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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