Area bounded by curve and horizontal or vertical line
Question 5 Deleted for CBSE Board 2025 Exams
Question 1 Deleted for CBSE Board 2025 Exams You are here
Question 9 Deleted for CBSE Board 2025 Exams
Question 9 Deleted for CBSE Board 2025 Exams
Question 2 Important Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Question 1 Deleted for CBSE Board 2025 Exams You are here
Question 2 Deleted for CBSE Board 2025 Exams
Ex 8.1, 4 (MCQ) Important
Misc 1 (i)
Question 3 Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Misc 3 Important
Example 4 Important
Example 3
Misc 4 (MCQ)
Misc 5 (MCQ) Important
Area bounded by curve and horizontal or vertical line
Last updated at April 16, 2024 by Teachoo
Question 1 Find the area of the region bounded by the curve π¦2 = π₯ and the lines π₯ = 1, π₯ = 4 and the π₯-axis in ο»Ώthe first quadrant.Let AB represent line π₯=1 CD represent line π₯=4 & CBOAD represent the curve π¦^2=π₯ Since we need area in the first quadrant We have to find area of BCFE Area of BCFE = β«_π^πβπ . π π So, we need to calculate β«_π^πβπ . π π We know that π¦^2=π₯ Taking square root on both sides β΄ π¦=Β±βπ₯ Since BCEF is in 1st Quadrant β΄ π=βπ Area of BCFE = β«_1^4βπ¦ . ππ₯ = β«_π^πββπ . π π = β«_1^4βγ(π₯)^(1/2) ππ₯γ = [π₯^(1/2+1)/(1/2 +1)]_1^4 = [ π₯^(3/2)/(3/2) ]_1^4 = π/π [π^(π/π) ]_π^π = 2/3 {(4)^(3/2)β(1)^(3/2) } = 2/3 {[(4)^(1/2) ]^3β1} = π/π {(π)^πβπ} = 2/3 [8β1] = 2/3 Γ 7 = 14/3 β΄ Thus Required Area = ππ/π square units