Ex 7.10, 7 - Chapter 7 Class 12 Integrals

Transcript

Ex7.10, 7 Evaluate the integrals using substitution −1﷮1 ﷮ 𝑑𝑥﷮ 𝑥﷮2﷯ + 2𝑥 + 5﷯﷯ we can write −1﷮1﷮ 𝑑𝑥﷮ 𝑥﷮2﷯ + 2𝑥 + 5﷯= −1﷮1﷮ 𝑑𝑥﷮ 𝑥 + 2𝑥 + 1﷯ + 4﷯﷯﷯ = −1﷮1﷮ 𝑑𝑥﷮ 𝑥 + 1﷯﷮2﷯ + 2﷮2﷯﷯﷯ Putting 𝑥+1=𝑡 Differentiating w.r.t.𝑥 𝑑﷮𝑑𝑥﷯ 𝑥+1﷯= 𝑑𝑡﷮𝑑𝑥﷯ 1= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥=𝑑𝑡 Hence when 𝑥 varies from – 1 to 1 then 𝑡 varies from 0 to 2 Therefore, −1﷮1﷮ 𝑑𝑥﷮ 𝑥+1﷯﷮2﷯ + 2﷮2﷯﷯= 0﷮2﷮ 𝑑𝑡﷮ 𝑡﷮2﷯ + 2﷮2﷯﷯﷯﷯ = 1﷮2﷯ tan﷮−1﷯﷮ 𝑡﷮2﷯﷯﷯﷮0﷮2﷯ = 1﷮2﷯ tan﷮−1﷯﷮ 2﷮2﷯− 1﷮2﷯ tan﷮−1﷯﷮ 0﷮2﷯﷯﷯ = 1﷮2﷯ tan﷮−1﷯﷮1− 1﷮2﷯ tan﷮−1﷯﷮0﷯﷯ = 1﷮2﷯ × 𝜋﷮4﷯−0 = 𝝅﷮𝟖﷯