oabdo = area of oabco - area of odbco. = the area enclosed between the straight line y = 2x, x-axis, x = 0 and X = 2 Minus the area enclosed between the curve y = x2, x-axis, x = 0 and x = 2. Step 3: solve the definite integral. square unitsExample - 4: Find the area bounded by the curves x2 = 4y and y2 = 4x. Solution: Step 1: Solve the given equations to find the points of intersection. (1) x2 = 4y, (2) y2 = 4x Squaring both sides of (1) gives us x4 = 16y2 Substituting y2 = 4x into this equation gives us x4 = 16(4x)
- 95. x4 = 64xx4 - 64x = 0x(x3 - 64) = 0x = 0 or x3 = 64x = 0 or x = 4Plugging x = 0 into x2 = 4y gives us 0 = 4y implies that y = 0Plugging x = 4 into x2 = 4y gives us 16 = 4y implies that y = 4therefore, the points of intersection are (0, 0) and (4, 4)Step 2: Sketch the graph.Step 3: Solve both equations for y and write the formula forfinding the area of the shaded region. Y2 = 4x Y = 2 since this is the equation of the top line, this will be the first part of our equation. X2 = 4y Y = x2since this is the equation of the bottom line, this will be thesecond part of our equation. (recall the formula )
- 96. Therefore, the area of the shaded portion Sq. UnitsArea Bounded by Two Curves: See Figure 12.3-8.Example 1Find the area of the region bounded by the graphs of f (x)=x3 andg(x )=x. (See Figure 12.3-9.)
- 97. Step 1. Sketch the graphs of f (x ) and g (x ).Step 2. Find the points of intersection.Step 3. Set up integrals.
- 98. Note: You can use the symmetry of the graphs and let areaAnalternate solution is to find the area using a calculator. Enter and obtain .Example 2Find the area of the region bounded by the curve y =ex, the y-axisand the line y =e2.Step 1. Sketch a graph. See Figure 12.3-10.Step 2. Find the point of intersection. Set e2 =ex x =2.Step 3. Set up an integral:Or using a calculator, enter and obtain (e2 +1).
- 99. Example 3Using a calculator, find the area of the region bounded by y = sin xand between 0≤ x ≤ π.Step 1. Sketch a graph. See Figure 12.3-11.Step 2. Find the points of intersection. Using the [Intersection] function of the calculator, the intersection points are x =0 and x =1.89549.Step 3. Enter nInt(sin(x ) &8211; .5x, x, 0, 1.89549) and obtain0.420798 ≈ 0.421. (Note: You could also use the function on your calculator and get the same result.)Example 4Find the area of the region bounded by the curve x y =1 and thelines y = –5, x =e, and x =e3.Step 1. Sketch a graph. See Figure 12.3-12.
- 100. Step 2. Set up an integral.Step 3. Evaluate the integral.
- 101. ASSIGNMENT OF INTEGRATIONQuestion 1 Evaluate: (i)** Integrate .[ Use the powersubstitution Put ] ** (iii) Integrate . [ Use the power substitution Put ] (iii) [answer is (2 - √2)/3 ](iv) ∫ dx[multiply÷ by sin(a-b)](v) dx[multiply & divide by ] (Vi)∫ dx [by partial fraction](v) dx [ use ∫ex(f(x)+f’(x))dx+ (vi) dx [put sinx= , cosx = , then put t=tanx/2. Answer is – ](vii) dx [ + = ∫+ve dx+∫ -ve dx ,answer is 5/2п- 1/п2] (viii) [ write sin2x = 1-cos2x answer isп/6] (ix) + dx * answer is √2 ] (x) dx [ putx=atan2Ѳ , answer is a/2(п-2) ] (xi) dx [ use property dx= dx , dx = dx ∵f(2a-x) = f(x) , then put t=tanx,answer is п²/2√2 ] (xii) dx , where f(x) =|x|+|x+2|+|x+5|.[ dx + dx , answer is 31.5 ] (xiii) Evaluate dx[use (f(x)+f’(x))dx Question 2 Using integration, find the area of theregions: (i) { (x,y): |x-1| ≤y ≤ }
- 102. (ii) *(x,y):0≤y≤x2+3; 0≤y≤2x+3; 0≤x≤3+[(i) A= dx- dx - dx = 5/2 [+ ] – ½ ] [(ii) dx + dx , answer is 50/3](iii) Find the area bounded by the curve x 2 = 4y & the line x = 4y – 2.[A = dx - dx = 9/8 sq. Unit.]**(iv) Sketch the graph of f(x) = ,evaluate dx[hint: dx = dx + dx = 62/3.]**Question 3 evaluate dx [ mult. & divide by , put 1+x=A.(d/dx)(x2+x)+B ,find A=B=1/2, integrate]
- 103. Definite integral as the limit of a sum , use formula : dx , where nh=b-a & n→∞ Question 4 Evaluate ) dx (ii) dx[ use = 1 for part (i) , use formulas of special sequences, answeris 6]Some special case :(1) Evaluate: [ put x+1=t²] (2) [ put x+1 = t² ](3) Evaluate: (4) Evaluate: [ put x=1/t for both](5) Evaluate: [ divide Nr. & Dr. By x2 , then write x²+1/x²=(x-1/x)²+2 according to Nr. , let x-1/x=t](6) Evaluate dx [ let x=A(d/dx) ( 1+x-x²) +B](7) Integrating by parts evaluate =(8) Evaluate dx = dx [ putsinx=Ad/dx(sinx+cosx)+B(sinx+cosx)+CIf Nr. Is constant term then use formulas of sinx,cosx as Ques. No. 1 (vi) part]

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