integral from-1 to 1 of |3^x-2^x|

解

\int_{- 1}^{1} ∣ 3^{x} - 2^{x} ∣ d x

\frac{8 ln ( 2 ) - 3 ln ( 3 )}{6 ln ( 3 ) ln ( 2 )}

十進法表記

0.49230 \dots

解答ステップ

\int_{- 1}^{1} ∣ 3^{x} - 2^{x} ∣ d x

絶対数を排除する

= \int_{- 1}^{0} - 3^{x} + 2^{x} d x + \int_{0}^{1} 3^{x} - 2^{x} d x

\int_{- 1}^{0} - 3^{x} + 2^{x} d x = - \frac{2}{3 ln ( 3 )} + \frac{1}{2 ln ( 2 )}

\int_{0}^{1} 3^{x} - 2^{x} d x = \frac{2}{ln ( 3 )} - \frac{1}{ln ( 2 )}

= - \frac{2}{3 ln ( 3 )} + \frac{1}{2 ln ( 2 )} + \frac{2}{ln ( 3 )} - \frac{1}{ln ( 2 )}

簡素化

- \frac{2}{3 ln ( 3 )} + \frac{1}{2 ln ( 2 )} + \frac{2}{ln ( 3 )} - \frac{1}{ln ( 2 )} : \frac{8 ln ( 2 ) - 3 ln ( 3 )}{6 ln ( 3 ) ln ( 2 )}

= \frac{8 ln ( 2 ) - 3 ln ( 3 )}{6 ln ( 3 ) ln ( 2 )}