integral from 0 to 4 of (x+2)^2

解

\int_{0}^{4} (x + 2)^{2} d x

\frac{208}{3}

十進法表記

69.33333 \dots

解答ステップ

\int_{0}^{4} (x + 2)^{2} d x

拡張

(x + 2)^{2} : x^{2} + 4 x + 4

= \int_{0}^{4} x^{2} + 4 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_{0}^{4} x^{2} d x + \int_{0}^{4} 4 x d x + \int_{0}^{4} 4 d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 4 x d x = 32

\int_{0}^{4} 4 d x = 16

= \frac{64}{3} + 32 + 16

簡素化

\frac{64}{3} + 32 + 16 : \frac{208}{3}

= \frac{208}{3}