integral from 0 to 4 of (2^x-x^e)

解

\int_{0}^{4} (2^{x} - x^{e}) d x

\frac{15}{ln ( 2 )} - \frac{4 ^{e + 1}}{e + 1}

十進法表記

- 24.94889 \dots

解答ステップ

\int_{0}^{4} 2^{x} - x^{e} d x

総和規則を適用する:

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

= \int_{0}^{4} 2^{x} d x - \int_{0}^{4} x^{e} d x

\int_{0}^{4} 2^{x} d x = \frac{15}{ln ( 2 )}

\int_{0}^{4} x^{e} d x = \frac{4 ^{e + 1}}{e + 1}

= \frac{15}{ln ( 2 )} - \frac{4 ^{e + 1}}{e + 1}