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Form 5 Chapter 3 integration

Form 5 Chapter 3 integration

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Integration Integration Presentation Transcript

  • Integration Learning Objectives : In this chapter, you will learn about
    • the concept of indefinite integral
    Learning Outcomes:
    • Determine integrals by reversing differentiation
  • 3.1 Indefinite Integral
  • 3.1.2 Integration of algebraic expressions Integrate (a) 8 (b) 3.5 (c) 3.1.2 (a) Integral of Constant
  • 3.1.2 Integration 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 of Differentiation Integration
  • Examples 1: Integrate each of the following with respect of x: (a) (b)
  • Examples 2: If the derivative of a function is given as find the function y.
  • 3.1.3 Determine the constant of Integration
  • Examples 1: Subsitute x=3 and y=5 into (1) If and y=5 when x=3, find the value of y when x=5
  • 3.1.4 Equations 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 is The gradient of a curve passing through the point (-1, 2) is given by
  • Examples 2: . Find the value of k. The gradient function of a curve passing through the point (-1, 2) and (0,k) is
  • The curve pass through (-1, 2) Therefore, the equation of the curve is At point (0, k),
  • Exercise 3-1-09 t0 6-1-09
    • Given that and that y=5 when x= -1, find the value of y
    • when x=2
    • Given that and that v=2 when t = 1, find the value of v
    • when t = 2.
  • 3.1.5 Integrate by substitution Find the integration by substitution
  • 3.1.5 Integrate by substitution Find the integration by substitution
  • 3.1.5 (a) Integral of
  • The gradient of the curve, Integrate with respect to x, we have Since the curve passes through (4, -3) The equation of the curve is The slope of 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)
  • Area under a curve
  • 3.2.2(b) Area under a curve bounded by x= a and 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
  • 3.2.2(b) Area under a curve bounded by x= a and x=b 1 2
  • Step (1) Find x-intercept a b On the x-axis, y =0
  • Area under a curve bounded by
  • Area under a curve bounded by
  • The area under a curve which is enclosed by y = a and y = b is
  • The area under a curve is 1 2
  • Area under a curve bounded by curve
  • Area under a curve bounded
  • Exercise 19-2-09 23-2-09 1 2
  • Exercise Text Book Page 71 23-2-09 10- ( a) ( b ) ( c ) 11- ( a) ( b ) ( c ) 12- ( a) ( b )
  • Exercise Text Book Page 72 13- ( a) ( b ) ( c ) 14 17 ( a ) (b ) ( c )
  • Volume of Revolutions
  • The resulting solid is a cone To find this volume, we could take slices (the yellow disk shown above), each dx wide and radius y :
  • The volume of a cylinder is given by V = π r 2 h Because radius = r = y and each disk is dx high, we notice that the volume of each slice is: V = π y 2 dx Adding the volumes of the disks (with infinitely small dx ), we obtain the formula: y = f ( x ) is the equation of the curve whose area is being rotated a and b are the limits of the area being rotated dx show that the area is being rotated abount the x-axis.
  •  
  • Example 2 Find the volume if the area bounded by the curve y = x 3 + 1, the x- axis and the limits of x = 0 and x = 3 is rotated around the x -axis..
  • When the shaded area is rotated 360° about the x -axis, we again observe that a volume is generated:
  •