hi-guyz-please-i-need-ur-help-I-1-5-xe-x-3-dx-

Question Number 136211 by greg_ed last updated on 19/Mar/21

hi, guyz!

please, i need ur help!

I = \int_{1}^{5} {x e}^{x^{3}} d x

Answered by Olaf last updated on 19/Mar/21

I = \int_{1}^{5} {x e}^{x^{3}} d x

I = \int_{1}^{125} u^{\frac{1}{3}} e^{u} (\frac{1}{3} u^{- \frac{2}{3}} d u)

I = \frac{1}{3} \int_{1}^{125} u^{- \frac{1}{3}} e^{u} d u

I = \frac{1}{3} \int_{1}^{125} u^{- \frac{1}{3}} \underset{n = 0}{\sum^{\infty}} \frac{u^{n}}{n!} d u

I = \frac{1}{3} {[\underset{n = 0}{\sum^{\infty}} \frac{u^{n + \frac{2}{3}}}{(n + \frac{2}{3}) n!}]}_{1}^{125}

I = \frac{1}{3} (\underset{n = 0}{\sum^{\infty}} \frac{125^{n + \frac{2}{3}} - 1}{(n + \frac{2}{3}) n!})

I = \underset{n = 0}{\sum^{\infty}} \frac{5^{3 n + 2} - 1}{(3 n + 2) n!}

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