integral from 0 to 4 of (x^2-sqrt(x))

解答

\int_{0}^{4} (x^{2} - x) d x

16

求解步骤

\int_{0}^{4} x^{2} - 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_{0}^{4} x^{2} d x - \int_{0}^{4} x d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} x d x = \frac{16}{3}

= \frac{64}{3} - \frac{16}{3}

化简

\frac{64}{3} - \frac{16}{3} : 16

= 16

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