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人気のある微分積分 >

integral from 0 to 5 of 6e^{2x}

解

\int_{0}^{5} 6 e^{2 x} d x

解

3 (e^{10} - 1)

+1

十進法表記

66076.39738 \dots

解答ステップ

\int_{0}^{5} 6 e^{2 x} d x

定数を除く:

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

= 6 \cdot \int_{0}^{5} e^{2 x} d x

u置換積分法を適用する

= 6 \cdot \int_{0}^{10} e^{u} \frac{1}{2} d u

定数を除く:

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

= 6 \cdot \frac{1}{2} \cdot \int_{0}^{10} e^{u} d u

共通積分を使用する:

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

= 6 \cdot \frac{1}{2} [e^{u}]_{0}^{10}

簡素化

6 \cdot \frac{1}{2} [e^{u}]_{0}^{10} : 3 [e^{u}]_{0}^{10}

= 3 [e^{u}]_{0}^{10}

限度を計算する:

e^{10} - 1

= 3 (e^{10} - 1)

グラフ

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