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Question Number 170889 by 0731619 last updated on 02/Jun/22

Commented by Tawa11 last updated on 03/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 02/Jun/22

$$x>0,x\ne 1$$$${\int }_0^2{x}^{t}dt$$$$={\int }_0^2{e}^{t\ln x}dt$$$$=\frac{1}{\ln x}{\int }_0^2{e}^{t\ln x}d\left(t\ln x\right)$$$$=\frac{1}{\ln x}{\left[x^t\right]}_0^2$$$$=\frac{x^2-1}{\ln x}=3$$$$x^2-1=3\ln x$$$$x^3=e^{x^2-1}$$$$letu=x^2-1$$$$x={(u+1)}^{\frac{1}{2}}$$$${(u+1)}^{\frac{3}{2}}=e^u$$$$u+1=e^{\frac{2u}{3}}$$$$(u+1)e^{\frac{2}{3}}=e^{\frac{2(u+1)}{3}}$$$$\frac{2(u+1)}{3}{e}^{\frac{2}{3}}=\frac{2}{3}{e}^{\frac{2(u+1)}{3}}$$$$-\frac{2(u+1)}{3}{e}^{-\frac{2(u+1)}{3}}=-\frac{2}{3e^{\frac{2}{3}}}$$$$-\frac{2(u+1)}{3}=W(-\frac{2}{3e^{\frac{2}{3}}})$$$$u=-\frac{3}{2}W(-\frac{2}{3e^{\frac{2}{3}}})-1$$$$x^2-1=-\frac{3}{2}W(-\frac{2}{3e^{\frac{2}{3}}})-1$$$$x^2=-\frac{3}{2}W(-\frac{2}{3e^{\frac{2}{3}}})$$$$x=\sqrt{-\frac{3}{2}W(-\frac{2}{3e^{\frac{2}{3}}})}=\{\begin{array}{l}1.464251632\\ 1\left(rejected\right)\end{array}$$$$$$$$generally:$$$${\int }_0^n{x}^{t}dt=m$$$$\Rightarrow x=\sqrt{-\frac{m}{n}W(-\frac{n}{{me}^{\frac{n}{m}}})}$$

Commented by Tawa11 last updated on 03/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

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