Integration Full Chapter Explained - Integration Class 12 - Everything you need





Last updated at Dec. 20, 2019 by Teachoo
Transcript
Ex 7.8, 3 โซ1_2^3โใ๐ฅ2 ๐๐ฅใ โซ1_2^3โใ๐ฅ2 ๐๐ฅใ Putting ๐ =2 ๐ =3 โ=(๐ โ ๐)/๐ =(3 โ 2)/๐ =1/๐ ๐(๐ฅ)=๐ฅ^2 We know that โซ1_๐^๐โใ๐ฅ ๐๐ฅใ =(๐โ๐) (๐๐๐)โฌ(๐โโ) 1/๐ (๐(๐)+๐(๐+โ)+๐(๐+2โ)โฆ+๐(๐+(๐โ1)โ)) Hence we can write โซ1_2^3โใ๐ฅ2 ๐๐ฅใ =(3โ2) (๐๐๐)โฌ(๐โโ) 1/๐ (๐(2)+๐(2+โ)+๐(2+2โ)+ โฆ+๐(2+(๐โ1)โ)) =(๐๐๐)โฌ(๐โโ) 1/๐ (๐(2)+๐(2+โ)+๐(2+2โ)+ โฆ+๐(2+(๐โ1)โ)) Here, ๐(๐ฅ)=๐ฅ^2 ๐(2)=(2)^2=4 ๐(2+โ)=(2+โ)^2 ๐ (2+2โ)=(2+2โ)^2 โฆ ๐(2+(๐โ1)โ)=(2+(๐โ1)โ)^2 Hence, our equation becomes โซ1_2^3โใ๐ฅ2 ๐๐ฅใ " " =(๐๐๐)โฌ(๐โโ) 1/๐ (๐(2)+๐(2+โ)+๐(2+2โ)+ โฆ+๐(2+(๐โ1)โ)) =1 (๐๐๐)โฌ(๐โโ) 1/๐ ((2)^2+(2+โ)^2+(2+2โ)^2+ โฆ+(2+(๐โ1)โ)^2 ) =(๐๐๐)โฌ(๐โโ) 1/๐ (โ(2^2+(2^2+โ^2+4โ)+ใ(2ใ^2+(2โ)^2+8โ)+ โฆโฆ @ โฆ+(2^2+((๐โ1)โ)^2+4(๐โ1) โ) )) =(๐๐๐)โฌ(๐โโ) 1/๐ [2^2+2^2+ โฆ +2^2 ] + โ^2+(2โ)^2+ โฆ +(๐โ1)โ^2 + [4โ+8โ+ โฆ +4(๐โ1)โ] =(๐๐๐)โฌ(๐โโ) 1/๐ (ใ๐(2)ใ^2+[โ^2+(2)^2 . โ^2+ โฆ +(๐โ1)^2 โ^2 ] +[4โ+2ร4โ+ โฆ +(๐โ1)ร4โ] ) =(๐๐๐)โฌ(๐โโ) 1/๐(๐(2)^2+๐^2 [(1)^2+(2)^2+ โฆ+(๐โ1)^2 ] + ๐๐ [1+2+ โฆ+(๐โ1)]) =(๐๐๐)โฌ(๐โโ) 1/๐ (๐(2)^2+โ^2 [๐(๐ โ 1)(2๐ โ 1)/6]+4โ[๐(๐ โ 1)/2] ) We know that 1^2+2^2+ โฆ+๐^2= (๐ (๐ + 1)(2๐ + 1))/6 1^2+2^2+ โฆโฆ+(๐โ1)^2 = ((๐ โ 1) (๐ โ1 + 1)(2(๐ โ 1) + 1))/6 = ((๐ โ 1) ๐ (2๐ โ 2 + 1) )/6 = (๐ (๐ โ 1) (2๐ โ 1) )/6 We know that 1+2+3+ โฆโฆ+๐= (๐ (๐ + 1))/2 1+2+3+ โฆโฆ+(๐โ1) = ((๐ โ 1) (๐ โ 1 + 1))/2 = (๐ (๐ โ 1) )/2 =(๐๐๐)โฌ(๐โโ) 1/๐ (4๐+โ^2 [๐(๐ โ 1)(2๐ โ 1)]/6+2โ[๐(๐ โ 1)] ) =(๐๐๐)โฌ(๐โโ) (4๐/๐+โ^2 [๐(๐ โ 1)(2๐ โ 1)/6๐]+2โ[๐(๐ โ 1)/๐]) =(๐๐๐)โฌ(๐โโ) (4+โ^2 [(๐ โ 1)(2๐ โ 1)/6]+2โ[(๐ โ 1)]) =(๐๐๐)โฌ(๐โโ) (4+(1/๐)^2 (๐ โ 1)(2๐ โ 1)/6+2(1/๐)(๐ โ 1)) =(๐๐๐)โฌ(๐โโ) (4+1/๐^2 . (๐ โ 1)(2๐ โ 1)/6 +2(1 โ 1/๐)) =(๐๐๐)โฌ(๐โโ) (4+ (1 โ 1/๐)(2 โ 1/๐)/6 +2(1 โ 1/๐)) =4+ (1 โ 1/โ)(2 โ 1/โ)/6 +2(1 โ 1/โ) =4+ (1 โ 0)(2 โ 0)/6 +2(1 โ0) =4+ (1 . 2)/6 +2 .1 =4+ 1/3 +2 =6+ 1/3 = ๐๐/๐
Ex 7.8
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