Ex 7.9, 19 - Class 12

Transcript

Ex7.9, 19 0﷮2﷮ 6𝑥 + 3﷮ 𝑥﷮2﷯ + 4﷯﷯ 𝑑𝑥 Step 1 :- Let F 𝑥﷯= ﷮﷮ 6𝑥 + 3﷮ 𝑥﷮2﷯ + 4﷯ 𝑑𝑥﷯ = ﷮﷮ 6𝑥﷮ 𝑥﷮2﷯ + 4﷯ 𝑑𝑥+ ﷮﷮ 3﷮ 𝑥﷮2﷯ + 4﷯ 𝑑𝑥 ﷯﷯ Solving 𝑰𝟏 𝐼1= ﷮﷮ 6𝑥﷮ 𝑥﷮2﷯ + 4﷯ 𝑑𝑥﷯ Put 𝑥﷮2﷯ + 4=𝑡 Differentiating w.r.t.𝑥 𝑑﷮𝑑𝑥﷯ 𝑥﷮2﷯+4﷯= 𝑑𝑡﷮𝑑𝑥﷯ 2𝑥= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= 𝑑𝑡﷮2𝑥﷯ Therefore, ﷮﷮ 6𝑥﷮ 𝑥﷮2﷯ + 4﷯ 𝑑𝑥= ﷮﷮ 6𝑥﷮𝑡﷯ 𝑑𝑡﷮2𝑥﷯﷯﷯ = ﷮﷮ 3﷮𝑡﷯ 𝑑𝑡﷯ =3 𝑙𝑜𝑔 𝑡﷯ =3 𝑙𝑜𝑔 𝑥﷮2﷯+4﷯ Solving 𝑰𝟐 𝐼2= ﷮﷮ 3﷮ 𝑥﷮2﷯ + 4﷯ 𝑑𝑥﷯ =3 ﷮﷮ 1﷮ 𝑥﷮2﷯+4﷯﷯ 𝑑𝑥 =3 ﷮﷮ 1﷮ 𝑥﷮2﷯ + 2﷮2﷯﷯﷯ 𝑑𝑥 =3 × 1﷮2﷯ tan﷮−1﷯﷮ 𝑥﷮2﷯﷯ = 3﷮2﷯ tan﷮−1﷯﷮ 𝑥﷮2﷯﷯ Therefore F 𝑥﷯= 𝐼1+𝐼2 F 𝑥﷯=3𝑙𝑜𝑔 𝑥﷮2﷯+4﷯+ 3﷮2﷯ tan﷮−1﷯﷮ 𝑥﷮2﷯﷯ Step 2 :- 0﷮2﷮ 6𝑥 + 3﷮ 𝑥﷮2﷯ + 4﷯ 𝑑𝑥=𝐹 2﷯−𝐹 0﷯﷯ =3𝑙𝑜𝑔 2﷮2﷯+4﷯+ 3﷮2﷯ tan﷮−1﷯﷮ 2﷮2﷯−3𝑙𝑜𝑔 0+4﷯− 3﷮2﷯ tan﷮−1﷯﷮ 0﷮2﷯﷯﷯﷯ =3𝑙𝑜𝑔 4+4﷯+ 3﷮2﷯ tan﷮−1﷯﷮1−3𝑙𝑜𝑔 4﷯− 3﷮2﷯ × 0﷯ =3𝑙𝑜𝑔 8﷯−3𝑙𝑜𝑔 4﷯+ 3﷮2﷯ 𝜋﷮4﷯ =3 𝑙𝑜𝑔 8﷯−𝑙𝑜𝑔 4﷯﷯+ 3𝜋﷮8﷯ =3𝑙𝑜𝑔 8﷮4﷯﷯+ 3𝜋﷮8﷯ =𝟑 𝐥𝐨𝐠﷮𝟐+ 𝟑𝝅﷮𝟖﷯﷯