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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
Ex 8.1, 8 Important
Example 3
Misc 3
Ex 8.1, 13 (MCQ) Important
Misc 1 (i)
Ex 8.1, 3 You are here
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, 3 Find the area of the region bounded by π₯2= 4π¦ , π¦ = 2 , π¦=4 and the π¦-axis in the first quadrant. Here, The curve is π₯^2=4π¦ We have to find area between y = 2 and y = 4 β΄ We have to find area of BCFE Area of BCFE = β«_2^4βπ₯ ππ¦ We know that π₯^2=4π¦ Taking square root on both sides π₯="Β±" β4π¦β π₯="Β±" 2βπ¦ Since, BCEF is in 1st Quadrant We take positive value of x β΄ π₯=2βπ¦ Area of BCFE = β«_2^4βπ₯ ππ¦ = β«_2^4βγ2βπ¦γ ππ¦ = 2β«_2^4βγ(π¦)^(1/2) ππ¦γ = 2 [π¦^(3/2)/(3/2)]_2^4 = 2 Γ 2/3 [π¦^(3/2) ]_2^4 = 4/3 [(4)^(3/2 )β(2)^(3/2) ] = 4/3 [((4)^(1/2) )^3β((2)^(1/2) )^3 ] = 4/3 [(2)^3β(β2)^3 ] = 4/3 [8 β2β2] = (32 β 8β2)/3 Thus, Area = (ππ β πβπ)/π square units