Edexcel Core Mathematics 4 Integration

Transcription

1 Edecel Core Mathematics 4 Integration Edited by: K V Kumaran kumarmaths.weebly.com

2 Integration It might appear to be a bit obvious but you must remember all of your C work on differentiation if you are to succeed with this unit. Content By the end of this unit you should be able to: Integrate trigonometric, eponential functions and Find volumes of revolution Find integrals using partial fractions and integrating by parts. Find areas under curves and volumes generated from functions given in parametric form. kumarmaths.weebly.com

3 Standard Integrals Function (f()) n (a + b) n a b e Sin Cos Tan Cot Cosec Sec Sec Sec Tan -Cosec e a+b Integral f()d n c n ln c n (a b) c a(n ) ln a b c a e + c -Cos + c Sin + c ln (Sec ) + c ln (Sin ) + c -ln (Cosec + Cot ) + c ln (Sec + Tan ) + c Tan + c Sec + c Cot + c e a ab c Cos (a + b) Sin(a b) c a Sin n Cos n Sin c n Cos n Sin n Cos c n Sin Sin c Cos Sin c kumarmaths.weebly.com

4 Integration of trigonometric functions. To succeed with these types of questions you need to have a very good grasp of the trig identities from C and obviously the differentiation from the same unit. Integration questions involving trigonometry are either of a substitution type or areas and volumes from parametrics. The following eamples give a flavour but can t cover every eventuality. Eample By using the substitution u = sin or otherwise, find 8sin sind Giving your answer in terms of. From C you should remember that: Hence: If u = sin Therefore sin = sin cos 8sin sind 6sin cos d du = cosd (part of the integral!) 6sin cos d 6u du 4 4 4u c 4sin c kumarmaths.weebly.com 4

5 Eample Use the substitution = sin θ to find the eact value of 4 ( ) d If = sin θ then the denominator becomes ( sin θ) which is cos θ. When this is raised to the power of.5 the result is cos θ as a denominator. The s have now been substituted but you can t forget about the d. If = sin θ then d = cosθdθ. Therefore the integral becomes: 4 4 d ( ) ( sin ) cosd 4 4 cosd d cos cos 4 The net thing to spot is that 4sec cos 4sec d 4tan. Therefore the question becomes: The question isn t quite finished as the original limits were given in terms of. So for the lower limit of =, sin θ =, hence θ =. For the upper limit =.5, so sin θ =.5, hence θ =. The question asks for the eact value so the integral is finally. 4sec d 4tan 4 4 There are a few tricks in this question but practice makes perfect and you must remember all of your differentiation and trig work from C and C4. Use ideas outlined in the eamples above to carry out the net question. Use the substitution u = cosθ and integration to find Sin Cosd. (You will need to use implicit differentiation when dealing with the u.) kumarmaths.weebly.com 5

6 Integrating by Parts. This concept is guaranteed to appear on a C4 paper and is not too tricky. Integration by parts is used to integrate functions that are made up of products of functions. Not surprisingly the process is the reverse of differentiating a product. The idea is best eplained through an eample. Eample 5 Use integration by parts to find the eact value of The rule for integrating by parts is: dv du u d uv v d d d 5 lnd The thing to remember is that one part of the question becomes u and the other part becomes dv d. Since we can t integrate ln, it must be u and therefore 5 must equal dv du. On the right of the rule we need v and this can be achieved by d d integrating and differentiating respectively. Therefore: 5 lnd Let u = ln and du d dv d v So by substituting into the rule: dv du u d uv v d d d kumarmaths.weebly.com 6

7 ln d d d kumarmaths.weebly.com 7

8 Volume of Revolution The basic principle here is that we take very thin strips, of width δ (delta ), underneath a curve in the y plane and rotate them around the ais to generate a volume. The diagram on the left shows the graph of y = in D. The pink section is the area under the graph between the values of and. This pink section has been rotated through 6 degrees about the ais to form the green solid in the diagram on the right. If you look at the far end of the green solid it has a circular appearance. The area of the end is πy and therefore the volume of a very thin disc of width δ will be πy δ. We use integration to add up the volumes of all of the very thin discs to give the volume of revolution. πy δ = y d Curves can be rotated about the or y ais and therefore there are two formulae to calculate volumes. The volume generated by a curve as it rotates around the ais is given by: y d kumarmaths.weebly.com 8

