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人気のある微分積分 >

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

解

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

解

\frac{1}{3}

+1

十進法表記

0.33333 \dots

解答ステップ

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

定数を除く:

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

= \frac{1}{4} \cdot \int_{- 1}^{1} x (2 x + 2) d x

部分積分を適用する

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

\int x^{2} d x = \frac{x ^{3}}{3}

= \frac{1}{4} [x^{2} (x + 1) - \frac{x ^{3}}{3}]_{- 1}^{1}

限度を計算する:

\frac{4}{3}

= \frac{1}{4} \cdot \frac{4}{3}

簡素化

= \frac{1}{3}

グラフ

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