Question Number 136211 by greg_ed last updated on 19/Mar/21
$$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{guyz}}\:!\: \\ $$$$\boldsymbol{\mathrm{please}},\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{ur}}\:\boldsymbol{\mathrm{help}}\:! \\ $$$$\boldsymbol{\mathrm{I}}=\int_{\mathrm{1}} ^{\:\mathrm{5}} \boldsymbol{{xe}}^{\boldsymbol{{x}}^{\mathrm{3}} } \boldsymbol{{dx}} \\ $$
Answered by Olaf last updated on 19/Mar/21
$$\mathrm{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{5}} {xe}^{{x}^{\mathrm{3}} } {dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{125}} {u}^{\frac{\mathrm{1}}{\mathrm{3}}} {e}^{{u}} \left(\frac{\mathrm{1}}{\mathrm{3}}{u}^{−\frac{\mathrm{2}}{\mathrm{3}}} {du}\right) \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{125}} {u}^{−\frac{\mathrm{1}}{\mathrm{3}}} {e}^{{u}} {du} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{125}} {u}^{−\frac{\mathrm{1}}{\mathrm{3}}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{u}^{{n}} }{{n}!}{du} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left[\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{u}^{{n}+\frac{\mathrm{2}}{\mathrm{3}}} }{\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right){n}!}\right]_{\mathrm{1}} ^{\mathrm{125}} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{125}^{{n}+\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{1}}{\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right){n}!}\right) \\ $$$$\mathrm{I}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{5}^{\mathrm{3}{n}+\mathrm{2}} −\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{2}\right){n}!} \\ $$