integral from 0 to 3 of (x+2)sqrt(x+1)

解答

\int_{0}^{3} (x + 2) x + 1 d x

\frac{256}{15}

十进制

17.06666 \dots

求解步骤

\int_{0}^{3} (x + 2) x + 1 d x

使用换元积分法

= \int_{1}^{2} 2 u^{2} (u^{2} + 1) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{1}^{2} u^{2} (u^{2} + 1) d u

乘开

u^{2} (u^{2} + 1) : u^{4} + u^{2}

= 2 \cdot \int_{1}^{2} u^{4} + u^{2} d u

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

= 2 (\int_{1}^{2} u^{4} d u + \int_{1}^{2} u^{2} d u)

\int_{1}^{2} u^{4} d u = \frac{31}{5}

\int_{1}^{2} u^{2} d u = \frac{7}{3}

= 2 (\frac{31}{5} + \frac{7}{3})

化简

2 (\frac{31}{5} + \frac{7}{3}) : \frac{256}{15}

= \frac{256}{15}

\int \frac{x ^{14}}{x ^{5} - 1} d x

\frac{\partial}{\partial z} (e^{x yz})

\int (e^{3 x} - 4)^{2} d x

d er i v a t i v e \frac{x}{x ^{2} + 64}

t an g e n t y = 3 x^{3} - x^{2} + 9, (2, 29)