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≠1 ∫_0 ^2 x^t dt =∫_0 ^2 e^(tln x) dt =(1/(ln x))∫_0 ^2 e^(tln x) d(tln x) =(1/(ln x))[x^t ]_0 ^2 =((x^2 −1)/(ln x))=3 x^2 −1=3 ln x x^3 =e^(x^2 −1) let u=x^2 −1 x=(u+1)^(1/2) (u+1)^(3/2) =e^u u+1=e^((2u)/3) (u+1)e^(2/3) =e^((2(u+1))/3) ((2(u+1))/3)e^(2/3) =(2/3)e^((2(u+1))/3) −((2(u+1))/3)e^(−((2(u+1))/3)) =−(2/(3e^(2/3) )) −((2(u+1))/3)=W(−(2/(3e^(2/3) ))) u=−(3/2)W(−(2/(3e^(2/3) )))−1 x^2 −1=−(3/2)W(−(2/(3e^(2/3) )))−1 x^2 =−(3/2)W(−(2/(3e^(2/3) ))) x=(√(−(3/2)W(−(2/(3e^(2/3) )))))= { ((1.464251632)),((1 (rejected))) :} generally: ∫_0 ^n x^t dt=m ⇒x=(√(−(m/n)W(−(n/(me^(n/m) )))))](https://www.tinkutara.com/question/Q170896.png)
$${x}>\mathrm{0},\:{x}\neq\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} {x}^{{t}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}} {e}^{{t}\mathrm{ln}\:{x}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:{x}}\int_{\mathrm{0}} ^{\mathrm{2}} {e}^{{t}\mathrm{ln}\:{x}} {d}\left({t}\mathrm{ln}\:{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:{x}}\left[{x}^{{t}} \right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{ln}\:{x}}=\mathrm{3} \\ $$$${x}^{\mathrm{2}} −\mathrm{1}=\mathrm{3}\:\mathrm{ln}\:{x} \\ $$$${x}^{\mathrm{3}} ={e}^{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${let}\:{u}={x}^{\mathrm{2}} −\mathrm{1} \\ $$$${x}=\left({u}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left({u}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} ={e}^{{u}} \\ $$$${u}+\mathrm{1}={e}^{\frac{\mathrm{2}{u}}{\mathrm{3}}} \\ $$$$\left({u}+\mathrm{1}\right){e}^{\frac{\mathrm{2}}{\mathrm{3}}} ={e}^{\frac{\mathrm{2}\left({u}+\mathrm{1}\right)}{\mathrm{3}}} \\ $$$$\frac{\mathrm{2}\left({u}+\mathrm{1}\right)}{\mathrm{3}}{e}^{\frac{\mathrm{2}}{\mathrm{3}}} =\frac{\mathrm{2}}{\mathrm{3}}{e}^{\frac{\mathrm{2}\left({u}+\mathrm{1}\right)}{\mathrm{3}}} \\ $$$$−\frac{\mathrm{2}\left({u}+\mathrm{1}\right)}{\mathrm{3}}{e}^{−\frac{\mathrm{2}\left({u}+\mathrm{1}\right)}{\mathrm{3}}} =−\frac{\mathrm{2}}{\mathrm{3}{e}^{\frac{\mathrm{2}}{\mathrm{3}}} } \\ $$$$−\frac{\mathrm{2}\left({u}+\mathrm{1}\right)}{\mathrm{3}}={W}\left(−\frac{\mathrm{2}}{\mathrm{3}{e}^{\frac{\mathrm{2}}{\mathrm{3}}} }\right) \\ $$$${u}=−\frac{\mathrm{3}}{\mathrm{2}}{W}\left(−\frac{\mathrm{2}}{\mathrm{3}{e}^{\frac{\mathrm{2}}{\mathrm{3}}} }\right)−\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{1}=−\frac{\mathrm{3}}{\mathrm{2}}{W}\left(−\frac{\mathrm{2}}{\mathrm{3}{e}^{\frac{\mathrm{2}}{\mathrm{3}}} }\right)−\mathrm{1} \\ $$$${x}^{\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{2}}{W}\left(−\frac{\mathrm{2}}{\mathrm{3}{e}^{\frac{\mathrm{2}}{\mathrm{3}}} }\right) \\ $$$${x}=\sqrt{−\frac{\mathrm{3}}{\mathrm{2}}{W}\left(−\frac{\mathrm{2}}{\mathrm{3}{e}^{\frac{\mathrm{2}}{\mathrm{3}}} }\right)}=\begin{cases}{\mathrm{1}.\mathrm{464251632}}\\{\mathrm{1}\:\left({rejected}\right)}\end{cases} \\ $$$$ \\ $$$${generally}: \\ $$$$\int_{\mathrm{0}} ^{{n}} {x}^{{t}} {dt}={m} \\ $$$$\Rightarrow{x}=\sqrt{−\frac{{m}}{{n}}{W}\left(−\frac{{n}}{{me}^{\frac{{n}}{{m}}} }\right)} \\ $$
Commented by Tawa11 last updated on 03/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$