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人気のある微分積分 >

integral from 0 to 1 of 6xsqrt(x^2+1)

解

\int_{0}^{1} 6 x x^{2} + 1 d x

解

2 (22 - 1)

+1

十進法表記

3.65685 \dots

解答ステップ

\int_{0}^{1} 6 x x^{2} + 1 d x

定数を除く:

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

= 6 \cdot \int_{0}^{1} x x^{2} + 1 d x

u置換積分法を適用する

= 6 \cdot \int_{1}^{2} u^{2} d u

べき乗則を適用する

= 6 [\frac{u ^{3}}{3}]_{1}^{2}

限度を計算する:

\frac{2 2}{3} - \frac{1}{3}

= 6 (\frac{2 2}{3} - \frac{1}{3})

簡素化

= 2 (22 - 1)

グラフ

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