Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Tamadun Rom (Sejarah Form 4) by Azfar Zulkifli 19663 views
- Form 5 Additional Maths Note by Chek Wei Tan 6090 views
- Chapter 9 differentiation by atiqah ayie 37473 views
- Form 5 formulae and note by smktsj2 32316 views
- Ancient Rome by Atrayee SenGupta 12061 views
- Chapter 3 quadratc functions by atiqah ayie 12257 views

No Downloads

Total views

16,037

On SlideShare

0

From Embeds

0

Number of Embeds

3,203

Shares

0

Downloads

413

Comments

0

Likes

5

No embeds

No notes for slide

- 1. Integration Learning Objectives : In this chapter, you will learn about <ul><li>the concept of indefinite integral </li></ul>Learning Outcomes: <ul><li>Determine integrals by reversing differentiation </li></ul>
- 2. 3.1 Indefinite Integral
- 3. 3.1.2 Integration of algebraic expressions Integrate (a) 8 (b) 3.5 (c) 3.1.2 (a) Integral of Constant
- 4. 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
- 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 as find the function y.
- 7. 3.1.3 Determine the constant of Integration
- 8. 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
- 9. 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
- 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 is At point (0, k),
- 12. Exercise 3-1-09 t0 6-1-09 <ul><li>Given that and that y=5 when x= -1, find the value of y </li></ul><ul><li>when x=2 </li></ul><ul><li>Given that and that v=2 when t = 1, find the value of v </li></ul><ul><li>when t = 2. </li></ul>
- 13. 3.1.5 Integrate by substitution Find the integration by substitution
- 14. 3.1.5 Integrate 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 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)
- 17. Area under a curve
- 18. 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
- 19. 3.2.2(b) Area under a curve bounded by x= a and 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 bounded by
- 23. The area under a curve which is enclosed by y = a and y = b is
- 24. The area under a curve is 1 2
- 25. Area under a curve bounded by curve
- 26. Area under a curve bounded
- 27. Exercise 19-2-09 23-2-09 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 (the yellow disk shown above), each dx wide and radius y :
- 32. 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.
- 34. 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..
- 35. When the shaded area is rotated 360° about the x -axis, we again observe that a volume is generated:

No public clipboards found for this slide

Be the first to comment