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

解答

\int_{- 1}^{0} π (1 - x^{2})^{2} d x

π \frac{8}{15}

十进制

1.67551 \dots

求解步骤

\int_{- 1}^{0} π (1 - x^{2})^{2} d x

提出常数:

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

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

乘开

(1 - x^{2})^{2} : 1 - 2 x^{2} + x^{4}

= π \cdot \int_{- 1}^{0} 1 - 2 x^{2} + x^{4} 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

= π (\int_{- 1}^{0} 1 d x - \int_{- 1}^{0} 2 x^{2} d x + \int_{- 1}^{0} x^{4} d x)

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

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

\int_{- 1}^{0} x^{4} d x = \frac{1}{5}

= π (1 - \frac{2}{3} + \frac{1}{5})

化简

1 - \frac{2}{3} + \frac{1}{5} : \frac{8}{15}

= π \frac{8}{15}

t an g e n t f (x) = 2 x^{3} - x^{2} + 3, (2, 15)

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