integral from 1 to 9 of 2^{sqrt(x)}

解

\int_{1}^{9} 2^{x} d x

\frac{44 ln ( 2 ) - 12}{ln ^{2} ( 2 )}

十進法表記

38.50215 \dots

解答ステップ

\int_{1}^{9} 2^{x} d x

u置換積分法を適用する

= \int_{1}^{3} 2^{u + 1} u d u

部分積分を適用する

= [\frac{2 ^{u + 1} u}{ln ( 2 )} - \int \frac{2 ^{u + 1}}{ln ( 2 )} d u]_{1}^{3}

\int \frac{2 ^{u + 1}}{ln ( 2 )} d u = \frac{1}{ln ^{2} ( 2 )} \cdot 2^{1 + u}

= [\frac{2 ^{u + 1} u}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{1 + u}]_{1}^{3}

限度を計算する:

\frac{44 ln ( 2 ) - 12}{ln ^{2} ( 2 )}

= \frac{44 ln ( 2 ) - 12}{ln ^{2} ( 2 )}