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

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

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