integral from 0 to y of 1/4 (x+y)

解答

\int_{0}^{y} \frac{1}{4} (x + y) d x

\frac{3 y ^{2}}{8}

求解步骤

\int_{0}^{y} \frac{1}{4} (x + y) d x

提出常数:

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

= \frac{1}{4} \cdot \int_{0}^{y} 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}{4} (\int_{0}^{y} y d x + \int_{0}^{y} x d x)

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

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

= \frac{1}{4} (y^{2} + \frac{y ^{2}}{2})

化简

\frac{1}{4} (y^{2} + \frac{y ^{2}}{2}) : \frac{3 y ^{2}}{8}

= \frac{3 y ^{2}}{8}

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