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

解答

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

- \frac{20}{3}

十进制

- 6.66666 \dots

求解步骤

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

提出常数:

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

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

乘开

x^{2} (2 x^{3} - 1)^{3} : 8 x^{11} - 12 x^{8} + 6 x^{5} - x^{2}

= 2 \cdot \int_{- 1}^{0} 8 x^{11} - 12 x^{8} + 6 x^{5} - x^{2} 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

= 2 (\int_{- 1}^{0} 8 x^{11} d x - \int_{- 1}^{0} 12 x^{8} d x + \int_{- 1}^{0} 6 x^{5} d x - \int_{- 1}^{0} x^{2} d x)

\int_{- 1}^{0} 8 x^{11} d x = - \frac{2}{3}

\int_{- 1}^{0} 12 x^{8} d x = \frac{4}{3}

\int_{- 1}^{0} 6 x^{5} d x = - 1

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

= 2 (- \frac{2}{3} - \frac{4}{3} - 1 - \frac{1}{3})

化简

2 (- \frac{2}{3} - \frac{4}{3} - 1 - \frac{1}{3}) : - \frac{20}{3}

= - \frac{20}{3}

\int_{0}^{12} \frac{1}{x + 12} d x

\int_{- 2}^{1} π (x^{2} + x - 2)^{2} d x

\int_{0}^{5} 2 t d t

\int_{- 1}^{1} x^{2} + 2 x d x

\int_{0}^{25} y + 11 d y