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

解

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

- \frac{13}{12}

十進法表記

- 1.08333 \dots

解答ステップ

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

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

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

簡素化

\frac{1}{4} - \frac{4}{3} : - \frac{13}{12}

= - \frac{13}{12}