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
![I = ∫_1 ^5 xe^x^3 dx I = ∫_1 ^(125) u^(1/3) e^u ((1/3)u^(−(2/3)) du) I = (1/3)∫_1 ^(125) u^(−(1/3)) e^u du I = (1/3)∫_1 ^(125) u^(−(1/3)) Σ_(n=0) ^∞ (u^n /(n!))du I = (1/3)[Σ_(n=0) ^∞ (u^(n+(2/3)) /((n+(2/3))n!))]_1 ^(125) I = (1/3)(Σ_(n=0) ^∞ ((125^(n+(2/3)) −1)/((n+(2/3))n!))) I = Σ_(n=0) ^∞ ((5^(3n+2) −1)/((3n+2)n!))](Q136219.png)
$$\mathrm{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{5}} {xe}^{{x}^{\mathrm{3}} } {dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{125}} {u}^{\frac{\mathrm{1}}{\mathrm{3}}} {e}^{{u}} \left(\frac{\mathrm{1}}{\mathrm{3}}{u}^{−\frac{\mathrm{2}}{\mathrm{3}}} {du}\right) \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{125}} {u}^{−\frac{\mathrm{1}}{\mathrm{3}}} {e}^{{u}} {du} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{125}} {u}^{−\frac{\mathrm{1}}{\mathrm{3}}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{u}^{{n}} }{{n}!}{du} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left[\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{u}^{{n}+\frac{\mathrm{2}}{\mathrm{3}}} }{\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right){n}!}\right]_{\mathrm{1}} ^{\mathrm{125}} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{125}^{{n}+\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{1}}{\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right){n}!}\right) \\ $$$$\mathrm{I}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{5}^{\mathrm{3}{n}+\mathrm{2}} −\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{2}\right){n}!} \\ $$