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人気のある微分積分 >

integral from 0 to 3 of 1/2 e^{-x/2}

解

\int_{0}^{3} \frac{1}{2} e^{- \frac{x}{2}} d x

解

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

+1

十進法表記

0.77686 \dots

解答ステップ

\int_{0}^{3} \frac{1}{2} e^{- \frac{x}{2}} d x

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}^{3} e^{- \frac{x}{2}} d x

u置換積分法を適用する

= \frac{1}{2} \cdot \int_{0}^{- \frac{3}{2}} - 2 e^{u} d u

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

= \frac{1}{2} (- \int_{- \frac{3}{2}}^{0} - 2 e^{u} d u)

定数を除く:

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

= \frac{1}{2} (- (- 2 \cdot \int_{- \frac{3}{2}}^{0} e^{u} d u))

共通積分を使用する:

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

= \frac{1}{2} (- (- 2 [e^{u}]_{- \frac{3}{2}}^{0}))

簡素化

\frac{1}{2} (- (- 2 [e^{u}]_{- \frac{3}{2}}^{0})) : [e^{u}]_{- \frac{3}{2}}^{0}

= [e^{u}]_{- \frac{3}{2}}^{0}

限度を計算する:

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

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

グラフ

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