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人気のある微分積分 >

integral from 0 to 1 of 8e^{2x}

解

\int_{0}^{1} 8 e^{2 x} d x

解

4 (e^{2} - 1)

+1

十進法表記

25.55622 \dots

解答ステップ

\int_{0}^{1} 8 e^{2 x} d x

定数を除く:

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

= 8 \cdot \int_{0}^{1} e^{2 x} d x

u置換積分法を適用する

= 8 \cdot \int_{0}^{2} 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

= 8 \cdot \frac{1}{2} \cdot \int_{0}^{2} e^{u} d u

共通積分を使用する:

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

= 8 \cdot \frac{1}{2} [e^{u}]_{0}^{2}

簡素化

8 \cdot \frac{1}{2} [e^{u}]_{0}^{2} : 4 [e^{u}]_{0}^{2}

= 4 [e^{u}]_{0}^{2}

限度を計算する:

e^{2} - 1

= 4 (e^{2} - 1)

グラフ

人気の例

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