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

解

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

\frac{13382}{5}

十進法表記

2676.4

解答ステップ

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

\int_{1}^{3} (x + 4)^{4} d x = \frac{13682}{5}

\int_{1}^{3} (x + 1)^{3} d x = 60

= \frac{13682}{5} - 60

簡素化

\frac{13682}{5} - 60 : \frac{13382}{5}

= \frac{13382}{5}