9 Eample The finite region bounded by the curve with the equation y = 8 and the ais is rotated through 6º about the ais. Using integration find, in terms of π, the volume of its solid form. The volume generated by a curve as it rotates around the ais is given by: y d The mistake a number of candidates make is to state that y in this case is WRONG Y = (8 )( 8 ) = Therefore volume is given by : 4 ( )d The quadratic, y = 8, has solutions of = and = 8 and therefore these become the limits on the integral kumarmaths.weebly.com 9

10 Eample 4 Find the volume generated when the region bounded by the curve with equation 5 y, the ais and the lines = and = is rotated through 6 degrees about the ais. Volume = y d 5 5 d 4 d 5 4 ln 5 4 ln ln 75 ln9 Take care with the negative powers when integrating and the rules of logs. In most cases the question will specifically ask for answers in terms of a + bln9 for eample. A calculator answer will loose you the final answer mark. kumarmaths.weebly.com

11 More comple integrations by substitution. Eample 9 Use the substitution u =, or otherwise, find the eact value of 5 6 d From an earlier eample you should remember that there are three parts that need dealing with, 6, the d and the square root in the denominator If u = then udu = d ( by implicit differentiation). So d = udu. If u = then =.5u +.5. The integral becomes: 5 u 6 d udu u (u )du u u And now for the limits. If = then u = 5 and if = 5 then u = d u u 5 4 kumarmaths.weebly.com

12 Integration of Parametrics to Find Areas under curves. Eample The diagram shows a sketch of the curve C with parametric equations = 7tsint, y = 5sect, t < The point P(a, ) lies on C. a) Find the eact value of a. The region R is enclosed by the curve C, the aes and the line = a. b) Show that the area of R is given by 5 tant t dt c) Find the eact value of the area R. a) Find the eact value of a. The point P has y coordinate. Therefore: 5sect = If t = sect = = 7 sin cost =.5 = 7 6 kumarmaths.weebly.com

13 t = a = 7 6 b) When integrating to find the area under a parametric curve you need to integrate the following: d y dt dt The part of the parametric is a product and therefore we need to differentiate the first (7t) and multiply it second (sint) then add the first multiplied by the differential of the second. This gives: So to find the area R: d 7 sint 7tcost dt d y dt 5sect7 sin t 7tcost dt sint Re membering that sect = and that tan t cost cost 5 tan t t dt The limits are defined from the work in part (a). c) The integral from part (b) becomes: sint cost 5 tdt The integral of tant should be in your notes. The denominator nearly differentiates to give the numerator. Therefore the integral is given by the negative ln of cost. kumarmaths.weebly.com

14 sin t 5 t dt 5 tan t t dt cost 5 ln cos t.5t 8.45 = 4.5 If you are asked to find the volume generated by a parametric as it rotates about the ais then the integral to evaluate is: d y dt dt kumarmaths.weebly.com 4

15 Finally an integration question that has an eponential substitution! Eample Use the substitution u = to find the eact value of d The integral very quickly becomes ud u but we haven t dealt with the d. From eample 9 of the C4 differentiation notes the differential of u is: du d ln Therefore du u ln d and the integral becomes: 9 du ln ( u ) ln ln( u ) 9 ln 5 ln Remember that the question said eact answer so leave it in terms of logs. kumarmaths.weebly.com 5

16 . (a) Epress Past paper questions on Integration 5 in partial fractions. ( )( ) () (b) Hence find the eact value of 6 5 ( )( ) d, giving your answer as a single logarithm. (5) (C4, June 5 Q). Use the substitution = sin to find the eact value of ( ) d. (7) (C4, June 5 Q4). Using the substitution u =, or otherwise, find the eact value of 5 ( ) d. (8) (C4, Jan 6 Q) 4. Figure y y = e R O Figure shows the finite region R, which is bounded by the curve y = e, the line =, the line = and the -ais. kumarmaths.weebly.com 6

