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人気のある微分積分 >

integral from 16 to 25 of (sqrt(x-16))/x

解

\int_{16}^{25} \frac{x - 16}{x} d x

解

8 (- arctan (\frac{3}{4}) + \frac{3}{4})

+1

十進法表記

0.85199 \dots

解答ステップ

\int_{16}^{25} \frac{x - 16}{x} d x

u置換積分法を適用する

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

\frac{u ^{2}}{u ^{2} + 16} = - \frac{16}{u ^{2} + 16} + 1

= 2 \cdot \int_{0}^{3} - \frac{16}{u ^{2} + 16} + 1 d u

置換積分法を適用する

= 2 \cdot \int_{0}^{\frac{3}{4}} 4 (- \frac{1}{v ^{2} + 1} + 1) d v

定数を除く:

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

= 2 \cdot 4 \cdot \int_{0}^{\frac{3}{4}} - \frac{1}{v ^{2} + 1} + 1 d v

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 \cdot 4 (- \int_{0}^{\frac{3}{4}} \frac{1}{v ^{2} + 1} d v + \int_{0}^{\frac{3}{4}} 1 d v)

\int_{0}^{\frac{3}{4}} \frac{1}{v ^{2} + 1} d v = arctan (\frac{3}{4})

\int_{0}^{\frac{3}{4}} 1 d v = \frac{3}{4}

= 2 \cdot 4 (- arctan (\frac{3}{4}) + \frac{3}{4})

簡素化

= 8 (- arctan (\frac{3}{4}) + \frac{3}{4})

グラフ

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