integral from 0 to 1 of (x-1)^3-(x-1)

解答

\int_{0}^{1} (x - 1)^{3} - (x - 1) d x

\frac{1}{4}

十进制

0.25

求解步骤

\int_{0}^{1} (x - 1)^{3} - (x - 1) d x

化简

= \int_{0}^{1} (x - 1)^{3} - x + 1 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 - 1)^{3} d x - \int_{0}^{1} x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} (x - 1)^{3} d x = - \frac{1}{4}

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 1 d x = 1

= - \frac{1}{4} - \frac{1}{2} + 1

化简

- \frac{1}{4} - \frac{1}{2} + 1 : \frac{1}{4}

= \frac{1}{4}

in v erse l a pl a ce \frac{1}{s \cdot ( s ^{2} + \frac{9}{2} s + 15 )}

\int 2 - 1 (e - x + 2) d x

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

\frac{\partial}{\partial x} (2^{4 x} + 4 x^{2})

t an g e n t x^{2} - x y + 2 y - 2 = 0