Applications Of The Definite Integral

1. Applications Of The Definite Integral The Area under the curve of a function The area between two curves The Volume of the Solid of revolution

2. Applications of the Definite Integral • In calculus, the integralof a function is an extension of the concept of a sum. The process of finding integrals is called integration. The process is usually used to find a measure of totality such as area, volume, mass, displacement, etc.

3. The integral would be written . The∫ sign represents integration, a and b are the endpoints of the interval, f(x) is the function we are integratingknown as theintegrand, and dx is a notation for the variable of integration. Integrals discussed in thisproject are termeddefinite integrals.

4. We use definite integrals • basically, • Results:

5. Given:   evaluate,

6. Solution:

7. Area under a Curve To find the area under a curve. This expression gives us a definite value (a number) at the end of the calculation. When the curve is above the ‘x’ axis, the area is the same as the definite integral :

8. But when the graph line is below the ‘x’ axis, the definite integral is negative. The area is then given by:

9. (Positive) (Negative)

10. Example 1: let f (x)=2-x .Find the area bounded by the curve of f , the x-axis and the lines x=a and x=b for each of the following cases: a=-2 b=2 a=2 b=3 a=-2 b=3 The graph: Is a straight line y=2-x: F (x) is positive on the interval [-2, 2) F (x) is negative on the interval (2, 3] 2 2 3 -2

11. Case 1: The area A1 between f, the x-axis and the lines x=-2 and x=2 is: f(x)>0; x [-2,2) A1 2 3 -2 2

12. Case2 The areaA2 between f, the x-axis and the Lines x=2 and x=3 is: f(x)<0; x (2, 3] 2 -2 2 3

13. Case3: The area a between f, the X-axis and the lines X=-2 and X=3 is : 2 2 3 -2

14. Area Bounded by 2 Curves Say you have 2 curves y = f(x) and y = g(x) Area under f(x) = Area under g(x) =

15. Superimposing the two graphs, Area bounded by f(x) and g(x)

16. Example (2) Letf (x) =X, g (X) = Find the area between f and g from X=a to X=b Following cases a=-1 b=0 a=0 b=1 a=-1 b=1 g (X)>f (X) on (-1,0) and hence on this interval , we have: g (X) –f (X)>0 So |g (X) –f (X)| =g (X)-f (X)= -x

17. Case (1): The areaA1 between f and g from X= -1 and x=0 is: is:g (X)>f (X) on (-1,0) and hence on this interval , we have : g (X) –f (X)>0 So |g (X) –f (X)| =g (X)-f (X)= -x

18. Case (2) The area A between f and g from X = 0 to X=1 f(x) >g (X) on(0,1) and hence on this interval , we have F(X) –g (X)>0 so |g (X) –f (X)| =f (X) –g (X) =x-

19. Case (3) The area A between f and g from X = -1 to X=1

20. Volumes of Revolution :V=Π∫ (x)dx • A solid of revolution is formed when a region bounded by part of a curve is rotated about a straight line.

21. Rotation about x-axis: Rotation about y-axis:

22. Example: Find the volume of the solid generated by revolving the region bounded by the graph of y = x, y =0, x = 0 and x = 2. At the solid Solution:we shall now use definite integrals to find the volume defined above. If we let f(x) = x according to 1 above, the volume is given by the definite integral Volume

23. 1 Example1:1 Consider the area bounded by the graph of the function f(x) =x- and x-axis: The volume of solid is:

24. In conclusion, an integral is a mathematical object that can be interpreted as an area or a generalization of area. Integrals, together with derivatives, are the fundamental objects of calculus. Other words for integral include antiderivative and primitive.

25. Thank you for listening

26. The group members: • Sabrina Kamal ___________________ID:2004/58527 • Manal Alsaadi ____________________ID:2004/51562 • Taiba Mustafa ____________________ID:2005/50524 • Muneera Ahmed__________________ID:2004/550244 Math119 - Section (1) Fall 2006 Dr.F.K.Al-Muhannadi