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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
Example 5 Important
Misc 5 Important
Example 13 Important Deleted for CBSE Board 2023 Exams
Example 12
Misc 16 (MCQ)
Misc 17 (MCQ) Important You are here
Area bounded by curve and horizontal or vertical line
Last updated at Aug. 11, 2021 by Teachoo
Misc 17 The area bounded by the curve π¦ = π₯ |π₯| , π₯βππ₯ππ and the ordinates π₯ = β 1 and π₯=1 is given by (A) 0 (B) 1/3 (C) 2/3 (D) 4/3 [Hint : π¦=π₯2 if π₯ > 0 πππ π¦ =βπ₯2 if π₯ < 0] We know that |π₯|={β(π₯, π₯β₯0@&βπ₯, π₯<0)β€ Therefore, y = x|π₯|={β(π₯π₯, π₯β₯0@&π₯(βπ₯), π₯<0)β€ y ={β(π₯^2, π₯β₯0@&βπ₯^2, π₯<0)β€ Area Required = Area ABO + Area DCO Area ABO Area ABO =β«_(β1)^0βγπ¦ ππ₯γ Here, π¦=γβπ₯γ^2 Therefore, Area ABO =β«_(β1)^0βγγβπ₯γ^2 ππ₯γ γ=β[π₯^3/3]γ_(β1)^0 =β[0^3/3β(β1)^3/3] =(β1)/3 Since Area is always positive, Area ABO = 1/3 Area DCO Area DCO =β«_0^1βγπ¦ ππ₯γ Here, π¦=π₯^2 Therefore, Area DCO =β«_0^1βγπ₯^2 ππ₯γ γ=[π₯^3/3]γ_0^1 =1/3 [1^3β0^3 ] =1/3 [1β0] =1/3 β΄ Required Area = Area ABO + Area DCO =1/3+1/3 =2/3 So, Option C is Correct