Evaluate the integrals

${\int}_{0}^{\mathrm{ln}9}{e}^{\theta}{({e}^{\theta}-1)}^{\frac{1}{2}}d\theta$

arenceabigns
2021-10-22
Answered

Evaluate the integrals

${\int}_{0}^{\mathrm{ln}9}{e}^{\theta}{({e}^{\theta}-1)}^{\frac{1}{2}}d\theta$

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wheezym

Answered 2021-10-23
Author has **103** answers

We have to evaluate the integral

${\int}_{0}^{\mathrm{ln}9}{e}^{\theta}{({e}^{\theta}-1)}^{\frac{1}{2}}d\theta$

let${e}^{\theta}-1==t$

on differentiating with respect to t we get

${e}^{\theta}d\theta =dt$

$d\theta =\frac{dt}{{e}^{\theta}}$

now the limits

$t={e}^{\theta}-1$

$t={e}^{\mathrm{ln}9}-1$

$t=9-1$

$t=8$

also

$t={e}^{0}-1$

$t=1-1$

$t=0$

now the definite integral

$\int}_{0}^{8}{e}^{\theta}{t}^{\frac{1}{2}}\frac{dt}{{e}^{\theta}$

${\int}_{0}^{8}{t}^{\frac{1}{2}}dt$

$\Rightarrow {\left[\frac{{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_{0}^{8}$

$\Rightarrow \frac{2}{3}{\left[{t}^{\frac{3}{2}}\right]}_{0}^{8}$

$\Rightarrow \frac{2}{3}\left[{8}^{\frac{3}{2}-0}\right]$

$\Rightarrow \frac{2}{3}\left[{2}^{3\times \frac{3}{2}}\right]$

$\Rightarrow \frac{{2}^{\frac{11}{2}}}{3}=\frac{32\times {2}^{\frac{1}{2}}}{3}$

therefore

$\int}_{0}^{\mathrm{ln}9}{e}^{\theta}{({e}^{\theta}-1)}^{\frac{1}{2}}d\theta =\frac{32\times {2}^{\frac{1}{2}}}{3$

let

on differentiating with respect to t we get

now the limits

also

now the definite integral

therefore

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