Misc 11 - Chapter 8 Class 12 Application of Integrals (Term 2)
Last updated at Dec. 12, 2019 by Teachoo
Last updated at Dec. 12, 2019 by Teachoo
Transcript
Misc 11 Using the method of integration find the area bounded by the curve |๐ฅ|+|๐ฆ|=1 [Hint: The required region is bounded by lines ๐ฅ+๐ฆ= 1, ๐ฅ โ๐ฆ=1, โ๐ฅ+๐ฆ =1 and โ๐ฅ โ๐ฆ=1 ] We know that "โ" ๐ฅ"โ"={โ(๐ฅ, ๐ฅโฅ0@&โ๐ฅ, ๐ฅ<0)โค & "โ" ๐ฆ"โ"={โ(๐ฆ, ๐ฆโฅ0@&โ๐ฆ, ๐ฆ<0)โค So, we can write โ๐ฅ"โ+โ" ๐ฆ"โ"=1 as {โ(โ(โ( ๐ฅ+๐ฆ=1 ๐๐๐ ๐ฅ>0 , ๐ฆ>0@โ๐ฅ+๐ฆ=1 ๐๐๐ ๐ฅ<0 ๐ฆ>0)@โ( ๐ฅโ๐ฆ =1 ๐๐๐ ๐ฅ>0 , ๐ฆ<0@โ๐ฅโ๐ฆ=1 ๐๐๐ ๐ฅ<0 ๐ฆ<0)))โค For ๐+๐=๐ For โ๐+๐=๐ For โ๐โ๐=๐ For ๐โ๐=๐ Joining them, we get our diagram Since the Curve symmetrical about ๐ฅ & ๐ฆโ๐๐ฅ๐๐ Required Area = 4 ร Area AOB Area AOB Area AOB = โซ_0^1โใ๐ฆ ๐๐ฅใ where ๐ฅ+๐ฆ=1 ๐ฆ=1โ๐ฅ Therefore, Area AOB = โซ_0^1โใ(1โ๐ฅ) ๐๐ฅใ = [๐ฅโ๐ฅ^2/2]_0^1 =1โใ 1ใ^2/2โ(0โ0^2/2)^2 =1โ1/2 =1/2 Hence, Required Area = 4 ร Area AOB = 4 ร 1/2 = 2 square units
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