Find C=∫_0 ^1 (√(1+e^(−u) )) du Prove that ∫_1 ^(1+(1/n)) (√(1+x^n )) dx ∼


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Question Number 125737 by snipers237 last updated on 13/Dec/20

$F i n d C = \int_{0}^{1} \sqrt{1 + e^{- u}} d u$ $P r o v e t h a t \int_{1}^{1 + \frac{1}{n}} \sqrt{1 + x^{n}} d x \underset{\infty}{\sim} \frac{C}{n}$
	Answered by MJS_new last updated on 13/Dec/20

	$\int \sqrt{1 + e^{- u}} d u =$ $[v = \sqrt{1 + e^{- u}} \to d u = 2 \frac{v}{1 - v^{2}} d v]$ $= \int \frac{v^{2}}{1 - v^{2}} d v = \int (- 2 + \frac{1}{v + 1} - \frac{1}{v - 1}) d v =$ $= - 2 v + \ln ∣ \frac{v + 1}{v - 1} ∣ = . . .$ $= - 2 \sqrt{1 + e^{- u}} + 2 \ln (\sqrt{e^{u}} + \sqrt{1 + e^{u}}) + C$ $now insert borders$
	Answered by mathmax by abdo last updated on 13/Dec/20

	$C = \int_{0}^{1} \sqrt{1 + e^{- u}} du changement \sqrt{1 + e^{- u}} = t give 1 + e^{- u} = t^{2} \Rightarrow$ $e^{- u} = t^{2} - 1 \Rightarrow - u = \ln (t^{2} - 1) \Rightarrow u = - \ln (t^{2} - 1) \Rightarrow \frac{du}{dt} = - \frac{2 t}{t^{2} - 1} \Rightarrow$ $C = - 2 \int_{\sqrt{2}}^{\sqrt{1 + e^{- 1}}} \frac{t^{2}}{t^{2} - 1} dt = 2 \int_{\sqrt{1 + e^{- 1}}}^{\sqrt{2}} \frac{t^{2} - 1 + 1}{t^{2} - 1} dt$ $= 2 (\sqrt{2} - \sqrt{1 + e^{- 1}}) + \int_{\sqrt{1 + e^{- 1}}}^{\sqrt{2}} (\frac{1}{t - 1} - \frac{1}{t + 1}) dt$ $= 2 \sqrt{2} - 2 \sqrt{1 + e^{- 1}} + {[\ln ∣ \frac{t - 1}{t + 1} ∣]}_{\sqrt{1 + e^{- 1}}}^{\sqrt{2}}$ $= 2 \sqrt{2} - 2 \sqrt{1 + e^{- 1}} + \ln (\frac{\sqrt{2} - 1}{\sqrt{2} + 1}) - \ln (\frac{\sqrt{1 + e^{- 1}} - 1}{\sqrt{1 + e^{- 1}} + 1})$
	Answered by mathmax by abdo last updated on 13/Dec/20

	$A_{n} = \int_{1}^{1 + \frac{1}{n}} \sqrt{1 + x^{n}} dx changement x = t + 1 give$ $A_{n} = \int_{0}^{\frac{1}{n}} \sqrt{1 + {(t + 1)}^{n}} dt but n \to + \infty \Rightarrow t \to o and$ $1 + {(t + 1)}^{n} \sim 1 + nt \Rightarrow A_{n} \sim \int_{0}^{\frac{1}{n}} \sqrt{1 + nt} dt changement$ $\sqrt{1 + nt} = z give 1 + nt = z^{2} \Rightarrow nt = z^{2} - 1 \Rightarrow t = \frac{1}{n} (z^{2} - 1) \Rightarrow$ $\int_{0}^{\frac{1}{n}} \sqrt{1 + nt} dt = \int_{1}^{\sqrt{2}} z \frac{1}{n} (2 z) dz = \frac{2}{n} \int_{1}^{\sqrt{2}} z^{2} dz = \frac{2}{3 n} {[z^{3}]}_{1}^{\sqrt{2}}$ $= \frac{2}{3 n} (2 \sqrt{2} - 1) \Rightarrow A_{n} \sim \frac{4 \sqrt{2} - 1}{3 n} we can take C = \frac{4 \sqrt{2} - 1}{3}$
	Commented by mathmax by abdo last updated on 14/Dec/20
	$sorry 1+(t+1)^n ∼2+nt ⇒A_n ∼∫_0 ^(1/n) (√(2+nt))dt we do the changement (√(2+nt))= u ⇒2+nt=u^2 ⇒nt=u^2 −2 ⇒t=(1/n)(u^2 −2) ⇒ ∫_0 ^(1/n) (√(2+nt))dt =∫_(√2) ^(√3) u×(2/n) u du =(2/n)∫_(√2) ^(√3) u^2 du =(2/(3n))[u^3 ]_(√2) ^(√3) =(2/(3n)){3(√3)−2(√2)} ⇒A_n ∼((6(√3)−4(√2))/(3n)) ⇒C=((6(√3)−4(√2))/3)$
	$sorry 1 + {(t + 1)}^{n} \sim 2 + nt \Rightarrow A_{n} \sim \int_{0}^{\frac{1}{n}} \sqrt{2 + nt} dt we do the changement$ $\sqrt{2 + nt} = u \Rightarrow 2 + nt = u^{2} \Rightarrow nt = u^{2} - 2 \Rightarrow t = \frac{1}{n} (u^{2} - 2) \Rightarrow$ $\int_{0}^{\frac{1}{n}} \sqrt{2 + nt} dt = \int_{\sqrt{2}}^{\sqrt{3}} u \times \frac{2}{n} u du = \frac{2}{n} \int_{\sqrt{2}}^{\sqrt{3}} u^{2} du$ $= \frac{2}{3 n} {[u^{3}]}_{\sqrt{2}}^{\sqrt{3}} = \frac{2}{3 n} {3 \sqrt{3} - 2 \sqrt{2}} \Rightarrow A_{n} \sim \frac{6 \sqrt{3} - 4 \sqrt{2}}{3 n} \Rightarrow C = \frac{6 \sqrt{3} - 4 \sqrt{2}}{3}$
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