integral from 0 to 1 of-x^3+4x^2-3x

解

\int_{0}^{1} - x^{3} + 4 x^{2} - 3 x d x

- \frac{5}{12}

十進法表記

- 0.41666 \dots

解答ステップ

\int_{0}^{1} - x^{3} + 4 x^{2} - 3 x 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_{0}^{1} x^{3} d x + \int_{0}^{1} 4 x^{2} d x - \int_{0}^{1} 3 x d x

\int_{0}^{1} x^{3} d x = \frac{1}{4}

\int_{0}^{1} 4 x^{2} d x = \frac{4}{3}

\int_{0}^{1} 3 x d x = \frac{3}{2}

= - \frac{1}{4} + \frac{4}{3} - \frac{3}{2}

簡素化

- \frac{1}{4} + \frac{4}{3} - \frac{3}{2} : - \frac{5}{12}

= - \frac{5}{12}