17 The region R is rotated through 6 degrees about the -ais. Use integration by parts to find an eact value for the volume of the solid generated. (8) (C4, Jan 6 Q4) 5. Figure y R O The curve shown in Figure has parametric equations = t sin t, y = cos t, t. 5 (a) Show that the curve crosses the -ais where t = and t =. () The finite region R is enclosed by the curve and the -ais, as shown shaded in Figure. (b) Show that the area R is given by the integral 5 ( cos t ) dt. (c) Use this integral to find the eact value of the shaded area. 6. Figure () (7) (C4, Jan 6 Q8) kumarmaths.weebly.com 7

18 y O The curve with equation y = sin,, is shown in Figure. The finite region enclosed by the curve and the -ais is shaded. (a) Find, by integration, the area of the shaded region. () This region is rotated through radians about the -ais. (b) Find the volume of the solid generated. (6) (C4, June 6 Q) kumarmaths.weebly.com 8

19 7. Figure y The curve with equation y =, >, is shown in Figure. ( ) The region bounded by the lines = 4, =, the -ais and the curve is shown shaded in Figure. This region is rotated through 6 degrees about the -ais. (a) Use calculus to find the eact value of the volume of the solid generated. (5) Figure A B Figure shows a paperweight with ais of symmetry AB where AB = cm. A is a point on the top surface of the paperweight, and B is a point on the base of the paperweight. The paperweight is geometrically similar to the solid in part (a). (b) Find the volume of this paperweight. () (C4, Jan 7 Q) kumarmaths.weebly.com 9

20 8. Use the substitution u = to find the eact value of ( ) d. (6) (C4, June 7 Q) 9. (a) Find cos d. (4) (b) Hence, using the identity cos = cos, deduce cos d. () (C4, June 7 Q). (4 ) ( )( ) A + B ( ) + C. ( ) (a) Find the values of the constants A, B and C. (b) Hence show that the eact value of constant k. (4 ) ( )( ) (4) d is + ln k, giving the value of the (6) (C4, June 7 Q4) kumarmaths.weebly.com

21 . The curve shown in Figure has equation y =. The finite region bounded by the curve, ( ) the -ais and the lines = a and = b is shown shaded in Figure. This region is rotated through 6 about the -ais to generate a solid of revolution. Find the volume of the solid generated. Epress your answer as a single simplified fraction, in terms of a and b. (5) (C4, Jan 8 Q). (i) Find ln d. (ii) Find the eact value of sin 4 d. (4) (5) (C4, Jan 8 Q4) kumarmaths.weebly.com

22 . Figure The curve C has parametric equations = ln (t + ), y =, t >. ( t ) The finite region R between the curve C and the -ais, bounded by the lines with equations = ln and = ln 4, is shown shaded in Figure. (a) Show that the area of R is given by the integral (b) Hence find an eact value for this area. dt. ( t )( t ) (c) Find a cartesian equation of the curve C, in the form y = f(). (d) State the domain of values for for this curve. 4. (a) Use integration by parts to find e d. (4) (6) (4) () (C4, Jan 8 Q7) () (b) Hence find e d. () (C4, June 8 Q) kumarmaths.weebly.com

23 5. Figure Figure shows the curve C with parametric equations = 8 cos t, y = 4 sin t, t. The point P lies on C and has coordinates (4, ). (a) Find the value of t at the point P. () The line l is a normal to C at P. (b) Show that an equation for l is y = + 6. (6) The finite region R is enclosed by the curve C, the -ais and the line = 4, as shown shaded in Figure. (c) Show that the area of R is given by the integral 64 sin t cos t dt. (d) Use this integral to find the area of R, giving your answer in the form a + b, where a and b are constants to be determined. (4) (4) (C4, June 8 Q8) kumarmaths.weebly.com

