integral from 0 to 2 of (6-x-y)/8

解答

\int_{0}^{2} \frac{6 - x - y}{8} d x

\frac{- y + 5}{4}

求解步骤

\int_{0}^{2} \frac{6 - x - y}{8} d x

提出常数:

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

= \frac{1}{8} \cdot \int_{0}^{2} 6 - y - 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

= \frac{1}{8} (\int_{0}^{2} 6 d x - \int_{0}^{2} y d x - \int_{0}^{2} x d x)

\int_{0}^{2} 6 d x = 12

\int_{0}^{2} y d x = 2 y

\int_{0}^{2} x d x = 2

= \frac{1}{8} (12 - 2 y - 2)

化简

\frac{1}{8} (12 - 2 y - 2) : \frac{- y + 5}{4}

= \frac{- y + 5}{4}

\int_{x}^{x^{4}} t^{3} d t

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