1. Class 12
2. Important Question for exams Class 12

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 Step 1: Make Figure 2𝑥 + 𝑦 = 4 3𝑥 – 2𝑦 = 6 𝑥 – 3𝑦 + 5 = 0 Step 2: 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 A (4, 3) Finding area 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− 2 𝑥﷮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﷯ 3 𝑥﷮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

Class 12
Important Question for exams Class 12