Ex 7.9, 6 - Chapter 7 Class 12 Integrals

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Ex 7.9, 6 Evaluate the integrals using substitution ∫_0^(2 )▒𝑑𝑥/(𝑥 + 4 − 𝑥^2 ) We can write ∫_0^2▒〖𝑑𝑥/(𝑥 + 4 − 𝑥^2 )=∫_0^2▒𝑑𝑥/(−(𝑥^2 − 𝑥 − 4) )〗 =−∫_0^2▒𝑑𝑥/(𝑥^2 − 𝑥 − 4) =−∫_0^2▒𝑑𝑥/(𝑥^2 −2 × 1/2 × 𝑥 − 4) =−∫_0^2▒𝑑𝑥/(𝑥^2 −2 × 1/2 × 𝑥 + 1/2^2 − 1/2^2 − 4) =−∫_0^2▒𝑑𝑥/((𝑥 − 1/2)^2− 1/4 − 4) =−∫_0^2▒𝑑𝑥/((𝑥 − 1/2)^2− 17/4 ) =−∫_0^2▒𝑑𝑥/((𝑥 − 1/2)^2− (√17/4)^2 ) Let 𝑡=𝑥−1/2 Differentiating w.r.t.𝑥 𝑑𝑡/𝑑𝑥=1 𝑑𝑡=𝑑𝑥 When x varies from 0 to 2, then t varies from (−1)/2 to 3/2. Therefore, −∫_0^2▒〖𝑑𝑥/((𝑥 − 1/2)^2−(√17/2)^2 )=−∫_((−1)/2)^(3/2)▒𝑑𝑡/(𝑡 − (√17/2)^2 )〗 =−[1/2(√17/2) 𝑙𝑜𝑔|(𝑡 − √17/2)/(𝑡 + √17/2)|]_((−1)/( 2))^(3/2) =−1/√17 [𝑙𝑜𝑔|(3/2 − √17/2)/(3/2 + √17/2)|+𝑙𝑜𝑔|((−1)/( 2) − √17/2)/((−1)/( 2) + √17/2)|] =−1/√17 [𝑙𝑜𝑔|(3 − √17)/(3 + √17)|+𝑙𝑜𝑔|(−(1 + √17))/(−(1 − √17) )|] =−1/√17 𝑙𝑜𝑔|((3 − √17)/(3 + √17))/((1 + √17)/(1 − √17))| =−1/√17 𝑙𝑜𝑔|(3 − √17)/(3 + √17) ×(1 − √17)/(1 + √17)| =−1/√17 𝑙𝑜𝑔|(3+17 − 3√17 − √17)/(3 +17 + 3√17 + √17) | =−1/√17 𝑙𝑜𝑔|(20 − 4√17)/(20 + 4√17) | =−1/√17 𝑙𝑜𝑔|4(5 − √17)/4(5 + √17) | =−1/√17 𝑙𝑜𝑔|(5 − √17)/(5 + √17) | =1/√17 𝑙𝑜𝑔|(5 − √17)/(5 + √17) |^(−1) =1/√17 𝑙𝑜𝑔|(5 + √17)/(5 − √17)| =1/√17 𝑙𝑜𝑔|(5 + √17)/(5 − √17) ×(5 + √17)/(5 + √17)| =1/√17 𝑙𝑜𝑔|(5 − √17)^2/(5^2 − (√17)^2 ) | =1/√17 𝑙𝑜𝑔|(25 + 17 + 10√17)/(25 − 17) | =1/√17 𝑙𝑜𝑔|(42 + 10√17)/8 | =𝟏/√𝟏𝟕 𝒍𝒐𝒈|(𝟐𝟏 + 𝟓√𝟏𝟕)/𝟒 |