integral from 0 to 1 of xpi^x

解

\int_{0}^{1} x π^{x} d x

\frac{π}{ln ( π )} + \frac{- π + 1}{ln ^{2} ( π )}

十進法表記

1.11010 \dots

解答ステップ

\int_{0}^{1} x π^{x} d x

部分積分を適用する

= [\frac{π ^{x} x}{ln ( π )} - \int \frac{π ^{x}}{ln ( π )} d x]_{0}^{1}

\int \frac{π ^{x}}{ln ( π )} d x = \frac{1}{ln ^{2} ( π )} π^{x}

= [\frac{π ^{x} x}{ln ( π )} - \frac{1}{ln ^{2} ( π )} π^{x}]_{0}^{1}

限度を計算する:

\frac{π}{ln ( π )} - \frac{π}{ln ^{2} ( π )} + \frac{1}{ln ^{2} ( π )}

= \frac{π}{ln ( π )} - \frac{π}{ln ^{2} ( π )} + \frac{1}{ln ^{2} ( π )}

簡素化

= \frac{π}{ln ( π )} + \frac{- π + 1}{ln ^{2} ( π )}