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

解

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

15

解答ステップ

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

\int_{- 2}^{1} 2 x^{3} d x = - \frac{15}{2}

\int_{- 2}^{1} 4 x^{2} d x = 12

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

\int_{- 2}^{1} 3 d x = 9

= - \frac{15}{2} + 12 - (- \frac{3}{2}) + 9

簡素化

- \frac{15}{2} + 12 - (- \frac{3}{2}) + 9 : 15

= 15