Example 28 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at May 29, 2018 by Teachoo
Last updated at May 29, 2018 by Teachoo
Transcript
Example 28(Method 1) Evaluate −115𝑥4𝑥5+1 𝑑𝑥 Step 1 :- Let F𝑥=5𝑥4𝑥5+1𝑑𝑥 Putting 𝑡=𝑥5+1 Differentiating w.r.t.𝑥 𝑑𝑡𝑑𝑥=5𝑥4 𝑑𝑡5𝑥4=𝑑𝑥 Therefore we can write 5𝑥4𝑥5+1 𝑑𝑥=5𝑥4 𝑡 . 𝑑𝑡5𝑥4 =𝑡 𝑑𝑡 =𝑡12 𝑑𝑡 =𝑡 12 +112 +1 =23𝑡32 Putting back 𝑡=𝑥5+1 =23𝑥5+132 Hence , F𝑥=23𝑥5+132 Step 2 :- −115𝑥4𝑥5+1 𝑑𝑥=𝐹1−𝐹−1 =2315+132−23−15+132 =231+132−23−1+132 =23232−0 =23 22 =𝟒𝟐𝟑 Example 28 (Method 2) Evaluate −115𝑥4𝑥5+1 𝑑𝑥 Put 𝑡=𝑥5+1 Differentiating w.r.t. 𝑥 𝑑𝑡𝑑𝑥=𝑑𝑑𝑥𝑥5+1 𝑑𝑡𝑑𝑥=5𝑥4 𝑑𝑡5𝑥4=𝑑𝑥 Hence when 𝑥 varies from 𝑥=−1 to 1, 𝑡 varies from 0 to 2 Therefore, −115𝑥41+𝑥5 𝑑𝑥=025𝑥4 𝑡 𝑑𝑡5𝑥4 =02𝑡 𝑑𝑡 =𝑡12 + 112 +102 =𝑡323202 =23𝑡3202 =23232−032 =23 232 =23 ×22 =𝟒𝟑 𝟐
Definite Integration - By Substitution
Definite Integration - By Substitution
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