24 6. Figure Figure shows part of the curve y =. The region R is bounded by the curve, the -ais, ( 4) and the lines = and =, as shown shaded in Figure. (a) Use integration to find the area of R. (4) The region R is rotated 6 about the -ais. (b) Use integration to find the eact value of the volume of the solid formed. (5) (C4, Jan 9 Q) 7. (a) Find tan d. (b) Use integration by parts to find ln d. (c) Use the substitution u = + e to show that () (4) e d = e e e + ln ( + e ) + k, where k is a constant. (7) (C4, Jan9 Q6) kumarmaths.weebly.com 4

25 8. f() = 4 ( )( )( ) = A ( ) + B ( ) + C ( ). (a) Find the values of the constants A, B and C. (4) (b) (i) Hence find f ( ) d. () (ii) Find f ( ) d in the form ln k, where k is a constant. () 9. (a) Find ( 5 ) d. (C4, June 9 Q) () Figure Figure shows a sketch of the curve with equation y = ( ) (5 ), 5 (b) (i) Using integration by parts, or otherwise, find ( ) (5 ) d. (4) (ii) Hence find 5 ( ) (5 ) d.. () (C4, June 9 Q6) kumarmaths.weebly.com 5

26 . (a) Using the identity cos θ = sin θ, find sin d. () Figure 4 Figure 4 shows part of the curve C with parametric equations = tan θ, y = sin θ, θ <. The finite shaded region S shown in Figure 4 is bounded by C, the line = and the -ais. This shaded region is rotated through radians about the -ais to form a solid of revolution. (b) Show that the volume of the solid of revolution formed is given by the integral where k is a constant. k 6 sin d, (c) Hence find the eact value for this volume, giving your answer in the form p + q, where p and q are constants. () (5) (C4, June 9 Q8) kumarmaths.weebly.com 6

27 . Figure Figure shows a sketch of the curve C with parametric equations = 5t 4, y = t(9 t ) The curve C cuts the -ais at the points A and B. (a) Find the -coordinate at the point A and the -coordinate at the point B. () The region R, as shown shaded in Figure, is enclosed by the loop of the curve. (b) Use integration to find the area of R. (6) (C4, Jan Q7) kumarmaths.weebly.com 7

28 . (a) Using the substitution = cos u, or otherwise, find the eact value of (4 ) d. (7) Figure Figure shows a sketch of part of the curve with equation y = 4 (4 ) 4, < <. The shaded region S, shown in Figure, is bounded by the curve, the -ais and the lines with equations = and =. The shaded region S is rotated through π radians about the -ais to form a solid of revolution. (b) Using your answer to part (a), find the eact volume of the solid of revolution formed. () (C4, Jan Q8). Using the substitution u = cos +, or otherwise, show that e cos sin d = e(e ). (6) (C4, June Q) kumarmaths.weebly.com 8

29 4. f(θ) = 4 cos θ sin θ (a) Show that f(θ) = + 7 cos θ. () (b) Hence, using calculus, find the eact value of f( ) d. (7) (C4, June Q6) 5. Use integration to find the eact value of sin d. (6) 6. The curve C has parametric equations (C4, Jan Q) = ln t, y = t, t >. Find (a) an equation of the normal to C at the point where t =, (b) a cartesian equation of C. (6) () Figure The finite area R, shown in Figure, is bounded by C, the -ais, the line = ln and the line = ln 4. The area R is rotated through 6 about the -ais. (c) Use calculus to find the eact volume of the solid generated. (6) (C4, June Q6) kumarmaths.weebly.com 9

30 7. Figure Figure shows part of the curve C with parametric equations = tan, y = sin, <. The point P lies on C and has coordinates (a) Find the value of at the point P.,. The line l is a normal to C at P. The normal cuts the -ais at the point Q. (b) Show that Q has coordinates (k, ), giving the value of the constant k. () (6) The finite shaded region S shown in Figure is bounded by the curve C, the line = and the - ais. This shaded region is rotated through radians about the -ais to form a solid of revolution. (c) Find the volume of the solid of revolution, giving your answer in the form p + q, where p and q are constants. (7) (C4, June Q7) 8. (a) Use integration by parts to find sin d. (b) Using your answer to part (a), find cos d. () () 9. (C4, June Q) kumarmaths.weebly.com

