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Ex 7.10, 8 - Evaluate integral (1/x - 1/ 2x2) e2x dx - Definate Integration - By Formulae

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex7.10, 8 (Method 1) Evaluate the integrals using substitution 1﷮2 ﷮ 1﷮𝑥﷯ − 1﷮2 𝑥﷮2﷯﷯﷯﷯ 𝑒﷮2𝑥﷯ 𝑑𝑥 Let 𝑡=2𝑥 𝑑𝑡﷮𝑑𝑥﷯=2 𝑑𝑡﷮2﷯=𝑑𝑥 Thus, when x varies from 1 to 2, t varies from 2 to 4 Substituting, = 2﷮4﷮ 1﷮ 𝑡﷮2﷯﷯− 1﷮2 𝑡﷮2﷯﷯﷮2﷯﷯﷯﷯ 𝑑𝑡﷮2﷯ = 2﷮4﷮ 2﷮𝑡﷯− 4﷮2 𝑡﷮2﷯﷯﷯﷯ 𝑑𝑡﷮2﷯ = 2﷮4﷮ 1﷮𝑡﷯− 2﷮ 𝑡﷮2﷯﷯﷯﷯𝑑𝑡 Now, Let f(x) = 1﷮𝑡﷯ f’(x) = −1﷮ 𝑡﷮2﷯﷯ and we know that ﷮﷮ 𝑒﷮𝑥﷯ 𝑓 𝑥﷯+ 𝑓﷮′﷯ 𝑥﷯﷯𝑑𝑥= 𝑒﷮𝑥﷯𝑓(𝑥)﷯ Hence, our equation becomes = 𝑒﷮𝑡﷯× 1﷮𝑡﷯﷯﷮4﷯﷮2﷯ = 𝑒﷮4﷯﷮4﷯− 𝑒﷮2﷯﷮2﷯﷯ = 𝒆﷮𝟐﷯( 𝒆﷮𝟐﷯ − 𝟐)﷮𝟒﷯ Ex7.10, 8 (Method 2) Evaluate the integrals using substitution 1﷮2 ﷮ 1﷮𝑥﷯ − 1﷮2 𝑥﷮2﷯﷯﷯﷯ 𝑒﷮2𝑥﷯ 𝑑𝑥 1﷮2 ﷮ 1﷮𝑥﷯ − 1﷮2 𝑥﷮2﷯﷯﷯﷯ 𝑒﷮2𝑥﷯ 𝑑𝑥 = 1﷮2﷮ 𝑒﷮2𝑥﷯﷮𝑥﷯﷯𝑑𝑥− 1﷮2﷯ ﷮﷮ 𝑒﷮2𝑥﷯﷮ 𝑥﷮2﷯﷯﷯𝑑𝑥 Solving 𝑰﷮𝟏﷯ Let 𝐼﷮1﷯= 1﷮2﷮ 𝑒﷮2𝑥﷯﷮𝑥﷯﷯𝑑𝑥 𝐼﷮1﷯ = 1﷮𝑥﷯ ﷮﷮ 𝑒﷮2𝑥﷯𝑑𝑥− ﷮﷮ 𝑑﷮𝑑𝑥﷯ 1﷮𝑥﷯﷯ ﷮﷮ 𝑒﷮2𝑥﷯𝑑𝑥﷯﷯𝑑𝑥﷯﷯﷯﷮2﷯﷮1﷯ 𝐼﷮1﷯ = 1﷮𝑥﷯ 𝑒﷮2𝑥﷯﷮2﷯− ﷮﷮ −1﷮ 𝑥﷮2﷯﷯ 𝑒﷮2𝑥﷯﷮2﷯﷯𝑑𝑥﷯﷮1﷮2﷯ 𝐼﷮1﷯ = 1﷮2𝑥﷯ 𝑒﷮2𝑥﷯﷯﷮1﷮2﷯+ 1﷮2﷯ ﷮﷮ 𝑒﷮2𝑥﷯﷮ 𝑥﷮2﷯﷯ 𝑑𝑥﷯ 𝐼﷮1﷯ = 𝑒﷮2 × 2﷯﷮2 × 2﷯− 𝑒﷮2 × 1﷯﷮2 × 1﷯+ 1﷮2﷯ ﷮﷮ 𝑒﷮2𝑥﷯﷮ 𝑥﷮2﷯﷯ 𝑑𝑥﷯ 𝐼﷮1﷯ = 𝑒﷮4﷯﷮4﷯− 𝑒﷮2﷯﷮𝑒﷯+ 1﷮2﷯ ﷮﷮ 𝑒﷮2𝑥﷯﷮ 𝑥﷮2﷯﷯﷯ Putting value of 𝐼﷮1﷯ in (1) ﷮﷮ 1﷮𝑥﷯− 1﷮ 2𝑥﷮2﷯﷯﷯﷯ 𝑒﷮2𝑥﷯ 𝑑𝑥= 𝑒﷮4﷯﷮4﷯− 𝑒﷮2﷯﷮2﷯+ 1﷮2﷯ 1﷮2﷮ 𝑒﷮2𝑥﷯﷮ 𝑥﷮2﷯﷯﷯ 𝑑𝑥− 1﷮2﷯ 1﷮2﷮ 𝑒﷮2𝑥﷯﷮ 𝑥﷮2﷯﷯﷯ 𝑑𝑥 = 𝑒﷮4﷯﷮4﷯− 𝑒﷮2﷯﷮2﷯ = 𝒆﷮𝟐﷯﷮𝟒﷯ ( 𝒆﷮𝟐﷯−𝟐)

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