Question 11 - Chapter 8 Class 12 Application of Integrals (Important Question)

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Question 11 Using the method of integration find the area of the region bounded by lines: 2π₯ + π¦ = 4, 3π₯β2π¦=6 and π₯β3π¦+5=0
Plotting the 3 lines on the graph
2π₯ + π¦ = 4
3π₯ β 2π¦ = 6
π₯ β 3π¦ + 5 = 0
Find intersecting Points A & B
Point A
Point A is intersection of lines
x β 3y + 5 = 0 & 2x + y = 4
Now,
x β 3y + 5 = 0
x = 3y β 5
Putting x = 3y β 5 in
2x + y = 4
2(3y β 5) + y = 4
6y β 10 + y = 4
7y = 14
y = 2
Putting y = 2 in
x β 3y + 5 = 0
x β 3(2) + 5 = 0
x β 6 + 5 = 0
x = 1
So, point A (1, 2)
Point B
Point B is intersection of lines
x β 3y + 5 = 0 & 3x β 2y = 6
Now,
x β 3y + 5 = 0
x = 3y β 5
Putting x = 3y β 5 in
3x β 2y = 6
3(3y β 5) β 2y = 6
9y β 15 β 2y = 6
7y = 21
y = 3
Putting y = 3 in
x β 3y + 5 = 0
x β 3(3) + 5 = 0
x β 9 + 5 = 0
x = 4
So, point B is (4, 3)
Finding area
Area Required = Area ABED β Area ACD β Area CBE
Area ABED
Area ABED =β«_1^4βγπ¦ ππ₯γ
π¦β Equation of AB
π₯ β 3π¦+5=0
π₯+5=3π¦
(π₯ + 5)/3=π¦
π¦=(π₯ + 5)/3
Therefore,
Area ABED =β«_1^4βγ((π₯+5)/3) ππ₯γ
=1/3 β«_1^4βγ(π₯+5) ππ₯γ
=1/3 [π₯^2/2+5π₯]_1^4
=1/3 [4^2/2+5.4β[1^2/2+5.1]]
=1/3 [8+20β1/2β5]
=1/3 [45/2]
=15/2
Area ACD
Area ACD =β«_1^2βγπ¦ ππ₯γ
π¦β Equation of line AC
2π₯+π¦=4
π¦=4β2π₯
Area ACD =β«_1^2βγ(4β2π₯" " ) ππ₯γ
=[4π₯β(2π₯^2)/2]_1^2
=[4π₯βπ₯^2 ]_1^2
=[4.2β2^2β[4.1β1^2 ]]
=[8β4β4+1]
= 1
Area CBE
Area CBE =β«_2^4βγπ¦ ππ₯γ
π¦β Equation of line BC
3π₯+2π¦=6
3π₯β6=2π¦
(3π₯ β 6)/2=π¦
π¦=(3π₯ β 6)/2
Therefore,
Area CBE =β«_2^4βγ((3π₯ β 6)/2) ππ₯γ
=1/2 β«_2^4βγ(3π₯β6) ππ₯γ
=1/2 [(3π₯^2)/2β6π₯]_2^4
=1/2 [γ3.4γ^2/2β6.4β[γ3.2γ^2/2β6.2]]
=1/2 [24β24β6+12]
=3
Hence
Area Required = Area ABED β Area ACD β Area CBE
=15/2β1β3
=15/2β4
=(15 β 8)/2
=π/π square units

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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