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人気のある微分積分 >

integral from 0 to 1 of 4xarctan(x)

解

\int_{0}^{1} 4 x arctan (x) d x

解

π - 2

+1

十進法表記

1.14159 \dots

解答ステップ

\int_{0}^{1} 4 x arctan (x) d x

定数を除く:

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

= 4 \cdot \int_{0}^{1} x arctan (x) d x

部分積分を適用する

= 4 [\frac{1}{2} x^{2} arctan (x) - \int \frac{x ^{2}}{2 ( x ^{2} + 1 )} d x]_{0}^{1}

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

= 4 [\frac{1}{2} x^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)]_{0}^{1}

簡素化

4 [\frac{1}{2} x^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)]_{0}^{1} : 4 [\frac{1}{2} (x^{2} arctan (x) + arctan (x) - x)]_{0}^{1}

= 4 [\frac{1}{2} (x^{2} arctan (x) + arctan (x) - x)]_{0}^{1}

限度を計算する:

\frac{π - 2}{4}

= 4 \cdot \frac{π - 2}{4}

簡素化

= π - 2

グラフ

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