31 Figure Figure shows the curve with equation y =,. 4 The finite region S, shown shaded in Figure, is bounded by the curve, the -ais and the line =. The region S is rotated 6 about the -ais. Use integration to find the eact value of the volume of the solid generated, giving your answer in the form k ln a, where k and a are constants. (5) (C4, Jan Q4). f() = ( ) A B = + ( ) C ( ) +. (a) Find the values of the constants A, B and C. (b) (i) Hence find f ( ) d. (4) (ii) Find f ( ) d, leaving your answer in the form a + ln b, where a and b are constants. (6) (C4, June Q) kumarmaths.weebly.com

32 . (a) Use integration to find. (b) Hence calculate ln d. ln d. (5) () (C4, Jan Q) Figure Figure shows a sketch of part of the curve with equation y = cos, where is measured in radians. The curve crosses the -ais at the point A and at the point B. (a) Find, in terms of, the coordinate of the point A and the coordinate of the point B. () The finite region S enclosed by the curve and the -ais is shown shaded in Figure. The region S is rotated through radians about the -ais. (b) Find, by integration, the eact value of the volume of the solid generated. (6) (C4, Jan Q4) kumarmaths.weebly.com

33 . (a) Find ed. (5) (b) Hence find the eact value of ed. () (C4, June Q) 4. (a) Use the substitution = u, u >, to show that (b) Hence show that d = ( ) du u(u) () 9 d = ln a ( ) b where a and b are integers to be determined. (7) (C4, June Q5) 5. Using the substitution u = + ( + ), or other suitable substitutions, find the eact value of 4 d ( ) giving your answer in the form A + ln B, where A is an integer and B is a positive constant. (8) (C4, June _R Q5) kumarmaths.weebly.com

34 6. Figure Figure shows a sketch of the curve C with parametric equations 7sec t, y tan t, t (a) Find the gradient of the curve C at the point where t =. (4) 6 (b) Show that the cartesian equation of C may be written in the form y ( 9), a b stating values of a and b. () Figure The finite region R which is bounded by the curve C, the -ais and the line = 5 is shown shaded in Figure. This region is rotated through π radians about the -ais to form a solid of revolution. (c) Use calculus to find the eact value of the volume of the solid of revolution. (5) (C4, June _R Q7) kumarmaths.weebly.com 4

35 7. (a) Epress 5 in partial fractions. (4) Figure Figure shows a sketch of part of the curve C with equation 5 y, >. The finite region R is bounded by the curve C, the -ais, the line with equation = and the line with equation = 4. This region is shown shaded in Figure. The region R is rotated through 6 about the -ais. (b) Use calculus to find the eact volume of the solid of revolution generated, giving your answer in the form a + bln c, where a, b and c are constants. (6) (C4, June 4_R Q4) kumarmaths.weebly.com 5

36 8. Figure 4 Figure 4 shows a sketch of part of the curve C with parametric equations = tan θ, y = 4cos θ, θ < The point P lies on C and has coordinates (, ). The line l is the normal to C at P. The normal cuts the -ais at the point Q. (a) Find the coordinate of the point Q. (6) The finite region S, shown shaded in Figure 4, is bounded by the curve C, the -ais, the y-ais and the line l. This shaded region is rotated π radians about the -ais to form a solid of revolution. (b) Find the eact value of the volume of the solid of revolution, giving your answer in the form pπ + qπ, where p and q are rational numbers to be determined. [You may use the formula V = πr h for the volume of a cone.] (9) (C4, June 4 Q7) kumarmaths.weebly.com 6

