Question Number 4807 by Dnilka228 last updated on 14/Mar/16
$${Find}\:\mathbb{X}\:{if}… \\ $$$$\underset{\mathrm{0}} {\overset{{e}} {\int}}{x}\neq\mathrm{0}\:{and}\:\frac{{xr}}{{r}^{{e}} }=\sqrt{{e}+{n}} \\ $$$${or} \\ $$$$\underset{{n}!} {\sum}{e}=\mathrm{0}\:{and}\:\frac{\mho{x}\neq{y}}{\mho{y}\neq{x}}=−\mathrm{1} \\ $$
Commented by Dnilka228 last updated on 14/Mar/16
$$\left.\mathrm{I}\left.'\left.\mathrm{m}\:\mathrm{know},\:\mathrm{im}\:\mathrm{crazy}\right)\right)\mathrm{0}\right) \\ $$
Commented by prakash jain last updated on 15/Mar/16
$$\mathrm{What}\:\mathrm{is}\:\mathrm{question}? \\ $$
Answered by FilupSmith last updated on 16/Mar/16
![((xr)/r^e )=(√(e+n)) xr^(1−e) =(e+n)^(1/2) x=r^(e−1) (e+n)^(1/2) let A=∫_0 ^( e) xdr (1) dr or dn is unspecified ∴A=∫_0 ^( e) r^(e−1) (e+n)^(1/2) dr =(e+n)^(1/2) ∫_0 ^( e) r^(1−e) dr =(e+n)^(1/2) (1/(2−e))[r^(2−e) ]_0 ^e A=(e+n)^(1/2) (1/(2−e))e^(2−e) A=((e^2 (√(e+n)))/((2−e)e^e )) let A=∫_0 ^( e) xdn (2) ∴A=∫_0 ^( e) r^(e−1) (e+n)^(1/2) dn A=r^(e−1) ∫_0 ^( e) (√(e+n))dn let u=e+n du=1dn ∴A=r^(e−1) ∫_0 ^( e) (√u)du ∴A=r^(e−1) [(2/3)u^(3/2) ]_0 ^e A=r^(e−1) [(2/3)(e+n)^(3/2) ]_0 ^e A=r^(e−1) (2/3)[(2e)^(3/2) −(e)^(3/2) ] A=r^(e−1) (2/3)(2^(3/2) e^(3/2) −e^(3/2) ) A=(2/3)r^(e−1) (√e^3 )((√8)−1)](https://www.tinkutara.com/question/Q4823.png)
$$\frac{{xr}}{{r}^{{e}} }=\sqrt{{e}+{n}} \\ $$$${xr}^{\mathrm{1}−{e}} =\left({e}+{n}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${x}={r}^{{e}−\mathrm{1}} \left({e}+{n}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$ \\ $$$${let}\:{A}=\int_{\mathrm{0}} ^{\:{e}} {xdr}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right)\:\:\:\:{dr}\:{or}\:{dn}\:{is}\:{unspecified} \\ $$$$\therefore{A}=\int_{\mathrm{0}} ^{\:{e}} {r}^{{e}−\mathrm{1}} \left({e}+{n}\right)^{\mathrm{1}/\mathrm{2}} {dr} \\ $$$$=\left({e}+{n}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:{e}} {r}^{\mathrm{1}−{e}} {dr} \\ $$$$=\left({e}+{n}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}−{e}}\left[{r}^{\mathrm{2}−{e}} \right]_{\mathrm{0}} ^{{e}} \\ $$$${A}=\left({e}+{n}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}−{e}}{e}^{\mathrm{2}−{e}} \\ $$$${A}=\frac{{e}^{\mathrm{2}} \sqrt{{e}+{n}}}{\left(\mathrm{2}−{e}\right){e}^{{e}} } \\ $$$$ \\ $$$${let}\:{A}=\int_{\mathrm{0}} ^{\:{e}} {xdn}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\therefore{A}=\int_{\mathrm{0}} ^{\:{e}} {r}^{{e}−\mathrm{1}} \left({e}+{n}\right)^{\mathrm{1}/\mathrm{2}} {dn} \\ $$$${A}={r}^{{e}−\mathrm{1}} \int_{\mathrm{0}} ^{\:{e}} \sqrt{{e}+{n}}{dn} \\ $$$${let}\:{u}={e}+{n} \\ $$$${du}=\mathrm{1}{dn} \\ $$$$ \\ $$$$\therefore{A}={r}^{{e}−\mathrm{1}} \int_{\mathrm{0}} ^{\:{e}} \sqrt{{u}}{du} \\ $$$$\therefore{A}={r}^{{e}−\mathrm{1}} \left[\frac{\mathrm{2}}{\mathrm{3}}{u}^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{{e}} \\ $$$${A}={r}^{{e}−\mathrm{1}} \left[\frac{\mathrm{2}}{\mathrm{3}}\left({e}+{n}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{{e}} \\ $$$${A}={r}^{{e}−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}\left[\left(\mathrm{2}{e}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left({e}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right] \\ $$$${A}={r}^{{e}−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} {e}^{\frac{\mathrm{3}}{\mathrm{2}}} −{e}^{\frac{\mathrm{3}}{\mathrm{2}}} \right) \\ $$$${A}=\frac{\mathrm{2}}{\mathrm{3}}{r}^{{e}−\mathrm{1}} \sqrt{{e}^{\mathrm{3}} }\left(\sqrt{\mathrm{8}}−\mathrm{1}\right) \\ $$