





Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Area between curve and curve
Question 10 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Misc 2 Important
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 13 (MCQ) Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 14 (MCQ) Important Deleted for CBSE Board 2024 Exams
Area between curve and curve
Last updated at May 29, 2023 by Teachoo
Question 7 Using integration find the area of region bounded by the triangle whose vertices are (1, 0), (2, 2) and (3, 1) Area of ∆ formed by point 1 , 0 , 2 ,2 & 3 , 1 Step 1: Draw the figure Area ABD Area ABD= 12𝑦 𝑑𝑥 𝑦→ equation of line AB Equation of line between A(1, 0) & B(2, 2) is 𝑦 − 0𝑥 − 1= 2 − 02 − 1 𝑦𝑥 − 1= 21 y = 2(x – 1) y = 2x – 2 Area ABD = 12𝑦 𝑑𝑥 = 122 𝑥−1 𝑑𝑥 = 2 𝑥22−𝑥12 =2 222−2− 122−1 =2 2−2− 12+1 =2 12 = 1 Area BDEC Area BDEC = 23𝑦 𝑑𝑥 𝑦→ equation of line BC Equation of line between B(2, 2) & C(3, 1) is 𝑦 − 2𝑥 − 2= 1 − 23 − 2 𝑦 − 2𝑥 − 2= −11 y – 2 = –1(x – 2) y – 2 = –x + 2 y = 4 – x Area BDEC = 23𝑦 𝑑𝑥 = 23 4−𝑥 𝑑𝑥 =4 23𝑑𝑥− 23𝑥𝑑𝑥 =4 𝑥23− 𝑥2223 =4 3−2− 12 32− 22 =4 ×1− 12 9−4 =4− 12 ×5 = 4− 52 = 8 − 52 = 32 Area ACE Area ACE= 13𝑦 𝑑𝑥 𝑦→ equation of line AC Equation of line between A(1, 0) & C(3, 1) is 𝑦 − 0𝑥 − 1= 1 − 03 − 1 𝑦𝑥 − 1= 12 y = 12 (x – 1) Area ACE = 13𝑦 𝑑𝑥 = 13 12 𝑥−1 𝑑𝑥 = 12 13 𝑥−1 𝑑𝑥 = 12 𝑥22−𝑥13 = 12 322−3− 122−1 = 12 92−3− 12+1 = 12 42 =1 Hence Area Required = Area ABD + Area BDEC – Area ACE = 1 + 32−1 = 32