integral from 0 to 3 of (4^x-x^e)

解答

\int_{0}^{3} (4^{x} - x^{e}) d x

\frac{63}{2 ln ( 2 )} - \frac{3 ^{e + 1}}{e + 1}

十进制

29.45929 \dots

求解步骤

\int_{0}^{3} 4^{x} - x^{e} 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}^{3} 4^{x} d x - \int_{0}^{3} x^{e} d x

\int_{0}^{3} 4^{x} d x = \frac{63}{2 ln ( 2 )}

\int_{0}^{3} x^{e} d x = \frac{3 ^{e + 1}}{e + 1}

= \frac{63}{2 ln ( 2 )} - \frac{3 ^{e + 1}}{e + 1}

\int_{50}^{100} \frac{x}{24 + 5 x ^{2}} d x

\int_{5}^{\infty} \frac{1}{x ^{5}} d x

\int_{0}^{2} ∣ 2 x - 2 ∣ d x

\int_{1}^{2} 4^{3 x} d x

\int_{1}^{2} \frac{z ^{2} + 3}{z} d z