integral from-1 to 1 of pi(4-4x^2)^2

解答

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

π \frac{256}{15}

十进制

53.61651 \dots

求解步骤

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

乘开

(4 - 4 x^{2})^{2} : 16 - 32 x^{2} + 16 x^{4}

= π \cdot \int_{- 1}^{1} 16 - 32 x^{2} + 16 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}^{1} 16 d x - \int_{- 1}^{1} 32 x^{2} d x + \int_{- 1}^{1} 16 x^{4} d x)

\int_{- 1}^{1} 16 d x = 32

\int_{- 1}^{1} 32 x^{2} d x = \frac{64}{3}

\int_{- 1}^{1} 16 x^{4} d x = \frac{32}{5}

= π (32 - \frac{64}{3} + \frac{32}{5})

化简

32 - \frac{64}{3} + \frac{32}{5} : \frac{256}{15}

= π \frac{256}{15}

t an g e n t y = x^{4} + 2 x - 1, (1, 2)

\frac{d}{d x} (x^{5} - 5^{x})

t an g e n t f (x) = e^{x} cos (x)

t an g e n t \frac{x}{x + 6}

\int \frac{x ^{2} - 1}{x ( x + 1 ) ^{3}} d x