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

解答

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

\frac{125}{6}

十进制

20.83333 \dots

求解步骤

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

\int_{- 1}^{4} 3 x d x = \frac{45}{2}

\int_{- 1}^{4} 4 d x = 20

\int_{- 1}^{4} x^{2} d x = \frac{65}{3}

= \frac{45}{2} + 20 - \frac{65}{3}

化简

\frac{45}{2} + 20 - \frac{65}{3} : \frac{125}{6}

= \frac{125}{6}

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