integral from 1 to 3 of \sqrt[3]{x}(x-4)

解

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

\frac{27 3 3 - 3}{7} - 933 + 3

十進法表記

- 4.84585 \dots

解答ステップ

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

拡張

3 x (x - 4) : x^{\frac{4}{3}} - 4 3 x

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

\int_{1}^{3} x^{\frac{4}{3}} d x = \frac{27 3 3 - 3}{7}

\int_{1}^{3} 4 3 x d x = 933 - 3

= \frac{27 3 3 - 3}{7} - (933 - 3)

簡素化

= \frac{27 3 3 - 3}{7} - 933 + 3