integral from 0 to 1 of 3^xe^xsin(x)

解

\int_{0}^{1} 3^{x} e^{x} sin (x) d x

\frac{3 e ln ^{2} ( 3 e ) sin ( 1 ) - 3 e ln ( 3 e ) cos ( 1 ) + ln ( 3 e )}{ln ^{3} ( 3 e ) + ln ( 3 e )}

十進法表記

2.03448 \dots

解答ステップ

\int_{0}^{1} 3^{x} e^{x} sin (x) d x

部分積分を適用する

= [\frac{sin ( x ) ( 3 e ) ^{x}}{ln ( 3 e )} - \int \frac{cos ( x ) ( 3 e ) ^{x}}{ln ( 3 e )} d x]_{0}^{1}

\int \frac{cos ( x ) ( 3 e ) ^{x}}{ln ( 3 e )} d x = \frac{1}{ln ( 3 e ) + ln ^{3} ( 3 e )} sin (x) (3 e)^{x} + \frac{1}{1 + ln ^{2} ( 3 e )} cos (x) (3 e)^{x}

= [\frac{sin ( x ) ( 3 e ) ^{x}}{ln ( 3 e )} - (\frac{1}{ln ( 3 e ) + ln ^{3} ( 3 e )} sin (x) (3 e)^{x} + \frac{1}{1 + ln ^{2} ( 3 e )} cos (x) (3 e)^{x})]_{0}^{1}

簡素化

[\frac{sin ( x ) ( 3 e ) ^{x}}{ln ( 3 e )} - (\frac{1}{ln ( 3 e ) + ln ^{3} ( 3 e )} sin (x) (3 e)^{x} + \frac{1}{1 + ln ^{2} ( 3 e )} cos (x) (3 e)^{x})]_{0}^{1} : [\frac{1}{ln ^{3} ( 3 e ) + ln ( 3 e )} ln (3 e) \cdot 3^{x} e^{x} (ln (3 e) sin (x) - cos (x))]_{0}^{1}

= [\frac{1}{ln ^{3} ( 3 e ) + ln ( 3 e )} ln (3 e) \cdot 3^{x} e^{x} (ln (3 e) sin (x) - cos (x))]_{0}^{1}

限度を計算する:

\frac{3 e ln ^{2} ( 3 e ) sin ( 1 ) - 3 e ln ( 3 e ) cos ( 1 ) + ln ( 3 e )}{ln ^{3} ( 3 e ) + ln ( 3 e )}

= \frac{3 e ln ^{2} ( 3 e ) sin ( 1 ) - 3 e ln ( 3 e ) cos ( 1 ) + ln ( 3 e )}{ln ^{3} ( 3 e ) + ln ( 3 e )}