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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 = \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 by mathmax by abdo last updated on 09/Jun/20

thank you sir .

Commented by niroj 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 by mathmax 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} - 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}}}