integral from 1 to 5 of (e^{x/2})/(x^3)

解

\int_{1}^{5} \frac{e ^{\frac{x}{2}}}{x ^{3}} d x

\frac{- 14 e ^{\frac{5}{2}} + 25 Ei ( \frac{5}{2} )}{200} - \frac{- 6 e ^{\frac{1}{2}} + Ei ( \frac{1}{2} )}{8}

解答ステップ

\int_{1}^{5} \frac{e ^{\frac{x}{2}}}{x ^{3}} d x

部分積分を適用する

= [- \frac{e ^{\frac{x}{2}}}{2 x ^{2}} - \int - \frac{e ^{\frac{x}{2}}}{4 x ^{2}} d x]_{1}^{5}

\int - \frac{e ^{\frac{x}{2}}}{4 x ^{2}} d x = - \frac{1}{4} (- \frac{e ^{\frac{x}{2}}}{x} + \frac{1}{2} Ei (\frac{x}{2}))

= [- \frac{e ^{\frac{x}{2}}}{2 x ^{2}} - (- \frac{1}{4} (- \frac{e ^{\frac{x}{2}}}{x} + \frac{1}{2} Ei (\frac{x}{2})))]_{1}^{5}

簡素化

= [- \frac{e ^{\frac{x}{2}}}{2 x ^{2}} + \frac{1}{4} (- \frac{e ^{\frac{x}{2}}}{x} + \frac{1}{2} Ei (\frac{x}{2}))]_{1}^{5}

限度を計算する:

\frac{- 14 e ^{\frac{5}{2}} + 25 Ei ( \frac{5}{2} )}{200} - \frac{- 6 e ^{\frac{1}{2}} + Ei ( \frac{1}{2} )}{8}

= \frac{- 14 e ^{\frac{5}{2}} + 25 Ei ( \frac{5}{2} )}{200} - \frac{- 6 e ^{\frac{1}{2}} + Ei ( \frac{1}{2} )}{8}