Definite Integration - By Substitution

Chapter 7 Class 12 Integrals
Concept wise

### Transcript

Example 28(Method 1) Evaluate ﷐−1﷮1﷮5﷐𝑥﷮4﷯﷐﷮﷐𝑥﷮5﷯+1﷯﷯ 𝑑𝑥 Step 1 :- Let F﷐𝑥﷯=﷐﷮﷮5﷐𝑥﷮4﷯﷐﷮﷐𝑥﷮5﷯+1﷯﷯𝑑𝑥 Putting 𝑡=﷐𝑥﷮5﷯+1 Differentiating w.r.t.𝑥 ﷐𝑑𝑡﷮𝑑𝑥﷯=5﷐𝑥﷮4﷯ ﷐𝑑𝑡﷮5﷐𝑥﷮4﷯﷯=𝑑𝑥 Therefore we can write ﷐﷮﷮5﷐𝑥﷮4﷯﷐﷮﷐𝑥﷮5﷯+1﷯ 𝑑𝑥=﷐﷮﷮5﷐𝑥﷮4﷯ ﷐﷮𝑡﷯ . ﷐𝑑𝑡﷮5﷐𝑥﷮4﷯﷯﷯﷯ =﷐﷮﷮﷐﷮𝑡﷯﷯ 𝑑𝑡 =﷐﷮﷮﷐𝑡﷮﷐1﷮2﷯﷯ 𝑑𝑡﷯ =﷐﷐𝑡 ﷮﷐1﷮2﷯ +1﷯﷮﷐1﷮2﷯ +1﷯ =﷐2﷮3﷯﷐𝑡﷮﷐3﷮2﷯﷯ Putting back 𝑡=﷐𝑥﷮5﷯+1 =﷐2﷮3﷯﷐﷐﷐𝑥﷮5﷯+1﷯﷮﷐3﷮2﷯﷯ Hence , F﷐𝑥﷯=﷐2﷮3﷯﷐﷐﷐𝑥﷮5﷯+1﷯﷮﷐3﷮2﷯﷯ Step 2 :- ﷐−1﷮1﷮5﷐𝑥﷮4﷯﷯﷐﷮﷐𝑥﷮5﷯+1﷯ 𝑑𝑥=𝐹﷐1﷯−𝐹﷐−1﷯ =﷐2﷮3﷯﷐﷐﷐1﷮5﷯+1﷯﷮﷐3﷮2﷯﷯−﷐2﷮3﷯﷐﷐﷐﷐−1﷯﷮5﷯+1﷯﷮﷐3﷮2﷯﷯ =﷐2﷮3﷯﷐﷐1+1﷯﷮﷐3﷮2﷯﷯−﷐2﷮3﷯﷐﷐−1+1﷯﷮﷐3﷮2﷯﷯ =﷐2﷮3﷯﷐﷐2﷯﷮﷐3﷮2﷯﷯−0 =﷐2﷮3﷯ 2﷐﷮2﷯ =﷐𝟒﷐﷮𝟐﷯﷮𝟑﷯ Example 28 (Method 2) Evaluate ﷐−1﷮1﷮5﷐𝑥﷮4﷯﷐﷮﷐𝑥﷮5﷯+1﷯﷯ 𝑑𝑥 Put 𝑡=﷐𝑥﷮5﷯+1 Differentiating w.r.t. 𝑥 ﷐𝑑𝑡﷮𝑑𝑥﷯=﷐𝑑﷮𝑑𝑥﷯﷐﷐𝑥﷮5﷯+1﷯ ﷐𝑑𝑡﷮𝑑𝑥﷯=5﷐𝑥﷮4﷯ ﷐𝑑𝑡﷮5﷐𝑥﷮4﷯﷯=𝑑𝑥 Hence when 𝑥 varies from 𝑥=−1 to 1, 𝑡 varies from 0 to 2 Therefore, ﷐−1﷮1﷮5﷐𝑥﷮4﷯﷐﷮1+﷐𝑥﷮5﷯﷯ 𝑑𝑥=﷐0﷮2﷮5﷐𝑥﷮4﷯ ﷐﷮𝑡﷯ ﷐𝑑𝑡﷮5﷐𝑥﷮4﷯﷯﷯﷯ =﷐0﷮2﷮﷐﷮𝑡﷯ 𝑑𝑡﷯ =﷐﷐﷐﷐𝑡﷮﷐1﷮2﷯ + 1﷯﷮﷐1﷮2﷯ +1﷯﷯﷮0﷮2﷯ =﷐﷐﷐﷐𝑡﷮﷐3﷮2﷯﷯﷮﷐3﷮2﷯﷯﷯﷮0﷮2﷯ =﷐﷐﷐2﷮3﷯﷐𝑡﷮﷐3﷮2﷯﷯﷯﷮0﷮2﷯ =﷐2﷮3﷯﷐﷐2﷮﷐3﷮2﷯﷯−﷐0﷮﷐3﷮2﷯﷯﷯ =﷐2﷮3﷯ ﷐2﷮﷐3﷮2﷯﷯ =﷐2﷮3﷯ ×2﷐﷮2﷯ =﷐𝟒﷮𝟑﷯ ﷐﷮𝟐﷯

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.