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Question Number 27447 by raman last updated on 07/Jan/18

$${\underset{0}{\int }}^{\mathrm{\infty }}{\mathrm{v}}^4\mathrm{e}\frac{-{\mathrm{mv}}^2}{2\mathrm{KT}}\mathrm{dv}$$$$\mathrm{solve}\mathrm{it}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/May/18

$$letx=\frac{m}{2KT}{v}^{2}v=\frac{\sqrt{2KT}}{\sqrt{m}}\times x^{\frac{1}{2}}$$$$dx=\frac{m}{2KT}\times 2vdv$$$$=\frac{m}{KT}\times \frac{\sqrt{2KT}}{\sqrt{m}}\times x_{}^{\frac{1}{2}}dvsodv=\frac{KT}{m}\times \frac{\sqrt{m}}{\sqrt{2KT}}\times x^{-\frac{1}{2}}\times$$$$dx\frac{}{}$$$${\int }_0^{\mathrm{\infty }}\frac{{\left(2KT\right)}^2}{m^2}\times x^2\times e^{-x}\times \frac{\sqrt{KT}}{\sqrt{m}}\times \frac{1}{\sqrt{2}}\times x^{\frac{-1}{2}}\times dx$$$${\int }_0^{\mathrm{\infty }}\frac{2^{\frac{3}{2}}\times {\left(KT\right)}^{\frac{5}{2}}}{m^{\frac{5}{2}}}{e}^{-x}\times x^{\frac{3}{2}}\times dx$$$$={\left[\frac{8{\left(KT\right)}^5}{m^5}\right]}^{\frac{1}{2}}\times {\int }_0^{\mathrm{\infty }}{e}^{-x}\times x^{\frac{3}{2}}dx$$$$=ditto\times \lceil \left(\frac{5}{2}\right)\{{\int }_0^{\mathrm{\infty }}{e}^{-x}\times x^{n-1}=\lceil \left(n\right)\}$$$$now\lceil \left(\frac{5}{2}\right)=\frac{3}{2}\times \frac{1}{2}\times \sqrt{\mathrm{\Pi }}$$$$={\left\{\frac{8{\left(KT\right)}^5}{m^5}\right\}}^{\frac{1}{2}}\times \frac{3}{4}\times \sqrt{\mathrm{\Pi }}$$$$plscheck$$$$$$

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