integral of 64x^3(4x^2-1)

解答

\int 64 x^{3} (4 x^{2} - 1) d x

64 (\frac{2 x ^{6}}{3} - \frac{x ^{4}}{4}) + C

求解步骤

\int 64 x^{3} (4 x^{2} - 1) d x

提出常数:

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

= 64 \cdot \int x^{3} (4 x^{2} - 1) d x

乘开

x^{3} (4 x^{2} - 1) : 4 x^{5} - x^{3}

= 64 \cdot \int 4 x^{5} - x^{3} 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

= 64 (\int 4 x^{5} d x - \int x^{3} d x)

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

\int x^{3} d x = \frac{x ^{4}}{4}

= 64 (\frac{2 x ^{6}}{3} - \frac{x ^{4}}{4})

解答补常数

= 64 (\frac{2 x ^{6}}{3} - \frac{x ^{4}}{4}) + C

x \to - 7 lim (h (x))

a re a y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{20}

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

x \to 5 lim (\frac{x + 3}{5 - x})

d er i v a t i v e \frac{1}{3 x - 2}