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Last updated at March 16, 2023 by Teachoo
Ex 7.10, 4 Evaluate the integrals using substitution 02 𝑥 𝑥+2 𝑝𝑢𝑡 𝑥+2= 𝑡2 02 𝑥 𝑥+2 𝑑𝑥 Put 𝑥+2= 𝑡2 Differentiating w.r.t. 𝑥 𝑑 𝑥 + 2𝑑𝑥= 𝑑 𝑡2𝑑𝑡 × 𝑑𝑡𝑑𝑥 1=2𝑡 × 𝑑𝑡𝑑𝑥 𝑑𝑥=2𝑡 𝑑𝑡 Hence, when 𝑥 varies from 0 to 2, then t varies from 2 to 2 Therefore we can write 02𝑥 𝑥+2 𝑑𝑥 = 22 𝑡2−2 𝑡2 2𝑡 𝑑𝑡 = 22 𝑡2−2𝑡 ×2𝑡 𝑑𝑡 =2 22 𝑡2−2 𝑡2 𝑑𝑡 =2 22 𝑡4−2 𝑡2 𝑑𝑡 =2 𝑡4+14+1−2 𝑡2+12+1 22 =2 𝑡55−2 𝑡33 22 =2× 255− 23 23− 255− 23 23 =2× 𝑥52− 23 23− 252− 23 23 =2× 325− 163− 4 25+ 43 2 =2× 96 − 80 − 12 2 + 20 215 =2× 16 + 8 215 = 2 × 8 2 2 + 115 = 𝟏𝟔 𝟐 𝟐 + 𝟏𝟏𝟓