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人気のある微分積分 >

integral from 0 to 2 of 8e^{-3x}

解

\int_{0}^{2} 8 e^{- 3 x} d x

解

\frac{8 ( e ^{6} - 1 )}{3 e ^{6}}

+1

十進法表記

2.66005 \dots

解答ステップ

\int_{0}^{2} 8 e^{- 3 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}^{2} e^{- 3 x} d x

u置換積分法を適用する

= 8 \cdot \int_{0}^{- 6} - \frac{1}{3} e^{u} d u

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

= 8 (- \int_{- 6}^{0} - \frac{1}{3} e^{u} d u)

定数を除く:

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

= 8 (- (- \frac{1}{3} \cdot \int_{- 6}^{0} e^{u} d u))

共通積分を使用する:

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

= 8 (- (- \frac{1}{3} [e^{u}]_{- 6}^{0}))

簡素化

8 (- (- \frac{1}{3} [e^{u}]_{- 6}^{0})) : \frac{8}{3} [e^{u}]_{- 6}^{0}

= \frac{8}{3} [e^{u}]_{- 6}^{0}

限度を計算する:

1 - \frac{1}{e ^{6}}

= \frac{8}{3} (1 - \frac{1}{e ^{6}})

簡素化

= \frac{8 ( e ^{6} - 1 )}{3 e ^{6}}

グラフ

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