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

1. Chapter 7 Class 12 Integrals
2. Concept wise
3. Definite Integration - By Formulae

Transcript

Ex 7.9, 19 โซ_0^2โ(6๐ฅ + 3)/(๐ฅ^2 + 4) ๐๐ฅ Let F(๐ฅ)=โซ1โใ(6๐ฅ + 3)/(๐ฅ^2 + 4) ๐๐ฅใ =โซ1โใ6๐ฅ/(๐ฅ^2 + 4) ๐๐ฅ+โซ1โใ3/(๐ฅ^2 + 4) ๐๐ฅ ใใ Solving ๐ฐ๐ ๐ผ1=โซ1โใ6๐ฅ/(๐ฅ^2 + 4) ๐๐ฅใ Put ๐ฅ^2 + 4=๐ก Differentiating w.r.t.๐ฅ ๐/๐๐ฅ (๐ฅ^2+4)=๐๐ก/๐๐ฅ 2๐ฅ+0=๐๐ก/๐๐ฅ ๐๐ฅ=๐๐ก/2๐ฅ Therefore, โซ1โใ6๐ฅ/(๐ฅ^2 + 4) ๐๐ฅ=โซ1โใ6๐ฅ/๐ก ๐๐ก/2๐ฅใใ =โซ1โใ3/๐ก ๐๐กใ =3 ๐๐๐|๐ก| =3 ๐๐๐|๐ฅ^2+4| (โต๐ก=๐ฅ^2+4) Solving ๐ฐ๐ ๐ผ2=โซ1โใ3/(๐ฅ^2 + 4) ๐๐ฅใ =3โซ1โ1/(๐ฅ^2+4) ๐๐ฅ =3โซ1โ1/(๐ฅ^2 + 2^2 ) ๐๐ฅ =3 ร 1/2 tan^(โ1)โกใ๐ฅ/2ใ =3/2 tan^(โ1)โกใ๐ฅ/2ใ Therefore F(๐ฅ)= ๐ผ1+๐ผ2 F(๐ฅ)=3๐๐๐|๐ฅ^2+4|+3/2 tan^(โ1)โกใ๐ฅ/2ใ ( As โซ1โใ๐๐ฅ/(๐ฅ^2 + ๐^2 )=1/๐ tan^(โ1)โกใ๐ฅ/๐ใ ใ) Now, โซ_0^2โใ(6๐ฅ + 3)/(๐ฅ^2 + 4) ๐๐ฅ=๐น(2)โ๐น(0) ใ =3๐๐๐|2^2+4|+3/2 tan^(โ1)โกใ2/2โ3๐๐๐|0+4|โ3/2 tan^(โ1)โก(0/2) ใ =3๐๐๐|4+4|+3/2 tan^(โ1)โกใ1โ3๐๐๐|4|โ3/2 ร 0ใ =3๐๐๐|8|โ3๐๐๐|4|+3/2 ๐/4 =3(๐๐๐|8|โ๐๐๐|4|)+3๐/8 =3๐๐๐|8/4|+3๐/8 =๐ ๐ฅ๐จ๐ โกใ๐+๐๐/๐ใ (As log A โ log B = log A/B )

Definite Integration - By Formulae