integral from 0 to 1 of 6x^2(3-5x^3)^3

解答

\int_{0}^{1} 6 x^{2} (3 - 5 x^{3})^{3} d x

\frac{13}{2}

十进制

6.5

求解步骤

\int_{0}^{1} 6 x^{2} (3 - 5 x^{3})^{3} d x

提出常数:

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

= 6 \cdot \int_{0}^{1} x^{2} (3 - 5 x^{3})^{3} d x

乘开

x^{2} (3 - 5 x^{3})^{3} : 27 x^{2} - 135 x^{5} + 225 x^{8} - 125 x^{11}

= 6 \cdot \int_{0}^{1} 27 x^{2} - 135 x^{5} + 225 x^{8} - 125 x^{11} 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

= 6 (\int_{0}^{1} 27 x^{2} d x - \int_{0}^{1} 135 x^{5} d x + \int_{0}^{1} 225 x^{8} d x - \int_{0}^{1} 125 x^{11} d x)

\int_{0}^{1} 27 x^{2} d x = 9

\int_{0}^{1} 135 x^{5} d x = \frac{45}{2}

\int_{0}^{1} 225 x^{8} d x = 25

\int_{0}^{1} 125 x^{11} d x = \frac{125}{12}

= 6 (9 - \frac{45}{2} + 25 - \frac{125}{12})

化简

6 (9 - \frac{45}{2} + 25 - \frac{125}{12}) : \frac{13}{2}

= \frac{13}{2}

\int \frac{1}{x I n x} d x

d er i v a t i v e \frac{4}{3 x + 9}

\int \frac{1}{sinh ^{2} ( x ) + cosh ^{2} ( x )} d x

\int \frac{( ln ( x ) ) ^{23}}{x} d x

\frac{\partial}{\partial x} (2 \cdot ln (x) + y)