integral from 0 to 1/4 of yarctan(4y)

解

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

\frac{2 π - 4}{128}

十進法表記

0.01783 \dots

解答ステップ

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

部分積分を適用する

= [\frac{1}{2} (y^{2} arctan (4 y) - 2 \cdot \int \frac{2 y ^{2}}{16 y ^{2} + 1} d y)]_{0}^{\frac{1}{4}}

\int \frac{2 y ^{2}}{16 y ^{2} + 1} d y = \frac{1}{32} (- arctan (4 y) + 4 y)

= [\frac{1}{2} (y^{2} arctan (4 y) - 2 \cdot \frac{1}{32} (- arctan (4 y) + 4 y))]_{0}^{\frac{1}{4}}

2 \cdot \frac{1}{32} (- arctan (4 y) + 4 y) = \frac{1}{16} (- arctan (4 y) + 4 y)

= [\frac{1}{2} (y^{2} arctan (4 y) - \frac{1}{16} (- arctan (4 y) + 4 y))]_{0}^{\frac{1}{4}}

限度を計算する:

\frac{2 π - 4}{128}

= \frac{2 π - 4}{128}