integral from 2 to y of 1/2 (r-2)

解

\int_{2}^{y} \frac{1}{2} (r - 2) d r

\frac{y ^{2} - 4 y + 4}{4}

解答ステップ

\int_{2}^{y} \frac{1}{2} (r - 2) d r

定数を除く:

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

= \frac{1}{2} \cdot \int_{2}^{y} - 2 + r d r

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}{2} (- \int_{2}^{y} 2 d r + \int_{2}^{y} r d r)

\int_{2}^{y} 2 d r = 2 y - 4

\int_{2}^{y} r d r = \frac{y ^{2}}{2} - 2

= \frac{1}{2} (- (2 y - 4) + \frac{y ^{2}}{2} - 2)

簡素化

\frac{1}{2} (- (2 y - 4) + \frac{y ^{2}}{2} - 2) : \frac{y ^{2} - 4 y + 4}{4}

= \frac{y ^{2} - 4 y + 4}{4}