integral of (4-2x)^3

解答

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

- \frac{1}{8} (4 - 2 x)^{4} + C

求解步骤

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

使用换元积分法

= \int - \frac{u ^{3}}{2} d u

提出常数:

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

= - \frac{1}{2} \cdot \int u^{3} d u

使用幂法则

= - \frac{1}{2} \cdot \frac{u ^{4}}{4}

u = 4 - 2 x

代回

= - \frac{1}{2} \cdot \frac{( 4 - 2 x ) ^{4}}{4}

化简

- \frac{1}{2} \cdot \frac{( 4 - 2 x ) ^{4}}{4} : - \frac{1}{8} (4 - 2 x)^{4}

= - \frac{1}{8} (4 - 2 x)^{4}

解答补常数

= - \frac{1}{8} (4 - 2 x)^{4} + C

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