integral from 0 to 1 of x4^x

解

\int_{0}^{1} x 4^{x} d x

\frac{8 ln ( 2 ) - 3}{4 ln ^{2} ( 2 )}

十進法表記

1.32436 \dots

解答ステップ

\int_{0}^{1} x \cdot 4^{x} d x

部分積分を適用する

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

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

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

限度を計算する:

\frac{8 ln ( 2 ) - 3}{4 ln ^{2} ( 2 )}

= \frac{8 ln ( 2 ) - 3}{4 ln ^{2} ( 2 )}