Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month
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 You are here
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
Misc 19 (MCQ) Important Deleted for CBSE Board 2023 Exams
Area between curve and curve
Last updated at Dec. 12, 2019 by Teachoo
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₯ βπ¦=1, βπ₯+π¦ =1 and βπ₯ βπ¦=1 ] We know that "β" π₯"β"={β(π₯, π₯β₯0@&βπ₯, π₯<0)β€ & "β" π¦"β"={β(π¦, π¦β₯0@&βπ¦, π¦<0)β€ So, we can write βπ₯"β+β" π¦"β"=1 as {β(β(β( π₯+π¦=1 πππ π₯>0 , π¦>0@βπ₯+π¦=1 πππ π₯<0 π¦>0)@β( π₯βπ¦ =1 πππ π₯>0 , π¦<0@βπ₯βπ¦=1 πππ π₯<0 π¦<0)))β€ For π+π=π For βπ+π=π For βπβπ=π For πβπ=π Joining them, we get our diagram Since the Curve symmetrical about π₯ & π¦βππ₯ππ Required Area = 4 Γ Area AOB Area AOB Area AOB = β«_0^1βγπ¦ ππ₯γ where π₯+π¦=1 π¦=1βπ₯ Therefore, Area AOB = β«_0^1βγ(1βπ₯) ππ₯γ = [π₯βπ₯^2/2]_0^1 =1βγ 1γ^2/2β(0β0^2/2)^2 =1β1/2 =1/2 Hence, Required Area = 4 Γ Area AOB = 4 Γ 1/2 = 2 square units