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人気のある微分積分 >

integral from 0 to 1 of 2x^2sqrt(x^3+1)

解

\int_{0}^{1} 2 x^{2} x^{3} + 1 d x

解

\frac{2 ( 4 2 - 2 )}{9}

+1

十進法表記

0.81263 \dots

解答ステップ

\int_{0}^{1} 2 x^{2} x^{3} + 1 d x

定数を除く:

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

= 2 \cdot \int_{0}^{1} x^{2} x^{3} + 1 d x

u置換積分法を適用する

= 2 \cdot \int_{1}^{2} \frac{u}{3} d u

定数を除く:

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

= 2 \cdot \frac{1}{3} \cdot \int_{1}^{2} u d u

べき乗則を適用する

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

簡素化

2 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{2} : \frac{2}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{2}

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

限度を計算する:

\frac{4 2}{3} - \frac{2}{3}

= \frac{2}{3} (\frac{4 2}{3} - \frac{2}{3})

簡素化

= \frac{2 ( 4 2 - 2 )}{9}

グラフ

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