Ex 7.9

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.9, 4 Evaluate the integrals using substitution โซ_0^2โใ๐ฅโ(๐ฅ+2)ใโกใ (๐๐ข๐ก ๐ฅ+2=๐ก^2 )ใ โซ_0^2โใ๐ฅโ(๐ฅ+2)ใโกใ ๐๐ฅใ Put ๐ฅ+2=๐ก^2 Differentiating w.r.t. ๐ฅ ๐(๐ฅ + 2)/๐๐ฅ=๐(๐ก^2 )/๐๐ก ร๐๐ก/๐๐ฅ 1=2๐ก ร ๐๐ก/๐๐ฅ ๐๐ฅ=2๐ก ๐๐ก Hence, when ๐ฅ varies from 0 to 2, then t varies from โ2 to 2 Therefore we can write โซ_0^2โใ๐ฅโ(๐ฅ+2) ๐๐ฅ =โซ_(โ2)^2โใ(๐ก^2โ2) โ(๐ก^2 ) 2๐ก ๐๐กใใ =โซ_(โ2)^2โใ(๐ก^2โ2)๐ก ร2๐ก ๐๐กใ =2โซ_(โ2)^2โใ(๐ก^2โ2) ๐ก^2 ๐๐กใ =2โซ_(โ2)^2โใ(๐ก^4โ2๐ก^2 ) ๐๐กใ =2[๐ก^(4+1)/(4+1)โ2 ๐ก^(2+1)/(2+1)]_(โ2)^2 =2[๐ก^5/5โ2 ๐ก^3/3]_(โ2)^2 =2ร [2^5/5โ2/3 2^3โ((โ2)^5/5โ2/3 (โ2)^3 )] =2ร [๐ฅ^5/2โ2/3 2^3โ((โ2)^5/2โ2/3 (โ2)^3 )] =2ร [32/5โ16/3โ(4โ2)/5+4/3 โ2] =2ร [(96 โ 80 โ 12โ2 + 20โ2)/15] =2ร [(16 + 8โ2)/15] =(2 ร 8 โ2 (โ2 + 1))/15 =(๐๐โ(๐ ) (โ๐ + ๐))/๐๐