37 9. Figure Figure shows a sketch of part of the curve with equation y = 4 e,. The curve meets the -ais at the origin O and cuts the -ais at the point A. (a) Find, in terms of ln, the coordinate of the point A. () (b) Find e d. () The finite region R, shown shaded in Figure, is bounded by the -ais and the curve with equation y = 4 e,. (c) Find, by integration, the eact value for the area of R. Give your answer in terms of ln. () (C4, June 5, Q) kumarmaths.weebly.com 7

38 4. (i) Given that y >, find y 4 d y. y(y ) (ii) (a) Use the substitution = 4sin θ to show that d = λ 4 where λ is a constant to be determined. (b) Hence use integration to find sin d, (6) (5) d, 4 giving your answer in the form aπ + b, where a and b are eact constants. (C4, June 6, Q6) 4. (a) Find giving your answer in its simplest form. () d, (4) () Figure shows a sketch of part of the curve C with equation kumarmaths.weebly.com 8

39 DO NOT WRITE IN THIS AREA y 4,. The curve C cuts the line y = 8 at the point P with coordinates (k, 8), where k is a constant. (b) Find the value of k. () The finite region S, shown shaded in Figure, is bounded by the curve C, the -ais, the y-ais and the line y = 8. This region is rotated through π radians about the -ais to form a solid of revolution. (c) Find the eact value of the volume of the solid generated. (4) (C4, June 6, Q7) y Diagram not drawn to scale S O ln4 Figure The finite region S, shown shaded in Figure, is bounded by the y-ais, the -ais, the line with equation = ln4 and the curve with equation y = e + e, The region S is rotated through π radians about the -ais. Use integration to find the eact value of the volume of the solid generated. Give your answer in its simplest form. (C4, June 7, Q5) kumarmaths.weebly.com 9

40 DO NOT WRITE IN THIS AREA y C Diagram not drawn to scale P(k, 8) R O k Figure 4 Figure 4 shows a sketch of part of the curve C with parametric equations = θsinθ, y = sec θ, θ < π The point P(k, 8) lies on C, where k is a constant. (a) Find the eact value of k. () The finite region R, shown shaded in Figure 4, is bounded by the curve C, the y-ais, the -ais and the line with equation = k. (b) Show that the area of R can be epressed in the form l ò b a (q sec q + tanq sec q)dq where λ, α and β are constants to be determined. (c) Hence use integration to find the eact value of the area of R. (4) (6) (C4, June 7, Q8) kumarmaths.weebly.com 4

41 44. Given that (A + B)( + 4) + C + D (a) find the values of the constants A, B, C and D. (b) Hence find (4) d 4 giving your answer in the form p + ln q, where p and q are integers. (6) (C4, Jan 4, Q) Figure 4 Figure 4 shows a sketch of part of the curve with equation y = (sin + cos ), 4 The finite region R, shown shaded in Figure 4, is bounded by the curve, the -ais and the line. This shaded region is rotated through π radians about the -ais to form a solid of 4 revolution, with volume V. 4 (a) Assuming the formula for volume of revolution show that V ( sin )d. () (b) Hence using calculus find the eact value of V. You must show your working. (9) (C4, IAL Jan 4, Q) kumarmaths.weebly.com 4

42 46. Find (a) d 5 (b) d 4 () () (C4, IAL June 4, Q4) 47. Figure Figure shows a sketch of part of the curve with equation y e, >. The finite region R, shown shaded in Figure, is bounded by the curve, the -ais and the lines = 4 and = 9. (a) Use the trapezium rule, with 5 strips of equal width, to obtain an estimate for the area of R, giving your answer to decimal places. (4) (b) Use the substitution u = to find, by integrating, the eact value for the area of R. (7) (C4, IAL June 4, Q9) kumarmaths.weebly.com 4

43 48. Figure Figure shows a sketch of part of the curve C with parametric equations tan t, y sin t, t The finite region S, shown shaded in Figure, is bounded by the curve C, the line = and the -ais. This shaded region is rotated through π radians about the -ais to form a solid of revolution. (a) Show that the volume of the solid of revolution formed is given by 4 tan sin d t t t (b) Hence use integration to find the eact value for this volume. (6) (6) (C4, IAL June 4, Q) 49. Use the substitution = sin θ to find the eact value of 4 d (7) (C4, IAL Jan 5, Q4) kumarmaths.weebly.com 4

