Question:
Using the method of integration find the area of the region bounded by lines: 2x + y = 4, 3x - 2y = 6 and x - 3y + 5 = 0
Answer:
Given eqaution of lines are:
2x + y = 4 ..............1
3x - 2y = 6 ...............2
x - 3y + 5 = 0 ...........3
On solving equation 1 and 2, we get
x = 2, y =0
So the point is (2, 0)
On solving equation 2 and 3, we get
x = 4, y =3
So the point is (4, 3)
On solving equation 1 and 3, we get
x = 1, y =2
So the point is (1, 2)
Now area is
A = (Area enclosed by AC and x-axis) + (Area enclosed by AB and x-axis) + (Area enclosed by BC and x-axis)
=> A = 4∫1 (x+5)/3 dx + 1∫2 (4 - 2x)dx + 2∫4 (3x-6)/2 dx
=> A = 4∫1 (x+5)/3 dx - 2∫1 (4 - 2x)dx - 4∫2 (3x-6)/2 dx
=> A = 1/3 *[x2 /2 + 5x 1]4 - [ 4x - x2 1]2 - 1/3*[3x2 /2 - 6x 2]4
=> A = 1/3 * [8 + 20 - 1/2 - 5] - [8 - 4 - 4 + 1] - 1/3 *[24 - 24 - 6 + 12]
=> A = 1/3 * 45/2 - 1 - 1/2 *6
=> A = 15/2 - 1 - 3
=> A = 15/2 - 4
=> A = (15 - 8)/2
=> A = 7/2
So required area is 7/2 unit2
=> A = 17/2 - 1
=> A = 15/2
=> A = 7.5
Not what you are looking for? Go ahead and submit the question, we will get back to you.
