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

解答

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

\frac{128}{5}

十进制

25.6

求解步骤

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

提出常数:

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

= \frac{1}{2} \cdot \int_{- 2}^{2} 4^{2} - x^{4} d x

化简

= \frac{1}{2} \cdot \int_{- 2}^{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

= \frac{1}{2} (\int_{- 2}^{2} 16 d x - \int_{- 2}^{2} x^{4} d x)

\int_{- 2}^{2} 16 d x = 64

\int_{- 2}^{2} x^{4} d x = \frac{64}{5}

= \frac{1}{2} (64 - \frac{64}{5})

化简

\frac{1}{2} (64 - \frac{64}{5}) : \frac{128}{5}

= \frac{128}{5}

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

l a pl a ce t r an s f or m - 3 sin (3 t)

\frac{d}{d x} (sin^{- 2} (x))

x \to 6 + lim (\frac{1}{x - 6})

\int \frac{1}{( x + 4 ) ^{2} ( x - 2 )} d x