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

解

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

\frac{125}{6}

十進法表記

20.83333 \dots

解答ステップ

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

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

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

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

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

簡素化

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

= \frac{125}{6}