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人気のある微分積分 >

integral from 0 to 2 of (15sqrt(4x+1))

解

\int_{0}^{2} (15 4 x + 1) d x

解

65

解答ステップ

\int_{0}^{2} 15 4 x + 1 d x

定数を除く:

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

= 15 \cdot \int_{0}^{2} 4 x + 1 d x

u置換積分法を適用する

= 15 \cdot \int_{1}^{9} \frac{u}{4} d u

定数を除く:

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

= 15 \cdot \frac{1}{4} \cdot \int_{1}^{9} u d u

べき乗則を適用する

= 15 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

簡素化

15 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9} : \frac{15}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= \frac{15}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

限度を計算する:

\frac{52}{3}

= \frac{15}{4} \cdot \frac{52}{3}

簡素化

= 65

グラフ

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