integral from-1 to 4 of |6x^2-x^3-5x|

解

\int_{- 1}^{4} 6 x^{2} - x^{3} - 5 x d x

\frac{121}{4}

十進法表記

30.25

解答ステップ

\int_{- 1}^{4} 6 x^{2} - x^{3} - 5 x d x

絶対数を排除する

= \int_{- 1}^{0} 6 x^{2} - x^{3} - 5 x d x + \int_{0}^{1} - 6 x^{2} + x^{3} + 5 x d x + \int_{1}^{4} 6 x^{2} - x^{3} - 5 x d x

\int_{- 1}^{0} 6 x^{2} - x^{3} - 5 x d x = \frac{19}{4}

\int_{0}^{1} - 6 x^{2} + x^{3} + 5 x d x = \frac{3}{4}

\int_{1}^{4} 6 x^{2} - x^{3} - 5 x d x = \frac{99}{4}

= \frac{19}{4} + \frac{3}{4} + \frac{99}{4}

簡素化

\frac{19}{4} + \frac{3}{4} + \frac{99}{4} : \frac{121}{4}

= \frac{121}{4}