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人気のある微分積分 >

integral from 0 to 4 of 1/(1+sqrt(u))

解

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

解

2 (2 - ln (3))

+1

十進法表記

1.80277 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{3} \frac{2 ( v - 1 )}{v} d v

定数を除く:

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

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

拡張

\frac{v - 1}{v} : 1 - \frac{1}{v}

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

\int_{1}^{3} 1 d v = 2

\int_{1}^{3} \frac{1}{v} d v = ln (3)

= 2 (2 - ln (3))

グラフ

人気の例

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\int_{3}^{x} \frac{sin ( t )}{t} d t

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