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人気のある微分積分 >

integral from 0 to 1 of (2xe^{x^2})

解

\int_{0}^{1} (2 x e^{x^{2}}) d x

解

e - 1

+1

十進法表記

1.71828 \dots

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

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

定数を除く:

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

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

共通積分を使用する:

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

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

簡素化

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

= [e^{u}]_{0}^{1}

限度を計算する:

e - 1

= e - 1

グラフ

人気の例

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\int_{0}^{16} 1 d x

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