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Area between curve and curve
Misc 13
Ex 8.2, 4 Important Deleted for CBSE Board 2023 Exams
Ex 8.2, 5 Important Deleted for CBSE Board 2023 Exams
Misc 14 Important
Misc 4 Important
Misc 11 Important
Ex 8.2,1 Important Deleted for CBSE Board 2023 Exams
Misc 15 Deleted for CBSE Board 2023 Exams
Example 7 Important Deleted for CBSE Board 2023 Exams
Misc 18 (MCQ) Deleted for CBSE Board 2023 Exams
Example 15 Important
Ex 8.2, 2 Deleted for CBSE Board 2023 Exams
Example 10 Important Deleted for CBSE Board 2023 Exams
Example 6 Important Deleted for CBSE Board 2023 Exams You are here
Misc 19 (MCQ) Important Deleted for CBSE Board 2023 Exams
Area between curve and curve
Last updated at Dec. 12, 2019 by Teachoo
Example 6 Find the area of the region bounded by the two parabolas π¦=π₯2 and π¦2 = π₯ Drawing figure Here, we have parabolas π¦^2=π₯ π₯^2=π¦ Area required = Area OABC Finding Point of intersection B Solving π¦2 = π₯ π₯2 =π¦ Put (2) in (1) π¦2 = π₯ (π₯^2 )^2=π₯ π₯^4βπ₯=0 π₯(π₯^3β1)=0 Finding y β coordinate For π=π π¦=π₯^2=0^2= 0 So, coordinates are (0 , 0) For π=π π¦=π₯^2=1^2=1 So, coordinates are (1 , 1) Since point B lies in 1st quadrant So, co-ordinate of B is (1 , 1) Finding Area Area OABC = Area OABD β Area OCBD Finding Area OABD Area OABD =β«_0^1βγπ¦ ππ₯γ Here, π¦^2=π₯ π¦=Β±βπ₯ As OABD is in 1st quadrant, value of y is positive β΄ π¦=βπ₯ Area OBQP =β«_0^1βγβπ₯ ππ₯γ =β«_0^1βγβπ₯ ππ₯γ =β«_0^1βγπ₯^(1/2) ππ₯γ = [π₯^(1/2 + 1)/(1/2 + 1)]_0^1 = [π₯^(3/2)/(3/2)]_0^1 = 2/3 [π₯^(3/2) ]_0^1 =2/3 [(1)^(3/2)β(0)^(3/2) ] =2/3 [1β0] =2/3 Area OCBD Area OCBD =β«_0^1βγπ¦ ππ₯γ Here, π₯^2=π¦ π¦=π₯^2 Area OAQP =β«_0^1βγπ₯^2 ππ₯γ =[π₯^(2 + 1)/(2 + 1)]_0^1 =1/3 [π₯^3 ]_0^1 =1/3 [1^3β0^3 ] =π/π Therefore, Area OABC = Area OABD β Area OCBD = 2/3β1/3 = π/π square units