let f(x)=∫_x ^(x^2 +3) e^(−xt) ln(1+e^(−xt) )dt with x>0 1) calculate f(x) 2)find lim


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Question Number 74017 by mathmax by abdo last updated on 17/Nov/19

$l e t f (x) = \int_{x}^{x^{2} + 3} e^{- x t} l n (1 + e^{- x t}) d t w i t h x > 0$ $1) c a l c u l a t e f (x)$ $2) f i n d {l i m}_{x \to + \infty} f (x) .$
Commented by mathmax by abdo last updated on 17/Nov/19

$1) c a l c u l a t e f^{^{'}} (x)$
Commented by mind is power last updated on 17/Nov/19
$u=xt⇒du=xdt f(x)=(1/x)∫_x^2 ^(x^3 +3x) e^(−u) ln(1+e^(−u) )du ∫e^(−u) ln(1+e^(−u) )du =−e^(−u) ln(1+e^(−u) )−∫(e^(−2u) /(1+e^(−u) ))du =−e^(−u) ln(1+e^(−u) )−∫(((e^(−u) +1)(e^(−u) −1))/((1+e^(−u) )))−∫(e^u /(e^u +1))du =−e^(−u) ln(1+e^(−u) )+∫(−e^(−u) +1)du−ln(e^u +1) =−e^(−u) ln(1+e^(−u) )+e^(−u) +u−ln(e^u +1)+c f(x)=(1/x){[−e^(−(x^3 +3x)) ln(1+e^(−x^3 −3x) )+e^(−(x^3 +3x)) +(x^3 +3x)−ln(e^(x^3 +3x) +1)] +e^(−x^2 ) ln(1+e^(−x^2 ) )−e^(−x^2 ) −x^2 +ln(e^x^2 +1))] lim f(x)=0$
$u = xt \Rightarrow du = xdt$ $f (x) = \frac{1}{x} \int_{x^{2}}^{x^{3} + 3 x} e^{- u} \ln (1 + e^{- u}) du$ $\int e^{- u} \ln (1 + e^{- u}) du$ $= - e^{- u} \ln (1 + e^{- u}) - \int \frac{e^{- 2 u}}{1 + e^{- u}} du$ $= - e^{- u} \ln (1 + e^{- u}) - \int \frac{(e^{- u} + 1) (e^{- u} - 1)}{(1 + e^{- u})} - \int \frac{e^{u}}{e^{u} + 1} du$ $= - e^{- u} \ln (1 + e^{- u}) + \int (- e^{- u} + 1) du - \ln (e^{u} + 1)$ $= - e^{- u} \ln (1 + e^{- u}) + e^{- u} + u - \ln (e^{u} + 1) + c$ $f (x) = \frac{1}{x} {[- e^{- (x^{3} + 3 x)} \ln (1 + e^{- x^{3} - 3 x}) + e^{- (x^{3} + 3 x)} + (x^{3} + 3 x) - \ln (e^{x^{3} + 3 x} + 1)]$ $+ e^{- x^{2}} \ln (1 + e^{- x^{2}}) - e^{- x^{2}} - x^{2} + \ln (e^{x^{2}} + 1))]$ $\lim f (x) = 0$
Commented by mathmax by abdo last updated on 21/Nov/19
$1) we have f(x)=∫_x ^(x^2 +3) e^(−xt) ln(1+e^(−xt) )dt (x>0) changement xt=u give f(x)=(1/x)∫_x^2 ^(x^3 +3x) e^(−u) ln(1+e^(−u) )du ⇒ xf(x)=∫_x^2 ^(x^3 +3x) e^(−u) ln(1+e^(−u) )dy =_(bypsrts) [−e^(−u) ln(1+e^(−u) )]_x^2 ^(x^3 +3x) −∫_x^2 ^(x^3 +3x) (−e^(−u) )×((−e^(−u) )/(1+e^(−u) )) du =e^(−x^2 ) ln(1+e^(−x^2 ) ) −e^(−(x^3 +3x)) ln(1+e^(−(x^3 +3x)) ) +∫_x^2 ^(x^3 +3x) (e^(−2u) /(1+e^(−u) ))du we have ∫_x^2 ^(x^3 +3x) (e^(−2u) /(1+e^(−u) ))du =∫_x^2 ^(x^3 +3x) (1/(e^(2u) +e^u ))du =_(e^u =z) ∫_e^x^2 ^e^(x^3 +3x) (1/(z^2 +z))(dz/z) = ∫_e^x^2 ^e^(x^3 +3x) (dz/(z^2 (z+1))) decomposition of F(z)=(1/(z^2 (z+1))) F(z)=(a/z) +(b/z^2 ) +(c/(z+1)) b=1 , c=1 ⇒F(z)=(a/z) +(1/z^2 ) +(1/(z+1)) lim_(z→+∞) zF(z)=0=a+c ⇒a=−c=−1 ⇒F(z)=−(1/z)+(1/z^2 ) +(1/(z+1)) ∫_e^x^2 ^e^(x^3 +3x) (dz/(z^2 (z+1))) =[ln∣((z+1)/z)∣]_e^x^2 ^e^(x^3 +3x) +[−(1/z)]_e^x^2 ^e^(x^3 +3x) =ln(((1+e^(x^3 +3x) )/e^(x^3 +3x) ))−ln(((1+e^x^2 )/e^x^2 ))+(e^(−x^2 ) −e^(−(x^3 +3x)) ) ⇒ f(x)=(1/x){e^(−x^2 ) ln(1+e^(−x^2 ) )−e^(−(x^3 +3x)) ln(1+e^(−(x^3 +3x)) ) ln(((1+e^(x^3 +3x) )/e^(x^3 +3x) ))−ln(((1+e^x^2 )/e^x^2 )) +e^(−x^2 ) −e^(−(x^3 +3x)) } 2) lim_(x→+∞) f(x)=0$
$1) w e h a v e f (x) = \int_{x}^{x^{2} + 3} e^{- x t} l n (1 + e^{- x t}) d t (x > 0) c h a n g e m e n t$ $x t = u g i v e f (x) = \frac{1}{x} \int_{x^{2}}^{x^{3} + 3 x} e^{- u} l n (1 + e^{- u}) d u \Rightarrow$ $x f (x) = \int_{x^{2}}^{x^{3} + 3 x} e^{- u} l n (1 + e^{- u}) d y =_{b y p s r t s} {[- e^{- u} l n (1 + e^{- u})]}_{x^{2}}^{x^{3} + 3 x}$ $- \int_{x^{2}}^{x^{3} + 3 x} (- e^{- u}) \times \frac{- e^{- u}}{1 + e^{- u}} d u = e^{- x^{2}} l n (1 + e^{- x^{2}})$ $- e^{- (x^{3} + 3 x)} l n (1 + e^{- (x^{3} + 3 x)}) + \int_{x^{2}}^{x^{3} + 3 x} \frac{e^{- 2 u}}{1 + e^{- u}} d u w e h a v e$ $\int_{x^{2}}^{x^{3} + 3 x} \frac{e^{- 2 u}}{1 + e^{- u}} d u = \int_{x^{2}}^{x^{3} + 3 x} \frac{1}{e^{2 u} + e^{u}} d u =_{e^{u} = z} \int_{e^{x^{2}}}^{e^{x^{3} + 3 x}} \frac{1}{z^{2} + z} \frac{d z}{z}$ $= \int_{e^{x^{2}}}^{e^{x^{3} + 3 x}} \frac{d z}{z^{2} (z + 1)} d e c o m p o s i t i o n o f F (z) = \frac{1}{z^{2} (z + 1)}$ $F (z) = \frac{a}{z} + \frac{b}{z^{2}} + \frac{c}{z + 1}$ $b = 1, c = 1 \Rightarrow F (z) = \frac{a}{z} + \frac{1}{z^{2}} + \frac{1}{z + 1}$ ${l i m}_{z \to + \infty} z F (z) = 0 = a + c \Rightarrow a = - c = - 1 \Rightarrow F (z) = - \frac{1}{z} + \frac{1}{z^{2}} + \frac{1}{z + 1}$ $\int_{e^{x^{2}}}^{e^{x^{3} + 3 x}} \frac{d z}{z^{2} (z + 1)} = {[l n ∣ \frac{z + 1}{z} ∣]}_{e^{x^{2}}}^{e^{x^{3} + 3 x}} + {[- \frac{1}{z}]}_{e^{x^{2}}}^{e^{x^{3} + 3 x}}$ $= l n (\frac{1 + e^{x^{3} + 3 x}}{e^{x^{3} + 3 x}}) - l n (\frac{1 + e^{x^{2}}}{e^{x^{2}}}) + (e^{- x^{2}} - e^{- (x^{3} + 3 x)}) \Rightarrow$ $f (x) = \frac{1}{x} {e^{- x^{2}} l n (1 + e^{- x^{2}}) - e^{- (x^{3} + 3 x)} l n (1 + e^{- (x^{3} + 3 x)})$ $l n (\frac{1 + e^{x^{3} + 3 x}}{e^{x^{3} + 3 x}}) - l n (\frac{1 + e^{x^{2}}}{e^{x^{2}}}) + e^{- x^{2}} - e^{- (x^{3} + 3 x)}}$ $2) {l i m}_{x \to + \infty} f (x) = 0$
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