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Ex 7.11, 17 - Evaluate root x / root x + root a - x dx - Ex 7.11

Ex 7.11, 17 - Chapter 7 Class 12 Integrals - Part 2

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Ex 7.11, 17 By using the properties of definite integrals, evaluate the integrals : ∫_0^π‘Žβ–’βˆšπ‘₯/(√π‘₯ + √(π‘Ž βˆ’ π‘₯)) 𝑑π‘₯ Let I=∫_0^π‘Žβ–’βˆšπ‘₯/(√π‘₯ + √(π‘Ž βˆ’ π‘₯)) 𝑑π‘₯ ∴ I=∫_0^π‘Žβ–’γ€–βˆš(π‘Ž βˆ’ π‘₯)/(√(π‘Ž βˆ’ π‘₯) +√(π‘Ž βˆ’ (π‘Ž βˆ’ π‘₯) )) 𝑑π‘₯γ€— I= ∫_0^π‘Žβ–’γ€–βˆš(π‘Ž βˆ’ π‘₯)/(√(π‘Ž βˆ’ π‘₯) + √(π‘Ž βˆ’ π‘Ž + π‘₯)) 𝑑π‘₯γ€— I=∫_0^π‘Žβ–’γ€–βˆš(π‘Ž βˆ’ π‘₯)/(√(π‘Ž βˆ’ π‘₯) + √π‘₯) 𝑑π‘₯γ€— Adding (1) and (2) i.e (1) + (2) I + I = ∫_0^π‘Žβ–’γ€–βˆšπ‘₯/(√π‘₯ +√(π‘Ž βˆ’ π‘₯)) 𝑑π‘₯γ€—+∫_0^π‘Žβ–’γ€–βˆš(π‘Ž βˆ’ π‘₯)/(√(π‘Žβˆ’ π‘₯) + √( π‘₯)) 𝑑π‘₯γ€— 2I = ∫_0^π‘Žβ–’γ€–(√π‘₯ + √(π‘Ž βˆ’ π‘₯))/(√π‘₯ + √(π‘Ž βˆ’ π‘₯)) 𝑑π‘₯γ€— 2I =∫_0^π‘Žβ–’γ€–1.𝑑π‘₯γ€— 2I = [π‘₯]_0^π‘Ž 2I = [π‘Žβˆ’0] 𝐈 =𝒂/𝟐

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