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人気のある微分積分 >

integral from 0 to 1 of 2te^{-2t}

解

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

解

- \frac{- e ^{2} + 3}{2 e ^{2}}

+1

十進法表記

0.29699 \dots

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

= 2 \cdot \int_{0}^{- 2} \frac{e ^{u} u}{4} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{- 2}^{0} \frac{e ^{u} u}{4} d u)

定数を除く:

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

= 2 (- \frac{1}{4} \cdot \int_{- 2}^{0} e^{u} u d u)

部分積分を適用する

= 2 (- \frac{1}{4} [e^{u} u - \int e^{u} d u]_{- 2}^{0})

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

= 2 (- \frac{1}{4} [e^{u} u - e^{u}]_{- 2}^{0})

簡素化

2 (- \frac{1}{4} [e^{u} u - e^{u}]_{- 2}^{0}) : - \frac{1}{2} [e^{u} u - e^{u}]_{- 2}^{0}

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

限度を計算する:

- 1 + \frac{3}{e ^{2}}

= - \frac{1}{2} (- 1 + \frac{3}{e ^{2}})

簡素化

= - \frac{- e ^{2} + 3}{2 e ^{2}}

グラフ

人気の例

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