Integration

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Form 5 Chapter 3 integration

Transcript of Integration

  • 1. Integration LearningObjectives : In this chapter, you will learn about
    • the concept of indefinite integral

Learning Outcomes:

  • Determine integrals by reversing differentiation

2. 3.1 Indefinite Integral 3. 3.1.2Integration of algebraic expressions Integrate (a) 8(b) 3.5(c)3.1.2 (a) Integral of Constant 4. 3.1.2Integration of algebraic expressions During differentiation, we carry out two operations on each term in x: multiply the term with the index, and reduce the index by 1. 3.1.2 (b) Integral ofDifferentiation Integration 5. Examples 1: Integrate each of the following with respect of x: (a)(b) 6. Examples 2: If the derivative of a function is given asfind the function y. 7. 3.1.3Determine the constant of Integration 8. Examples 1: Subsitute x=3 and y=5into (1) Ifand y=5 when x=3, find the value of y when x=5 9. 3.1.4Equations of curve from functions of gradients Examples 1: Find the equation of curve. by integration, The curve passing through the point (-1,2) x=-1 when y=2 The equation of the curve isThe gradient of a curve passing through the point (-1,2) is given by 10. Examples 2: . Find the value of k. The gradient function of a curve passing through the point (-1, 2) and (0,k) is 11. The curve pass through (-1, 2) Therefore, the equation of the curve isAt point (0,k), 12. Exercise3-1-09 t0 6-1-09

  • Given thatand that y=5 when x= -1, find the value ofy
  • when x=2
  • Given thatand that v=2 when t = 1, find the value ofv
  • when t = 2.

13. 3.1.5Integrate by substitution Find the integration by substitution 14. 3.1.5Integrate by substitution Find the integration by substitution 15. 3.1.5 (a)Integral of 16. The gradient of the curve,Integrate with respect to x, we haveSince the curve passes through (4, -3) The equation of the curve isTheslopeof a curve at any point P(x, y) is given by. Find the equation of the curve given that its passes through the point ( 4,-3) 17. Area under a curve 18. 3.2.2(b) Area under a curve bounded by x=aand x=b The area A under a curve by y = f(x) bounded by the x-axis from x=a to x=b is given by Integration as Summation of Area 19. 3.2.2(b) Area under a curve bounded by x=aand x=b 1 2 20. Step (1) Find x-intercept a b On the x-axis, y =0 21. Area under a curve bounded by 22. Area under a curve boundedby 23. The area under a curve which is enclosedby y = a and y = b is 24. The area under a curve is1 2 25. Area under a curve bounded by curve 26. Area under a curve bounded 27. Exercise19-2-0923-2-09 1 2 28. Exercise Text Book Page 71 23-2-09 10- ( a)( b )( c ) 11- ( a)( b )( c ) 12- ( a)( b ) 29. Exercise Text Book Page 72 13- ( a)( b )( c ) 14 17 ( a )(b )( c ) 30. Volume of Revolutions 31. The resulting solid is a cone To find this volume, we could take slices (theyellowdisk shown above), eachdxwide and radiusy : 32. Thevolumeof a cylinder is given by V = r 2 h Because radius= r=yand each disk isdxhigh, we notice that the volume of each slice is: V = y 2 dx Adding the volumes of the disks (with infinitely smalldx ), we obtain the formula: y = f ( x ) is the equation of the curve whose area is being rotated aandbare the limits of the area being rotated dx show that the area is being rotated abount the x-axis. 33. 34. Example 2 Find the volume if the area bounded by the curvey=x 3+ 1, thex- axis and the limits ofx= 0 andx= 3 is rotated around thex -axis.. 35. When the shaded area is rotated 360 about thex -axis, we again observe that a volume is generated: 36.