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

解

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

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

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

\int_{- 1}^{4} 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}