Example 28 - Chapter 7 Class 12 Integrals

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﷯