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

解答

\int 1 x (4 - 2 (x - 3)) d x

- \frac{2 x ^{3}}{3} + 5 x^{2} + C

求解步骤

\int 1 \cdot x (4 - 2 (x - 3)) d x

化简

1 \cdot x (4 - 2 (x - 3)) : x (- 2 x + 10)

= \int x (- 2 x + 10) d x

乘开

x (- 2 x + 10) : - 2 x^{2} + 10 x

= \int - 2 x^{2} + 10 x 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

= - \int 2 x^{2} d x + \int 10 x d x

\int 2 x^{2} d x = \frac{2 x ^{3}}{3}

\int 10 x d x = 5 x^{2}

= - \frac{2 x ^{3}}{3} + 5 x^{2}

解答补常数

= - \frac{2 x ^{3}}{3} + 5 x^{2} + C

s l o p e (- 5, - 1), (- 1, 5)

x \to 3 - lim (\frac{2}{3 - x})

\frac{d}{d x} (2 x + 3 x)

\int 7 sin (\frac{x}{3}) d x

y^{^{'}} = \frac{x ^{4} y ^{3} + 2 x ^{4}}{y ^{2}}