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Misc 14 - Find area bounded by lines: 2x + y = 4, 3x-2y=6

Misc 14 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 14 - Chapter 8 Class 12 Application of Integrals - Part 3
Misc 14 - Chapter 8 Class 12 Application of Integrals - Part 4
Misc 14 - Chapter 8 Class 12 Application of Integrals - Part 5
Misc 14 - Chapter 8 Class 12 Application of Integrals - Part 6
Misc 14 - Chapter 8 Class 12 Application of Integrals - Part 7
Misc 14 - Chapter 8 Class 12 Application of Integrals - Part 8
Misc 14 - Chapter 8 Class 12 Application of Integrals - Part 9
Misc 14 - Chapter 8 Class 12 Application of Integrals - Part 10

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Transcript

Misc 14 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 Plotting the 3 lines on the graph 2𝑥 + 𝑦 = 4 3𝑥 – 2𝑦 = 6 𝑥 – 3𝑦 + 5 = 0 Find intersecting Points A & B Point A Point A is intersection of lines x – 3y + 5 = 0 & 2x + y = 4 Now, x – 3y + 5 = 0 x = 3y – 5 Putting x = 3y – 5 in 2x + y = 4 2(3y – 5) + y = 4 6y – 10 + y = 4 7y = 14 y = 2 Putting y = 2 in x – 3y + 5 = 0 x – 3(2) + 5 = 0 x – 6 + 5 = 0 x = 1 So, point A (1, 2) Point B Point B is intersection of lines x – 3y + 5 = 0 & 3x – 2y = 6 Now, x – 3y + 5 = 0 x = 3y – 5 Putting x = 3y – 5 in 3x – 2y = 6 3(3y – 5) – 2y = 6 9y – 15 – 2y = 6 7y = 21 y = 3 Putting y = 3 in x – 3y + 5 = 0 x – 3(3) + 5 = 0 x – 9 + 5 = 0 x = 4 So, point B is (4, 3) Finding area Area Required = Area ABED – Area ACD – Area CBE Area ABED Area ABED =∫_1^4▒〖𝑦 𝑑𝑥〗 𝑦→ Equation of AB 𝑥 – 3𝑦+5=0 𝑥+5=3𝑦 (𝑥 + 5)/3=𝑦 𝑦=(𝑥 + 5)/3 Therefore, Area ABED =∫_1^4▒〖((𝑥+5)/3) 𝑑𝑥〗 =1/3 ∫_1^4▒〖(𝑥+5) 𝑑𝑥〗 =1/3 [𝑥^2/2+5𝑥]_1^4 =1/3 [4^2/2+5.4−[1^2/2+5.1]] =1/3 [8+20−1/2−5] =1/3 [45/2] =15/2 Area ACD Area ACD =∫_1^2▒〖𝑦 𝑑𝑥〗 𝑦→ Equation of line AC 2𝑥+𝑦=4 𝑦=4−2𝑥 Area ACD =∫_1^2▒〖(4−2𝑥" " ) 𝑑𝑥〗 =[4𝑥−(2𝑥^2)/2]_1^2 =[4𝑥−𝑥^2 ]_1^2 =[4.2−2^2−[4.1−1^2 ]] =[8−4−4+1] = 1 Area CBE Area CBE =∫_2^4▒〖𝑦 𝑑𝑥〗 𝑦→ Equation of line BC 3𝑥+2𝑦=6 3𝑥−6=2𝑦 (3𝑥 − 6)/2=𝑦 𝑦=(3𝑥 − 6)/2 Therefore, Area CBE =∫_2^4▒〖((3𝑥 − 6)/2) 𝑑𝑥〗 =1/2 ∫_2^4▒〖(3𝑥−6) 𝑑𝑥〗 =1/2 [(3𝑥^2)/2−6𝑥]_2^4 =1/2 [〖3.4〗^2/2−6.4−[〖3.2〗^2/2−6.2]] =1/2 [24−24−6+12] =3 Hence Area Required = Area ABED – Area ACD – Area CBE =15/2−1−3 =15/2−4 =(15 − 8)/2 =𝟕/𝟐 square units

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.