Area between curve and curve

Chapter 8 Class 12 Application of Integrals
Concept wise

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### Transcript

Misc 14 Using the method of integration find the area of the region bounded by lines: 2π₯ + π¦ = 4, 3π₯β2π¦=6 and π₯β3π¦+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«_1^4βγπ¦ ππ₯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«_1^4βγ((π₯+5)/3) ππ₯γ =1/3 β«_1^4βγ(π₯+5) ππ₯γ =1/3 [π₯^2/2+5π₯]_1^4 =1/3 [4^2/2+5.4β[1^2/2+5.1]] =1/3 [8+20β1/2β5] =1/3 [45/2] =15/2 Area ACD Area ACD =β«_1^2βγπ¦ ππ₯γ π¦β Equation of line AC 2π₯+π¦=4 π¦=4β2π₯ Area ACD =β«_1^2βγ(4β2π₯" " ) ππ₯γ =[4π₯β(2π₯^2)/2]_1^2 =[4π₯βπ₯^2 ]_1^2 =[4.2β2^2β[4.1β1^2 ]] =[8β4β4+1] = 1 Area CBE Area CBE =β«_2^4βγπ¦ ππ₯γ π¦β Equation of line BC 3π₯+2π¦=6 3π₯β6=2π¦ (3π₯ β 6)/2=π¦ π¦=(3π₯ β 6)/2 Therefore, Area CBE =β«_2^4βγ((3π₯ β 6)/2) ππ₯γ =1/2 β«_2^4βγ(3π₯β6) ππ₯γ =1/2 [(3π₯^2)/2β6π₯]_2^4 =1/2 [γ3.4γ^2/2β6.4β[γ3.2γ^2/2β6.2]] =1/2 [24β24β6+12] =3 Hence Area Required = Area ABED β Area ACD β Area CBE =15/2β1β3 =15/2β4 =(15 β 8)/2 =π/π square units