integral from-3 to x of 1/3 (1+x/3)

解

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

\frac{1}{18} (x^{2} + 6 x + 9)

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{- 3}^{x} 1 + \frac{x}{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

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

\int_{- 3}^{x} 1 d x = x + 3

\int_{- 3}^{x} \frac{x}{3} d x = \frac{x ^{2} - 9}{6}

= \frac{1}{3} (x + 3 + \frac{x ^{2} - 9}{6})

簡素化

\frac{1}{3} (x + 3 + \frac{x ^{2} - 9}{6}) : \frac{1}{18} (x^{2} + 6 x + 9)

= \frac{1}{18} (x^{2} + 6 x + 9)