1) find f(λ) =∫_0 ^1 (dx/(2+e^(−λx) )) with λ>0 2)calculate ∫_0 ^1 (x/((2+e^(−λx) )^2 ))dx 3) find the value of ∫_0 ^1 (dx/(2 +e^(−x(√3)) ))dx and ∫


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Question Number 48715 by Abdo msup. last updated on 27/Nov/18

$1) f i n d f (λ) = \int_{0}^{1} \frac{d x}{2 + e^{- λ x}} w i t h λ > 0$ $2) c a l c u l a t e \int_{0}^{1} \frac{x}{{(2 + e^{- λ x})}^{2}} d x$ $3) f i n d t h e v a l u e o f \int_{0}^{1} \frac{d x}{2 + e^{- x \sqrt{3}}} d x a n d \int_{0}^{1} \frac{x}{{(2 + e^{- x \sqrt{3}})}^{2}} d x$
Commented bymaxmathsup by imad last updated on 01/Dec/18
$1) we have f(λ)=∫_0 ^1 (dx/(2+e^(−λx) )) changement e^(λx ) =t give λx=ln(t) ⇒ x=(1/λ)ln(t) ⇒f(λ)=∫_1 ^e^λ (1/(λ(2+t^(−1) ))) (dt/t) =(1/λ) ∫_1 ^e^λ (dt/(2t +1)) =(1/(2λ))[ln(2t+1)]_1 ^e^λ = (1/(2λ)){ln(2e^λ +1)−ln(3)} ⇒ f(λ)=(1/(2λ)){ln(2 e^λ +1)−ln(3)} .$
$1) w e h a v e f (λ) = \int_{0}^{1} \frac{d x}{2 + e^{- λ x}} c h a n g e m e n t e^{λ x} = t g i v e λ x = l n (t) \Rightarrow$ $x = \frac{1}{λ} l n (t) \Rightarrow f (λ) = \int_{1}^{e^{λ}} \frac{1}{λ (2 + t^{- 1})} \frac{d t}{t} = \frac{1}{λ} \int_{1}^{e^{λ}} \frac{d t}{2 t + 1}$ $= \frac{1}{2 λ} {[l n (2 t + 1)]}_{1}^{e^{λ}} = \frac{1}{2 λ} {l n (2 e^{λ} + 1) - l n (3)} \Rightarrow$ $f (λ) = \frac{1}{2 λ} {l n (2 e^{λ} + 1) - l n (3)} .$
Commented bymaxmathsup by imad last updated on 01/Dec/18
$2) we have f^′ (λ)=∫_0 ^1 −((−x e^(−λx) )/((2+e^(−λx) )^2 )) dx =∫_0 ^1 ((x e^(−λx) )/((2+e^(−λx) )^2 ))dx ⇒ ∫_0 ^1 ((x e^(−λx) )/((2+e^(−λx) )^2 ))dx =f^′ (λ) =(1/2){−(1/λ^2 )(ln(2e^λ +1)−ln(3) +(1/λ) ((2 e^λ )/(2e^(λ ) +1))} =−(1/(2λ^2 )){( ln(2 e^λ +1)−ln(3)) −((2λ e^λ )/(2e^λ +1))} ∫_0 ^1 ((x e^(−x(√3)) )/((2+e^(−x(√3)) )^2 ))dx = f^′ ((√3)) =−(1/6){ (ln(2 e^(√3) +1)−ln(3)) −((2(√3)e^(√3) )/(2 e^(√3) +1))} .$
$2) w e h a v e f^{^{'}} (λ) = \int_{0}^{1} - \frac{- x e^{- λ x}}{{(2 + e^{- λ x})}^{2}} d x = \int_{0}^{1} \frac{x e^{- λ x}}{{(2 + e^{- λ x})}^{2}} d x \Rightarrow$ $\int_{0}^{1} \frac{x e^{- λ x}}{{(2 + e^{- λ x})}^{2}} d x = f^{^{'}} (λ) = \frac{1}{2} {- \frac{1}{λ^{2}} (l n (2 e^{λ} + 1) - l n (3) + \frac{1}{λ} \frac{2 e^{λ}}{2 e^{λ} + 1}}$ $= - \frac{1}{2 λ^{2}} {(l n (2 e^{λ} + 1) - l n (3)) - \frac{2 λ e^{λ}}{2 e^{λ} + 1}}$ $\int_{0}^{1} \frac{x e^{- x \sqrt{3}}}{{(2 + e^{- x \sqrt{3}})}^{2}} d x = f^{^{'}} (\sqrt{3}) = - \frac{1}{6} {(l n (2 e^{\sqrt{3}} + 1) - l n (3)) - \frac{2 \sqrt{3} e^{\sqrt{3}}}{2 e^{\sqrt{3}} + 1}} .$
Commented bymaxmathsup by imad last updated on 01/Dec/18

$2) t h e Q . i s c a l c u l a t e \int_{0}^{1} \frac{x e^{- λ x}}{{(2 + e^{- λ x})}^{2}} d x$
Commented bymaxmathsup by imad last updated on 01/Dec/18
$) we have ∫_0 ^1 (dx/(2+e^(−x(√3)) )) =f((√3)) =(1/(2(√3))){ln(2 e^(√3) +1)−ln(3)}$
$) w e h a v e \int_{0}^{1} \frac{d x}{2 + e^{- x \sqrt{3}}} = f (\sqrt{3}) = \frac{1}{2 \sqrt{3}} {l n (2 e^{\sqrt{3}} + 1) - l n (3)}$
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