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人気のある微分積分 >

integral from 0 to 1 of x(4+2x^2)^3

解

\int_{0}^{1} x (4 + 2 x^{2})^{3} d x

解

65

解答ステップ

\int_{0}^{1} x (4 + 2 x^{2})^{3} d x

u置換積分法を適用する

= \int_{4}^{6} \frac{u ^{3}}{4} d u

定数を除く:

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

= \frac{1}{4} \cdot \int_{4}^{6} u^{3} d u

べき乗則を適用する

= \frac{1}{4} [\frac{u ^{4}}{4}]_{4}^{6}

限度を計算する:

260

= \frac{1}{4} \cdot 260

簡素化

= 65

グラフ

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