Definite Integration - By Substitution

Chapter 7 Class 12 Integrals
Concept wise

### Transcript

Example 26 (Method 1) Evaluate โซ_(โ1)^1โใ5๐ฅ^4 โ(๐ฅ^5+1)ใ ๐๐ฅ Step 1 :- Let F(๐ฅ)=โซ1โใ5๐ฅ^4 โ(๐ฅ^5+1)ใ ๐๐ฅ Putting ๐ก=๐ฅ^5+1 Differentiating w.r.t.๐ฅ ๐๐ก/๐๐ฅ=5๐ฅ^4 ๐๐ก/(5๐ฅ^4 )=๐๐ฅ Therefore we can write โซ1โใ5๐ฅ^4 โ(๐ฅ^5+1) ๐๐ฅ=โซ1โใ5๐ฅ^4 โ๐ก . ๐๐ก/(5๐ฅ^4 )ใใ =โซ1โโ๐ก ๐๐ก =โซ1โใ๐ก^(1/2) ๐๐กใ =ใ๐ก ใ^(1/2 +1)/(1/2 +1) =2/3 ๐ก^(3/2) Putting back ๐ก=๐ฅ^5+1 =2/3 (๐ฅ^5+1)^(3/2) Hence , F(๐ฅ)=2/3 (๐ฅ^5+1)^(3/2) Step 2 :- โซ_(โ1)^1โใ5๐ฅ^4 ใ โ(๐ฅ^5+1) ๐๐ฅ=๐น(1)โ๐น(โ1) =2/3 (1^5+1)^(3/2)โ2/3 ((โ1)^5+1)^(3/2) =2/3 (1+1)^(3/2)โ2/3 (โ1+1)^(3/2) =2/3 (2)^(3/2)โ0 =2/3 2โ2 =(๐โ๐)/๐ Example 26 (Method 2) Evaluate โซ_(โ1)^1โใ5๐ฅ^4 โ(๐ฅ^5+1)ใ ๐๐ฅ Put ๐ก=๐ฅ^5+1 Differentiating w.r.t. ๐ฅ ๐๐ก/๐๐ฅ=๐/๐๐ฅ (๐ฅ^5+1) ๐๐ก/๐๐ฅ=5๐ฅ^4 ๐๐ก/(5๐ฅ^4 )=๐๐ฅ Hence when ๐ฅ varies from ๐ฅ=โ1 to 1, ๐ก varies from 0 to 2 Therefore, โซ_(โ1)^1โใ5๐ฅ^4 โ(1+๐ฅ^5 ) ๐๐ฅ=โซ_0^2โใ5๐ฅ^4 โ๐ก ๐๐ก/(5๐ฅ^4 )ใใ =โซ1_0^2โใโ๐ก ๐๐กใ =[๐ก^(1/2 + 1)/(1/2 +1)]_0^2 =[๐ก^(3/2)/(3/2)]_0^2 =[2/3 ๐ก^(3/2) ]_0^2 =2/3 (2^(3/2)โ0^(3/2) ) =2/3 2^(3/2) =2/3 ร2โ2 =๐/๐ โ๐