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

解

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

\frac{125}{6}

十進法表記

20.83333 \dots

解答ステップ

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

総和規則を適用する:

\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}

簡素化

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

= \frac{125}{6}