1)calculste f(a)=∫_0 ^∞ (dx/(√(x^2 −x+a))) with a >1 2) calculate f^′ (a) at form of integral then find its value.


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Question Number 77755 by abdomathmax last updated on 09/Jan/20

$1) c a l c u l s t e f (a) = \int_{0}^{\infty} \frac{d x}{\sqrt{x^{2} - x + a}} w i t h a > 1$ $2) c a l c u l a t e f^{^{'}} (a) a t f o r m o f i n t e g r a l t h e n f i n d$ $i t s v a l u e .$
Commented by mathmax by abdo last updated on 11/Jan/20

$f o r g i v e f (a) = \int_{0}^{1} \frac{d x}{\sqrt{x^{2} - x + a}} w i t h a > 1$
Commented by mathmax by abdo last updated on 11/Jan/20
$f(a)=∫_0 ^1 (dx/(√(x^2 −x+a))) we have x^2 −x+a=x^2 −2(x/2)+(1/4)+a−(1/4) =(x−(1/2))^2 +((4a−1)/4) and ((4a−1)/4)>0 fhangement x−(1/2)=((√(4a−1))/2)u give f(a) =∫_(−(1/(√(4a−1)))) ^(1/(√(4a−1))) (1/(((√(4a−1))/2)((√(1+u^2 ))))×((√(4a−1))/2)du =∫_((−1)/(√(4a−1))) ^(1/(√(4a−1))) (du/(√(1+u^2 ))) =2 ∫_0 ^(1/(√(4a−1))) (du/(√(1+u^2 ))) =2[ln(u+(√(=u^2 +1)))]_0 ^(1/(√(4a−1))) =2{ln((1/(√(4a−1)))+(√((1/(4a−1 ))+1)))}=2 ln((1/(√(4a−1)))+((2(√a))/(√(4a−1)))) =2ln(((2(√a)+1)/(√(4a−1)))) =2ln(2(√a)+1)−ln(4a−1) ⇒f(a)=2ln(2(√a)+1)−ln(4a−1) 2) we have f^′ (a)=∫_0 ^1 (∂/∂a)((1/(√(x^2 −x+a))))dx =∫_0 ^1 (∂/∂a){( x^2 −x+a)^(−(1/2)) }dx =∫_0 ^1 −(1/2)(x^2 −x+a)^(−(3/2)) dx =−(1/2)∫_0 ^1 (dx/((x^2 −x+a)(√(x^2 −x+a)))) from another side f^′ (a) =2 ×((1/(√a))/(2(√a)+1))−(4/(4a−1)) =2×(1/((√a)(2(√a)+1)))−(4/(4a−1)) =(2/(2a+(√a)))−(4/(4a−1)) ⇒ −(1/2)∫_0 ^1 (dx/((x^2 −x+a)(√(x^2 −x+a)))) =(2/(2a+(√a)))−(4/(4a−1)) also we get ∫_0 ^1 (dx/((x^2 −x+a)(√(x^2 −x+a)))) =((−4)/(2a+(√a)))+(8/(4a−1))$
$f (a) = \int_{0}^{1} \frac{d x}{\sqrt{x^{2} - x + a}} w e h a v e x^{2} - x + a = x^{2} - 2 \frac{x}{2} + \frac{1}{4} + a - \frac{1}{4}$ $= {(x - \frac{1}{2})}^{2} + \frac{4 a - 1}{4} a n d \frac{4 a - 1}{4} > 0 f h a n g e m e n t x - \frac{1}{2} = \frac{\sqrt{4 a - 1}}{2} u$ $g i v e f (a) = \int_{- \frac{1}{\sqrt{4 a - 1}}}^{\frac{1}{\sqrt{4 a - 1}}} \frac{1}{\frac{\sqrt{4 a - 1}}{2} (\sqrt{1 + u^{2}}} \times \frac{\sqrt{4 a - 1}}{2} d u$ $= \int_{\frac{- 1}{\sqrt{4 a - 1}}}^{\frac{1}{\sqrt{4 a - 1}}} \frac{d u}{\sqrt{1 + u^{2}}} = 2 \int_{0}^{\frac{1}{\sqrt{4 a - 1}}} \frac{d u}{\sqrt{1 + u^{2}}} = 2 {[l n (u + \sqrt{= u^{2} + 1})]}_{0}^{\frac{1}{\sqrt{4 a - 1}}}$ $= 2 {l n (\frac{1}{\sqrt{4 a - 1}} + \sqrt{\frac{1}{4 a - 1} + 1})} = 2 l n (\frac{1}{\sqrt{4 a - 1}} + \frac{2 \sqrt{a}}{\sqrt{4 a - 1}})$ $= 2 l n (\frac{2 \sqrt{a} + 1}{\sqrt{4 a - 1}}) = 2 l n (2 \sqrt{a} + 1) - l n (4 a - 1)$ $\Rightarrow f (a) = 2 l n (2 \sqrt{a} + 1) - l n (4 a - 1)$ $2) w e h a v e f^{^{'}} (a) = \int_{0}^{1} \frac{\partial}{\partial a} (\frac{1}{\sqrt{x^{2} - x + a}}) d x$ $= \int_{0}^{1} \frac{\partial}{\partial a} {{(x^{2} - x + a)}^{- \frac{1}{2}}} d x = \int_{0}^{1} - \frac{1}{2} {(x^{2} - x + a)}^{- \frac{3}{2}} d x$ $= - \frac{1}{2} \int_{0}^{1} \frac{d x}{(x^{2} - x + a) \sqrt{x^{2} - x + a}} f r o m a n o t h e r s i d e$ $f^{^{'}} (a) = 2 \times \frac{\frac{1}{\sqrt{a}}}{2 \sqrt{a} + 1} - \frac{4}{4 a - 1} = 2 \times \frac{1}{\sqrt{a} (2 \sqrt{a} + 1)} - \frac{4}{4 a - 1}$ $= \frac{2}{2 a + \sqrt{a}} - \frac{4}{4 a - 1} \Rightarrow$ $- \frac{1}{2} \int_{0}^{1} \frac{d x}{(x^{2} - x + a) \sqrt{x^{2} - x + a}} = \frac{2}{2 a + \sqrt{a}} - \frac{4}{4 a - 1} a l s o w e g e t$ $\int_{0}^{1} \frac{d x}{(x^{2} - x + a) \sqrt{x^{2} - x + a}} = \frac{- 4}{2 a + \sqrt{a}} + \frac{8}{4 a - 1}$
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