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人気のある微分積分 >

integral from 0 to 1 of e^{2t}

解

\int_{0}^{1} e^{2 t} d t

解

\frac{e ^{2} - 1}{2}

+1

十進法表記

3.19452 \dots

解答ステップ

\int_{0}^{1} e^{2 t} d t

u置換積分法を適用する

= \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

= \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}

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

限度を計算する:

e^{2} - 1

= \frac{1}{2} (e^{2} - 1)

簡素化

= \frac{e ^{2} - 1}{2}

グラフ

人気の例

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\int_{1}^{2} 3^{x} d x

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