integral from 0 to 4 of x^2-8x+16

解

\int_{0}^{4} x^{2} - 8 x + 16 d x

\frac{64}{3}

十進法表記

21.33333 \dots

解答ステップ

\int_{0}^{4} x^{2} - 8 x + 16 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} 8 x d x + \int_{0}^{4} 16 d x

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

\int_{0}^{4} 8 x d x = 64

\int_{0}^{4} 16 d x = 64

= \frac{64}{3} - 64 + 64

簡素化

= \frac{64}{3}