integral from 0 to 8 of sqrt(2x)-(x-4)

解答

\int_{0}^{8} 2 x - (x - 4) d x

\frac{64}{3}

十进制

21.33333 \dots

求解步骤

\int_{0}^{8} 2 x - (x - 4) d x

化简

2 x - (x - 4) : 2 x - x + 4

= \int_{0}^{8} 2 x - 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}^{8} 2 x d x - \int_{0}^{8} x d x + \int_{0}^{8} 4 d x

\int_{0}^{8} 2 x d x = \frac{64}{3}

\int_{0}^{8} x d x = 32

\int_{0}^{8} 4 d x = 32

= \frac{64}{3} - 32 + 32

化简

= \frac{64}{3}

\int_{0}^{8} x^{2} - 6 x + 8 d x

\int_{e}^{e^{2}} (\frac{1}{x ln ( x )}) d x

\int_{0}^{1} 5 x^{4} (8 - x^{5}) d x

\int_{0}^{t} e^{- a t} d t

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