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

解答

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

- \frac{2}{3}

十进制

- 0.66666 \dots

求解步骤

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

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

\int_{1}^{3} 4 d x = 8

= - \frac{26}{3} + 8

化简

- \frac{26}{3} + 8 : - \frac{2}{3}

= - \frac{2}{3}

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