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Area between curve and curve
Misc 13
Ex 8.2, 4 Important Deleted for CBSE Board 2023 Exams
Ex 8.2, 5 Important Deleted for CBSE Board 2023 Exams
Misc 14 Important You are here
Misc 4 Important
Misc 11 Important
Ex 8.2,1 Important Deleted for CBSE Board 2023 Exams
Misc 15 Deleted for CBSE Board 2023 Exams
Example 7 Important Deleted for CBSE Board 2023 Exams
Misc 18 (MCQ) Deleted for CBSE Board 2023 Exams
Example 15 Important
Ex 8.2, 2 Deleted for CBSE Board 2023 Exams
Example 10 Important Deleted for CBSE Board 2023 Exams
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Misc 19 (MCQ) Important Deleted for CBSE Board 2023 Exams
Area between curve and curve
Last updated at Dec. 12, 2019 by Teachoo
Misc 14 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