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人気のある微分積分 >

integral from 0 to 1 of x 4/(pi(1+x^2))

解

\int_{0}^{1} x \frac{4}{π ( 1 + x ^{2} )} d x

解

\frac{2}{π} ln (2)

+1

十進法表記

0.44127 \dots

解答ステップ

\int_{0}^{1} x \frac{4}{π ( 1 + x ^{2} )} d x

定数を除く:

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

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

簡素化

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

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

u置換積分法を適用する

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

定数を除く:

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

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

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{4}{π} \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2}

簡素化

\frac{4}{π} \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2} : \frac{2}{π} [ln ∣ u ∣]_{1}^{2}

= \frac{2}{π} [ln ∣ u ∣]_{1}^{2}

限度を計算する:

ln (2)

= \frac{2}{π} ln (2)

グラフ

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