44 5. The curve C has parametric equations = ln (t + ), 4 y = t t > The finite region R, shown shaded in Figure, is bounded by the curve C, the -ais and the lines with equations = ln and = ln5. (a) Show that the area of R is given by the integral 4 dt t ( t ) (b) Hence find an eact value for the area of R. Write your answer in the form (a + ln b), where a and b are rational numbers. (c) Find a cartesian equation of the curve C in the form y = f(). (C4, IAL Jan 5, Q9) 5. (i) Find the coordinate of each point on the curve y,, at which the gradient is 4. (ii) Given that a t d t ln 7 a > a t find the eact value of the constant a. (4) (C4, IAL June 5, Q5) () (7) () (4) kumarmaths.weebly.com 44

45 5. The curve C with equation y =, (4 ) 4 is shown in Figure The region bounded by the curve, the -ais and the lines = and =, is shown shaded in Figure This region is rotated through 6 degrees about the -ais. (a) Use calculus to find the eact value of the volume of the solid generated. (5) Figure shows a candle with ais of symmetry AB where AB = 5 cm. A is a point at the centre of the top surface of the candle and B is a point at the centre of the base of the candle. The candle is geometrically similar to the solid generated in part (a). (b) Find the volume of this candle. () (C4, IAL Jan 6, Q4) kumarmaths.weebly.com 45

46 5. f(θ) = 9cos θ + sin θ (a) Show that f(θ) = a + bcosθ, where a and b are integers which should be found. (b) Using your answer to part (a) and integration by parts, find the eact value of f( )d 54. Use integration by parts to find the eact value of d 55. Write your answer as a single simplified fraction. () (6) (C4, IAL Jan 6, Q8) (C4, IAL June 6, Q5) Figure shows a sketch of the curve with parametric equations = sint, y = sint, t The finite region S, shown shaded in Figure, is bounded by the curve, the -ais and the line with equation =. The shaded region S is rotated through π radians about the -ais to form a solid of revolution. (a) Show that the volume of the solid of revolution is given by a sin cos d k t t t where k and a are constants to be given in terms of π. (5) kumarmaths.weebly.com 46

47 DO NOT WRITE IN THIS AREA (b) Use the substitution u = sin t, or otherwise, to find the eact value of this volume, giving your answer in the form p where p and q are integers. q 56. (Solutions based entirely on graphical or numerical methods are not acceptable) y (6) (C4, IAL June 6, Q) Diagram not drawn to scale C R O Figure (a) By using the substitution u = +, show that The curve C has equation d = ( ) ln- 9 (7) y = 9 ( + ), > The finite region R, shown shaded in Figure, is bounded by the curve C, the -ais and the line with equation =. The region R is rotated through π radians about the -ais to form a solid of revolution. (b) Use the result of part (a) to find the eact value of the volume of the solid generated. () (C4, IAL Jan 7, Q9) kumarmaths.weebly.com 47

48 DO NOT WRITE IN THIS AR 57. Use integration by parts to find the eact value of ò e ln d Write your answer in the form a + b e, where a and b are integers. (C4, IAL June 7, Q) y l P S C Figure 6 shows a sketch of the curve C with parametric equations O Figure 6 = 8cos θ, y = 6sin θ, θ p Given that the point P lies on C and has parameter θ = p (a) find the coordinates of P. The line l is the normal to C at P. (b) Show that an equation of l is y = +.5 () (5) The finite region S, shown shaded in Figure 6, is bounded by the curve C, the line l, the y-ais and the -ais. (c) Show that the area of S is given by 4 44 (sin cos sin 4 cos )d (d) Hence, by integration, find the eact area of S. (6) () (C4, IAL June 7, Q4) kumarmaths.weebly.com 48