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人気のある微分積分 >

integral from 0 to 10 of te^{-0.2t}

解

\int_{0}^{10} t e^{- 0.2 t} d t

解

- 25 (- 1 + \frac{3}{e ^{2}})

+1

十進法表記

14.84985 \dots

解答ステップ

\int_{0}^{10} t e^{- 0.2 t} d t

u置換積分法を適用する

= \int_{0}^{- 2} \frac{5}{0.2} e^{u} u d u

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

= - \int_{- 2}^{0} \frac{5}{0.2} e^{u} u d u

定数を除く:

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

= - \frac{5}{0.2} \cdot \int_{- 2}^{0} e^{u} u d u

簡素化

= - 25 \cdot \int_{- 2}^{0} e^{u} u d u

部分積分を適用する

= - 25 [e^{u} u - \int e^{u} d u]_{- 2}^{0}

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

= - 25 [e^{u} u - e^{u}]_{- 2}^{0}

限度を計算する:

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

= - 25 (- 1 + \frac{3}{e ^{2}})

グラフ

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