Example 28 - Chapter 7 Class 12 Integrals
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
Examples
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
Example 8
Example 9 Important
Example 10 Important
Example 11
Example 12
Example 13
Example 14
Example 15 Important
Example 16
Example 17
Example 18
Example 19
Example 20 Important
Example 21
Example 22 Important
Example 23
Example 24
Example 25 Important
Example 26
Example 27 Important
Example 28 You are here
Example 29
Example 30 Important
Example 31
Example 32
Example 33
Example 34 Important
Example 35
Example 36 Important
Example 37
Example 38 Important
Example 39
Example 40 Important
Example 41 Important
Example 42
Example 43 Important
Example 44 Important
Example 25 (Supplementary NCERT)
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.