Find-C-0-1-1-e-u-du-Prove-that-1-1-1-n-1-x-n-dx-C-n-

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} ∣ = \dots

= - 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} \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}