integral from 0 to 4 of (16-x^2)^2

解答

\int_{0}^{4} (16 - x^{2})^{2} d x

\frac{8192}{15}

十进制

546.13333 \dots

求解步骤

\int_{0}^{4} (16 - x^{2})^{2} d x

乘开

(16 - x^{2})^{2} : 256 - 32 x^{2} + x^{4}

= \int_{0}^{4} 256 - 32 x^{2} + x^{4} 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}^{4} 256 d x - \int_{0}^{4} 32 x^{2} d x + \int_{0}^{4} x^{4} d x

\int_{0}^{4} 256 d x = 1024

\int_{0}^{4} 32 x^{2} d x = \frac{2048}{3}

\int_{0}^{4} x^{4} d x = \frac{1024}{5}

= 1024 - \frac{2048}{3} + \frac{1024}{5}

化简

1024 - \frac{2048}{3} + \frac{1024}{5} : \frac{8192}{15}

= \frac{8192}{15}

d er i v a t i v e y = (2 x + 3)^{2}

x \to 0 lim (e^{x} ln (x + 1))

t an g e n t \frac{x - 1}{x + 1}

\frac{d}{d x} (e^{x} (x - 5))

t a y l or 2 e^{z}