Area bounded by curve and horizontal or vertical line

Chapter 8 Class 12 Application of Integrals
Concept wise

Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month

### Transcript

Misc 17 The area bounded by the curve π¦ = π₯ |π₯| , π₯βππ₯ππ  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 πππ π¦ =βπ₯2 if π₯ < 0] We know that |π₯|={β(π₯, π₯β₯0@&βπ₯, π₯<0)β€ Therefore, y = x|π₯|={β(π₯π₯, π₯β₯0@&π₯(βπ₯), π₯<0)β€ y ={β(π₯^2, π₯β₯0@&βπ₯^2, π₯<0)β€ Area Required = Area ABO + Area DCO Area ABO Area ABO =β«_(β1)^0βγπ¦ ππ₯γ Here, π¦=γβπ₯γ^2 Therefore, Area ABO =β«_(β1)^0βγγβπ₯γ^2 ππ₯γ γ=β[π₯^3/3]γ_(β1)^0 =β[0^3/3β(β1)^3/3] =(β1)/3 Since Area is always positive, Area ABO = 1/3 Area DCO Area DCO =β«_0^1βγπ¦ ππ₯γ Here, π¦=π₯^2 Therefore, Area DCO =β«_0^1βγπ₯^2 ππ₯γ γ=[π₯^3/3]γ_0^1 =1/3 [1^3β0^3 ] =1/3 [1β0] =1/3 β΄ Required Area = Area ABO + Area DCO =1/3+1/3 =2/3 So, Option C is Correct