integral from 1 to 2 of (sqrt(x-1))/x

解

\int_{1}^{2} \frac{x - 1}{x} d x

2 (- \frac{π}{4} + 1)

十進法表記

0.42920 \dots

解答ステップ

\int_{1}^{2} \frac{x - 1}{x} d x

u置換積分法を適用する

= \int_{0}^{1} \frac{2 u ^{2}}{u ^{2} + 1} d u

定数を除く:

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

= 2 \cdot \int_{0}^{1} \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= 2 \cdot \int_{0}^{1} - \frac{1}{u ^{2} + 1} + 1 d u

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

= 2 (- \int_{0}^{1} \frac{1}{u ^{2} + 1} d u + \int_{0}^{1} 1 d u)

\int_{0}^{1} \frac{1}{u ^{2} + 1} d u = \frac{π}{4}

\int_{0}^{1} 1 d u = 1

= 2 (- \frac{π}{4} + 1)