Examples

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Example 26 (Method 1) Evaluate ∫_(−1)^1▒〖5𝑥^4 √(𝑥^5+1)〗 𝑑𝑥 Step 1 :- Let F(𝑥)=∫1▒〖5𝑥^4 √(𝑥^5+1)〗 𝑑𝑥 Putting 𝑡=𝑥^5+1 Differentiating w.r.t.𝑥 𝑑𝑡/𝑑𝑥=5𝑥^4 𝑑𝑡/(5𝑥^4 )=𝑑𝑥 Therefore we can write ∫1▒〖5𝑥^4 √(𝑥^5+1) 𝑑𝑥=∫1▒〖5𝑥^4 √𝑡 . 𝑑𝑡/(5𝑥^4 )〗〗 =∫1▒√𝑡 𝑑𝑡 =∫1▒〖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▒〖5𝑥^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 26 (Method 2) Evaluate ∫_(−1)^1▒〖5𝑥^4 √(𝑥^5+1)〗 𝑑𝑥 Put 𝑡=𝑥^5+1 Differentiating w.r.t. 𝑥 𝑑𝑡/𝑑𝑥=𝑑/𝑑𝑥 (𝑥^5+1) 𝑑𝑡/𝑑𝑥=5𝑥^4 𝑑𝑡/(5𝑥^4 )=𝑑𝑥 Hence when 𝑥 varies from 𝑥=−1 to 1, 𝑡 varies from 0 to 2 Therefore, ∫_(−1)^1▒〖5𝑥^4 √(1+𝑥^5 ) 𝑑𝑥=∫_0^2▒〖5𝑥^4 √𝑡 𝑑𝑡/(5𝑥^4 )〗〗 =∫1_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

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.