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Example 28 - Evaluate integral 5x4 root x5 + 1 dx - Examples

Example 28 - Chapter 7 Class 12 Integrals - Part 2
Example 28 - Chapter 7 Class 12 Integrals - Part 3 Example 28 - Chapter 7 Class 12 Integrals - Part 4 Example 28 - Chapter 7 Class 12 Integrals - Part 5 Example 28 - Chapter 7 Class 12 Integrals - Part 6

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Transcript

Example 28(Method 1) Evaluate ﷐−1﷮1﷮5﷐𝑥﷮4﷯﷐﷮﷐𝑥﷮5﷯+1﷯﷯ 𝑑𝑥 Step 1 :- Let F﷐𝑥﷯=﷐﷮﷮5﷐𝑥﷮4﷯﷐﷮﷐𝑥﷮5﷯+1﷯﷯𝑑𝑥 Putting 𝑡=﷐𝑥﷮5﷯+1 Differentiating w.r.t.𝑥 ﷐𝑑𝑡﷮𝑑𝑥﷯=5﷐𝑥﷮4﷯ ﷐𝑑𝑡﷮5﷐𝑥﷮4﷯﷯=𝑑𝑥 Therefore we can write ﷐﷮﷮5﷐𝑥﷮4﷯﷐﷮﷐𝑥﷮5﷯+1﷯ 𝑑𝑥=﷐﷮﷮5﷐𝑥﷮4﷯ ﷐﷮𝑡﷯ . ﷐𝑑𝑡﷮5﷐𝑥﷮4﷯﷯﷯﷯ =﷐﷮﷮﷐﷮𝑡﷯﷯ 𝑑𝑡 =﷐﷮﷮﷐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﷮5﷐𝑥﷮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 28 (Method 2) Evaluate ﷐−1﷮1﷮5﷐𝑥﷮4﷯﷐﷮﷐𝑥﷮5﷯+1﷯﷯ 𝑑𝑥 Put 𝑡=﷐𝑥﷮5﷯+1 Differentiating w.r.t. 𝑥 ﷐𝑑𝑡﷮𝑑𝑥﷯=﷐𝑑﷮𝑑𝑥﷯﷐﷐𝑥﷮5﷯+1﷯ ﷐𝑑𝑡﷮𝑑𝑥﷯=5﷐𝑥﷮4﷯ ﷐𝑑𝑡﷮5﷐𝑥﷮4﷯﷯=𝑑𝑥 Hence when 𝑥 varies from 𝑥=−1 to 1, 𝑡 varies from 0 to 2 Therefore, ﷐−1﷮1﷮5﷐𝑥﷮4﷯﷐﷮1+﷐𝑥﷮5﷯﷯ 𝑑𝑥=﷐0﷮2﷮5﷐𝑥﷮4﷯ ﷐﷮𝑡﷯ ﷐𝑑𝑡﷮5﷐𝑥﷮4﷯﷯﷯﷯ =﷐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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.