integral from 0 to 1 of (x^{19}+19^x)

解答

\int_{0}^{1} (x^{19} + 1 9^{x}) d x

\frac{1}{20} + \frac{18}{ln ( 19 )}

十进制

6.16321 \dots

求解步骤

\int_{0}^{1} x^{19} + 1 9^{x} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} x^{19} d x + \int_{0}^{1} 1 9^{x} d x

\int_{0}^{1} x^{19} d x = \frac{1}{20}

\int_{0}^{1} 1 9^{x} d x = \frac{18}{ln ( 19 )}

= \frac{1}{20} + \frac{18}{ln ( 19 )}

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