1) findf(a)= ∫_0 ^1 (√(x^2 −x+a))dx with a>(1/2) 2)explicite g(a) =∫_0 ^1 (dx/(√(x^2 −x+a))) 3) calculate ∫


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Question Number 97800 by abdomathmax last updated on 09/Jun/20

$1) findf (a) = \int_{0}^{1} \sqrt{x^{2} - x + a} dx with a > \frac{1}{2}$ $2) explicite g (a) = \int_{0}^{1} \frac{dx}{\sqrt{x^{2} - x + a}}$ $3) calculate \int_{0}^{1} \frac{dx}{\sqrt{x^{2} - x + 3}}$
	Answered by niroj last updated on 09/Jun/20
	$(3).I= ∫_0 ^1 (dx/(√(x^2 −x+3))) = ∫_0 ^( 1) (1/(√(x^2 −2.x.(1/2)+(1/4)−(1/4)+3)))dx = ∫_0 ^1 (1/(√((x−(1/2))^2 −(((1−12)/4)))))dx = ∫_0 ^1 (1/(√((x−(1/2))^2 +((( (√(11)))/2))^2 )))dx Put ,x−(1/2)=t dx=dt if x=1, t=(1/2) if x= 0, t=−(1/2) = ∫_(−(1/2)) ^(1/2) (( 1)/(√((t)^2 +(((√(11))/2))^2 )))dt = [log (t+(√(t^2 +((11)/4))) )]^(1/2) _(−(1/2)) = {log ((1/2)+(√((1/4)+((11)/4))) )−log(−(1/2)+(√((1/4)+((11)/4))) )} = log(((1/2)+((2(√3))/2))/(−(1/2)+((2(√3))/2)))= log (( (√3)+ (1/2))/( (√3)− (1/2))) = log ((2(√3) +1)/( 2(√3) −1)) //.$
	$(3) . I = \int_{0}^{1} \frac{dx}{\sqrt{x^{2} - x + 3}}$ $= \int_{0}^{1} \frac{1}{\sqrt{x^{2} - 2 . x . \frac{1}{2} + \frac{1}{4} - \frac{1}{4} + 3}} dx$ $= \int_{0}^{1} \frac{1}{\sqrt{{(x - \frac{1}{2})}^{2} - (\frac{1 - 12}{4})}} dx$ $= \int_{0}^{1} \frac{1}{\sqrt{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}}} dx$ $Put, x - \frac{1}{2} = t$ $dx = dt$ $if x = 1, t = \frac{1}{2}$ $if x = 0, t = - \frac{1}{2}$ $= \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{\sqrt{{(t)}^{2} + {(\frac{\sqrt{11}}{2})}^{2}}} dt$ $Missing \left or extra \right$ $= {\log (\frac{1}{2} + \sqrt{\frac{1}{4} + \frac{11}{4}}) - \log (- \frac{1}{2} + \sqrt{\frac{1}{4} + \frac{11}{4}})}$ $= \log \frac{\frac{1}{2} + \frac{2 \sqrt{3}}{2}}{- \frac{1}{2} + \frac{2 \sqrt{3}}{2}} = \log \frac{\sqrt{3} + \frac{1}{2}}{\sqrt{3} - \frac{1}{2}}$ $= \log \frac{2 \sqrt{3} + 1}{2 \sqrt{3} - 1} / / .$
	Commented bymathmax by abdo last updated on 09/Jun/20

	$thank you sir .$
	Commented byniroj last updated on 09/Jun/20
	you welcome sir��
	Answered by 1549442205 last updated on 10/Jun/20

	$1) \int_{0}^{1} \sqrt{x^{2} - x + a} dx = \int_{0}^{1} \sqrt{{(x - \frac{1}{2})}^{2} + a - \frac{1}{4}} dx = \int_{0}^{1} \sqrt{u^{2} + m} du (with u = x - \frac{1}{2}, m = a - \frac{1}{4})$ $Putting \frac{m}{u^{2}} + 1 = v^{2} \Rightarrow \frac{- mdu}{u^{3}} = vdv . We get$ $F = \int \frac{- {vu}^{3} \times uv}{m} dv \Rightarrow \frac{F}{m} = - \int \frac{v^{2} mdv}{{(v^{2} - 1)}^{2}} = - \frac{1}{4} \int (\frac{2 v - 1}{v^{2} - 2 v + 1} + \frac{2 v + 1}{v^{2} + 2 v + 1}) dv$ $= - \frac{1}{4} \int (\frac{d (v^{2} - 2 v + 1)}{v^{2} - 2 v + 1} - \int \frac{d (v^{2} + 2 v + 1)}{v^{2} + 2 v + 1} - \frac{1}{4} \int \frac{dv}{{(v - 1)}^{2}} + \frac{1}{4} \int \frac{dv}{{(v + 1)}^{2}}$ $= \frac{- 1}{4} \ln ∣ v^{2} - 2 v + 1 ∣ - \frac{1}{4} \ln ∣ v^{2} + 2 v + 1 ∣$ $+ \frac{1}{4 (v - 1)} - \frac{1}{4 (v + 1)} = - \frac{1}{4} \ln {(v^{2} - 1)}^{2} + \frac{1}{2 (v^{2} - 1)}$ $Hence, F = \frac{- m}{2} \ln (v^{2} - 1) + \frac{m}{2 (v^{2} - 1)} = \frac{\frac{1}{4} - a}{2} \ln ∣ \frac{\frac{1}{4} - a}{{(x - \frac{1}{2})}^{2}} ∣ + \frac{a - \frac{1}{4}}{2 {(x - \frac{1}{2})}^{2}}$ $\int_{0}^{1} \sqrt{x^{2} - x + a} dx = F (1) - F (0) = 0 since F (1) = F (0)$ $2 / \int_{0}^{1} \frac{dx}{\sqrt{x^{2} - x + a}} = \int_{0}^{1} \frac{d (x - \frac{1}{2})}{\sqrt{{(x - \frac{1}{2})}^{2} + \frac{4 a - 1}{4}}} = \ln ∣ x - \frac{1}{2} + \sqrt{x^{2} - x + a} ∣_{0}^{1} =$ $= \ln ∣ \frac{1}{2} + \sqrt{a} ∣ - \ln ∣ \frac{- 1}{2} + \sqrt{a} ∣= \ln ∣ \frac{1 + 2 \sqrt{a}}{2 \sqrt{a} - 1} ∣$
	Commented bymathmax by abdo last updated on 11/Jun/20

