Area Using Integration

In this page area using integration we are going to see how to find area bounded by the region.

Area using integration- Examples

Example 1:

find the area of the region bounded by the straight line 3x -2y + 6 = 0, x =  2 , x = 3 and  x-axis.

Solution:

First we need to draw the rough graph for the given line. For that let us find the x and y-intercepts for the given line.

3x -2y + 6 = 0

 2 y = 3x + 6

    y = (3x + 6)/2

    y = (3/2) (x+2)

To find y-intercept                                        To find x-intercept

Put x = 0                                                     put y = 0

3 (0) - 2y = 6                                            3x - 2(0) = 6 

        - 2y = 6                                                      3x = 6

             y = 6/(-2)                                                x =6/3

            y = -3                                                      x = 2

If we write this as point.                           If we write this as point. 

We will get (0,-3)                                    We will get (2,0)   

By using these two points we have drawn the graph.

From the above diagram we can decide that the required area is above the x-axis. So the formula to find the required area will be

Now let us calculate the required area using the formula 

From the calculation we are getting -6.75. Area will not be negative. So the required area will be 6.75 square units.

Let us see another example.

Example 2:

Find the area bounded by the line x - y = 1, x = 2 , x= 4 and x-axis.

Solution:

First let us draw the graph for the given line.For that let us find the x and y-intercepts for the given line.

x + y = 1               

x -intercept                                    y -intercept            

Put y = 0                                        put x = 0

 x + 0 = 1                                           0 + y = 1

       x = 1                                                y = 1

If we write this as point                          If we write this as point

we will get (1,0)                                     we will get (0,1)

Now let us calculate the required area using the formula