integral from 0 to 2 of (2x^2+1)

解

\int_{0}^{2} (2 x^{2} + 1) d x

\frac{22}{3}

十進法表記

7.33333 \dots

解答ステップ

\int_{0}^{2} 2 x^{2} + 1 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}^{2} 2 x^{2} d x + \int_{0}^{2} 1 d x

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

\int_{0}^{2} 1 d x = 2

= \frac{16}{3} + 2

簡素化

\frac{16}{3} + 2 : \frac{22}{3}

= \frac{22}{3}