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Example 26 - Chapter 7 Class 12 Integrals
Last updated at August 2, 2026 by Teachoo
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Example 1 (i)
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Chapter 7 Class 12 Integrals
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Transcript
Example 26 (Method 1) Evaluate ā«_(ā1)^1āć5š„^4 ā(š„^5+1)ć šš„ Step 1 :- Let F(š„)=ā«1āć5š„^4 ā(š„^5+1)ć šš„ Putting š”=š„^5+1 Differentiating w.r.t.š„ šš”/šš„=5š„^4 šš”/(5š„^4 )=šš„ Therefore we can write ā«1āć5š„^4 ā(š„^5+1) šš„=ā«1āć5š„^4 āš” . šš”/(5š„^4 )ćć =ā«1āāš” šš” =ā«1āćš”^(1/2) šš”ć =ćš” ć^(1/2 +1)/(1/2 +1) =2/3 š”^(3/2) Putting back š”=š„^5+1 =2/3 (š„^5+1)^(3/2) Hence , F(š„)=2/3 (š„^5+1)^(3/2) Step 2 :- ā«_(ā1)^1āć5š„^4 ć ā(š„^5+1) šš„=š¹(1)āš¹(ā1) =2/3 (1^5+1)^(3/2)ā2/3 ((ā1)^5+1)^(3/2) =2/3 (1+1)^(3/2)ā2/3 (ā1+1)^(3/2) =2/3 (2)^(3/2)ā0 =2/3 2ā2 =(šāš)/š Example 26 (Method 2) Evaluate ā«_(ā1)^1āć5š„^4 ā(š„^5+1)ć šš„ Put š”=š„^5+1 Differentiating w.r.t. š„ šš”/šš„=š/šš„ (š„^5+1) šš”/šš„=5š„^4 šš”/(5š„^4 )=šš„ Hence when š„ varies from š„=ā1 to 1, š” varies from 0 to 2 Therefore, ā«_(ā1)^1āć5š„^4 ā(1+š„^5 ) šš„=ā«_0^2āć5š„^4 āš” šš”/(5š„^4 )ćć =ā«1_0^2āćāš” šš”ć =[š”^(1/2 + 1)/(1/2 +1)]_0^2 =[š”^(3/2)/(3/2)]_0^2 =[2/3 š”^(3/2) ]_0^2 =2/3 (2^(3/2)ā0^(3/2) ) =2/3 2^(3/2) =2/3 Ć2ā2 =š/š āš
