Example 28 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at May 29, 2018 by
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
Examples
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Example 1 (ii)
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Example 2 (i)
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Example 25 Important Deleted for CBSE Board 2022 Exams
Example 26 Deleted for CBSE Board 2022 Exams
Example 27 (i)
Example 27 (ii) Important
Example 27 (iii)
Example 27 (iv) Important
Example 28 You are here
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Example 41 Important
Example 42 Important
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Example 44 Important
Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams
Chapter 7 Class 12 Integrals (Term 2)
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.
