integral from 0 to 2+y of 1-3z^2

解答

\int_{0}^{2 + y} 1 - 3 z^{2} d z

y + 2 - (y + 2)^{3}

求解步骤

\int_{0}^{2 + y} 1 - 3 z^{2} d z

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_{0}^{2 + y} 1 d z - \int_{0}^{2 + y} 3 z^{2} d z

\int_{0}^{2 + y} 1 d z = y + 2

\int_{0}^{2 + y} 3 z^{2} d z = (y + 2)^{3}

= y + 2 - (y + 2)^{3}

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