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人気のある微分積分 >

integral from 1 to e of (x^2+1)/(2x)

解

\int_{1}^{e} \frac{x ^{2} + 1}{2 x} d x

解

\frac{e ^{2} + 1}{4}

+1

十進法表記

2.09726 \dots

解答ステップ

\int_{1}^{e} \frac{x ^{2} + 1}{2 x} d x

定数を除く:

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

= \frac{1}{2} \cdot \int_{1}^{e} \frac{x ^{2} + 1}{x} d x

拡張

\frac{x ^{2} + 1}{x} : x + \frac{1}{x}

= \frac{1}{2} \cdot \int_{1}^{e} x + \frac{1}{x} 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}{2} (\int_{1}^{e} x d x + \int_{1}^{e} \frac{1}{x} d x)

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

\int_{1}^{e} \frac{1}{x} d x = 1

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

簡素化

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

= \frac{e ^{2} + 1}{4}

グラフ

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