integral from-2 to 1 of (6-x^2)-(x+4)

解答

\int_{- 2}^{1} (6 - x^{2}) - (x + 4) d x

\frac{9}{2}

十进制

4.5

求解步骤

\int_{- 2}^{1} 6 - x^{2} - (x + 4) d x

化简

= \int_{- 2}^{1} - x^{2} - x + 2 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_{- 2}^{1} x^{2} d x - \int_{- 2}^{1} x d x + \int_{- 2}^{1} 2 d x

\int_{- 2}^{1} x^{2} d x = 3

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

\int_{- 2}^{1} 2 d x = 6

= - 3 - (- \frac{3}{2}) + 6

化简

- 3 - (- \frac{3}{2}) + 6 : \frac{9}{2}

= \frac{9}{2}

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