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Area bounded by curve and horizontal or vertical line
Ex 8.1, 7
Ex 8.1, 1
Ex 8.1, 11
Example 11
Ex 8.1, 2 Important You are here
Ex 8.1, 8 Important
Example 3
Misc 3
Ex 8.1, 13 (MCQ) Important
Misc 1 (i)
Ex 8.1, 3
Example 5 Important
Misc 5 Important
Example 13 Important Deleted for CBSE Board 2023 Exams
Example 12
Misc 16 (MCQ)
Misc 17 (MCQ) Important
Area bounded by curve and horizontal or vertical line
Last updated at Dec. 12, 2019 by Teachoo
Ex 8.1, 2 Find the area of the region bounded by y2 = 9π₯, π₯ = 2, π₯ = 4 and the π₯-axis in the first quadrant. Given curve π¦^2=9π₯ We have to find area between x = 2 and x = 4 β΄ We have to find area of BCFE Area of BCFE = β«_2^4βγπ¦ .γ ππ₯ We know that π¦^2=9π₯ Taking square root on both sides π¦=Β±β9π₯ π¦=Β±3βπ₯ Since BCFE is in 1st Quadrant We take positive value of y β΄ π¦=3βπ₯ Area of BCFE = β«_2^4βγπ¦ .γ ππ₯ = 3β«_2^4ββπ₯ ππ₯ = 3β«_2^4βγ(π₯)^(1/2) ππ₯γ = 3 [π₯^(1/2 + 1)/(1/2 + 1 )]_2^4 = 3 [π₯^(3/2 )/(3/2)]_2^4 = 3 Γ 2/3 [π₯^(3/2) ]_2^4 = 2 [(4)^(3/2 )β(2)^(3/2) ] = 2 [((4)^(1/2) )^3β((2)^(1/2) )^3 ] = 2 [(2)^3β(β2)^3 ] = 2 [8 β2β2] = 16 β 4β2 Thus, Area = 16 β 4βπ square units