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人気のある微分積分 >

integral from-1 to 0 of 9x^2(2x^3+4)^2

解

\int_{- 1}^{0} 9 x^{2} (2 x^{3} + 4)^{2} d x

解

28

解答ステップ

\int_{- 1}^{0} 9 x^{2} (2 x^{3} + 4)^{2} d x

定数を除く:

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

= 9 \cdot \int_{- 1}^{0} x^{2} (2 x^{3} + 4)^{2} d x

u置換積分法を適用する

= 9 \cdot \int_{2}^{4} \frac{u ^{2}}{6} d u

定数を除く:

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

= 9 \cdot \frac{1}{6} \cdot \int_{2}^{4} u^{2} d u

べき乗則を適用する

= 9 \cdot \frac{1}{6} [\frac{u ^{3}}{3}]_{2}^{4}

簡素化

9 \cdot \frac{1}{6} [\frac{u ^{3}}{3}]_{2}^{4} : \frac{3}{2} [\frac{u ^{3}}{3}]_{2}^{4}

= \frac{3}{2} [\frac{u ^{3}}{3}]_{2}^{4}

限度を計算する:

\frac{56}{3}

= \frac{3}{2} \cdot \frac{56}{3}

簡素化

= 28

グラフ

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