Area between curve and curve

Chapter 8 Class 12 Application of Integrals
Concept wise

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### Transcript

Misc 11 Using the method of integration find the area bounded by the curve |π₯|+|π¦|=1 [Hint: The required region is bounded by lines π₯+π¦= 1, π₯ βπ¦=1, βπ₯+π¦ =1 and βπ₯ βπ¦=1 ] We know that "β" π₯"β"={β(π₯, π₯β₯0@&βπ₯, π₯<0)β€ & "β" π¦"β"={β(π¦, π¦β₯0@&βπ¦, π¦<0)β€ So, we can write βπ₯"β+β" π¦"β"=1 as {β(β(β( π₯+π¦=1 πππ π₯>0 , π¦>0@βπ₯+π¦=1 πππ π₯<0 π¦>0)@β( π₯βπ¦ =1 πππ π₯>0 , π¦<0@βπ₯βπ¦=1 πππ π₯<0 π¦<0)))β€ For π+π=π For βπ+π=π For βπβπ=π For πβπ=π Joining them, we get our diagram Since the Curve symmetrical about π₯ & π¦βππ₯ππ  Required Area = 4 Γ Area AOB Area AOB Area AOB = β«_0^1βγπ¦ ππ₯γ where π₯+π¦=1 π¦=1βπ₯ Therefore, Area AOB = β«_0^1βγ(1βπ₯) ππ₯γ = [π₯βπ₯^2/2]_0^1 =1βγ 1γ^2/2β(0β0^2/2)^2 =1β1/2 =1/2 Hence, Required Area = 4 Γ Area AOB = 4 Γ 1/2 = 2 square units