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

解答

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

\frac{58}{5}

十进制

11.6

求解步骤

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

提出常数:

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

= \frac{3}{2} \cdot \int_{0}^{3} x x + 1 d x

使用换元积分法

= \frac{3}{2} \cdot \int_{1}^{4} (u - 1) u d u

乘开

(u - 1) u : u^{\frac{3}{2}} - u

= \frac{3}{2} \cdot \int_{1}^{4} u^{\frac{3}{2}} - u 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

= \frac{3}{2} (\int_{1}^{4} u^{\frac{3}{2}} d u - \int_{1}^{4} u d u)

\int_{1}^{4} u^{\frac{3}{2}} d u = \frac{62}{5}

\int_{1}^{4} u d u = \frac{14}{3}

= \frac{3}{2} (\frac{62}{5} - \frac{14}{3})

化简

\frac{3}{2} (\frac{62}{5} - \frac{14}{3}) : \frac{58}{5}

= \frac{58}{5}

\int_{2}^{8} - \frac{x}{2} d x

\int \frac{cos ( ln ( 9 t ) )}{t} d t

\frac{d}{d x} (e^{2 c o s (3 x)})

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 9

a re a y^{2} - 6 x = 1, x - y = 1