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人気のある微分積分 >

integral from 0 to 1 of 8/(x^2+1)

解

\int_{0}^{1} \frac{8}{x ^{2} + 1} d x

解

2 π

+1

十進法表記

6.28318 \dots

解答ステップ

\int_{0}^{1} \frac{8}{x ^{2} + 1} d x

定数を除く:

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

= 8 \cdot \int_{0}^{1} \frac{1}{x ^{2} + 1} d x

共通積分を使用する:

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

= 8 [arctan (x)]_{0}^{1}

限度を計算する:

\frac{π}{4}

= 8 \cdot \frac{π}{4}

簡素化

= 2 π

グラフ

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