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

解

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

- \frac{2}{3}

十進法表記

- 0.66666 \dots

解答ステップ

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

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

\int_{1}^{2} 4 x^{2} d x = \frac{28}{3}

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

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

= - (\frac{15}{2} - \frac{28}{3} - \frac{3}{2} + 4)

結合

\frac{15}{2} - \frac{28}{3} - \frac{3}{2} + 4 : \frac{2}{3}

= - \frac{2}{3}