integral from 1 to 3 of (x^2-4x)

解答

\int_{1}^{3} (x^{2} - 4 x) d x

- \frac{22}{3}

十进制

- 7.33333 \dots

求解步骤

\int_{1}^{3} x^{2} - 4 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_{1}^{3} x^{2} d x - \int_{1}^{3} 4 x d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 4 x d x = 16

= \frac{26}{3} - 16

化简

\frac{26}{3} - 16 : - \frac{22}{3}

= - \frac{22}{3}

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