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人気のある微分積分 >

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

解

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

解

\frac{1}{2} arctan (\frac{1}{2})

+1

十進法表記

0.23182 \dots

解答ステップ

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

置換積分法を適用する

= \int_{0}^{\frac{1}{2}} \frac{1}{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

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

共通積分を使用する:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{2} [arctan (u)]_{0}^{\frac{1}{2}}

限度を計算する:

arctan (\frac{1}{2})

= \frac{1}{2} arctan (\frac{1}{2})

グラフ

人気の例

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x \to - 5 lim (\frac{2 x ^{2} + 7 x - 15}{x ^{2} - 25})

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