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

Applications of the Definite Integral In calculus, the integral of 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.

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 integrating known as the integrand, and dx is a notation for the variable of integration. Integrals discussed in this project are termed definite integrals.

We use definite integrals basically, Results:

Given: evaluate,

Solution:

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 :

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

(Positive) (Negative)

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]

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

f(x)<0; x (2, 3] Case2 The area A2 between f, the x-axis and the Lines x=2 and x=3 is:

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

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) =

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

Example (2) Let f (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

Case (1): The area A1 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

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-

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

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.

Rotation about x-axis: Rotation about y-axis:

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

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

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.

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