	$thanks sir .$
	Answered by mathmax by abdo last updated on 11/Jun/20
	$1) we hsve x^2 −x+a =x^2 −2x×(1/2) +(1/4) +a−(1/4) =(x−(1/2))^2 +((4a−1)/4) so we do the changement x−(1/2) =((√(4a−1))/2)t ⇒ f(a) =∫_(−(1/(√(4a−1)))) ^(1/(√(4a−1))) ((√(4a−1))/2)(√(1+t^2 ))((√(4a−1))/2)dt =((4a−1)/2) ∫_0 ^(1/(√(4a−1))) (√(1+t^2 ))dt =_(t =shu) ∫_0 ^(argsh((1/(√(4a−1))))) chu chu du =(1/2)∫_0 ^(argsh((1/(√(4a−1))))) (1+ch(2u))du =(1/2) argsh((1/(√(4a−1)))) +(1/4) [sh(2u)]_0 ^(ln((1/(√(4a−1)))+(√(1+(1/(4a−1)))))) =(1/2)ln((1/(√(4a−1))) +(√(1+(1/(4a−1))))) +(1/8)[ e^(2u) −e^(−2u) ]_0 ^(ln((1/(√(4a−1)))+((2(√a))/(√(4a−1))))) =(1/2)ln((1/(√(4a−1))) +((2(√a))/(√(4a−1)))) +(1/8){ (((1+2(√a))/(√(4a−1))))^2 −(((√(4a−1))/(1+2(√a))))^2 } f(a)=(1/2)ln(((1+2(√a))/(√(4a−1)))) +(1/8){ (((1+2(√a))^2 )/(4a−1))−((4a−1)/((1+2(√a))^2 ))}$
	$1) we hsve x^{2} - x + a = x^{2} - 2 x \times \frac{1}{2} + \frac{1}{4} + a - \frac{1}{4} = {(x - \frac{1}{2})}^{2} + \frac{4 a - 1}{4}$ $so we do the changement x - \frac{1}{2} = \frac{\sqrt{4 a - 1}}{2} t \Rightarrow$ $f (a) = \int_{- \frac{1}{\sqrt{4 a - 1}}}^{\frac{1}{\sqrt{4 a - 1}}} \frac{\sqrt{4 a - 1}}{2} \sqrt{1 + t^{2}} \frac{\sqrt{4 a - 1}}{2} dt = \frac{4 a - 1}{2} \int_{0}^{\frac{1}{\sqrt{4 a - 1}}} \sqrt{1 + t^{2}} dt$ $=_{t = shu} \int_{0}^{argsh (\frac{1}{\sqrt{4 a - 1}})} chu chu du = \frac{1}{2} \int_{0}^{argsh (\frac{1}{\sqrt{4 a - 1}})} (1 + ch (2 u)) du$ $= \frac{1}{2} argsh (\frac{1}{\sqrt{4 a - 1}}) + \frac{1}{4} {[sh (2 u)]}_{0}^{\ln (\frac{1}{\sqrt{4 a - 1}} + \sqrt{1 + \frac{1}{4 a - 1}})}$ $= \frac{1}{2} \ln (\frac{1}{\sqrt{4 a - 1}} + \sqrt{1 + \frac{1}{4 a - 1}}) + \frac{1}{8} {[e^{2 u} - e^{- 2 u}]}_{0}^{\ln (\frac{1}{\sqrt{4 a - 1}} + \frac{2 \sqrt{a}}{\sqrt{4 a - 1}})}$ $= \frac{1}{2} \ln (\frac{1}{\sqrt{4 a - 1}} + \frac{2 \sqrt{a}}{\sqrt{4 a - 1}}) + \frac{1}{8} {{(\frac{1 + 2 \sqrt{a}}{\sqrt{4 a - 1}})}^{2} - {(\frac{\sqrt{4 a - 1}}{1 + 2 \sqrt{a}})}^{2}}$ $f (a) = \frac{1}{2} \ln (\frac{1 + 2 \sqrt{a}}{\sqrt{4 a - 1}}) + \frac{1}{8} {\frac{{(1 + 2 \sqrt{a})}^{2}}{4 a - 1} - \frac{4 a - 1}{{(1 + 2 \sqrt{a})}^{2}}}$
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