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人気のある微分積分 >

integral from 0 to 1 of 4e^{6sqrt(x)}

解

\int_{0}^{1} 4 e^{6 x} d x

解

\frac{2 ( 5 e ^{6} + 1 )}{9}

+1

十進法表記

448.47643 \dots

解答ステップ

\int_{0}^{1} 4 e^{6 x} d x

定数を除く:

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

= 4 \cdot \int_{0}^{1} e^{6 x} d x

u置換積分法を適用する

= 4 \cdot \int_{0}^{6} \frac{e ^{u} u}{18} d u

定数を除く:

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

= 4 \cdot \frac{1}{18} \cdot \int_{0}^{6} e^{u} u d u

部分積分を適用する

= 4 \cdot \frac{1}{18} [e^{u} u - \int e^{u} d u]_{0}^{6}

\int e^{u} d u = e^{u}

= 4 \cdot \frac{1}{18} [e^{u} u - e^{u}]_{0}^{6}

簡素化

4 \cdot \frac{1}{18} [e^{u} u - e^{u}]_{0}^{6} : \frac{2}{9} [e^{u} u - e^{u}]_{0}^{6}

= \frac{2}{9} [e^{u} u - e^{u}]_{0}^{6}

限度を計算する:

5 e^{6} + 1

= \frac{2}{9} (5 e^{6} + 1)

簡素化

= \frac{2 ( 5 e ^{6} + 1 )}{9}

グラフ

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