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人気のある微分積分 >

integral from-1 to 0 of 2x^{17}e^{1+x^9}

解

\int_{- 1}^{0} 2 x^{17} e^{1 + x^{9}} d x

解

\frac{2 ( - e + 2 )}{9}

+1

十進法表記

- 0.15961 \dots

解答ステップ

\int_{- 1}^{0} 2 x^{17} e^{1 + x^{9}} d x

定数を除く:

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

= 2 \cdot \int_{- 1}^{0} x^{17} e^{1 + x^{9}} d x

x^{17} e^{1 + x^{9}} = e^{1} x^{17} e^{x^{9}}

= 2 \cdot \int_{- 1}^{0} e^{1} x^{17} e^{x^{9}} d x

定数を除く:

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

= 2 e^{1} \cdot \int_{- 1}^{0} x^{17} e^{x^{9}} d x

簡素化

= 2 e \cdot \int_{- 1}^{0} x^{17} e^{x^{9}} d x

u置換積分法を適用する

= 2 e \cdot \int_{- 1}^{0} \frac{e ^{u^{\frac{9}{17}}} u ^{\frac{1}{17}}}{17} d u

定数を除く:

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

= 2 e \frac{1}{17} \cdot \int_{- 1}^{0} e^{u^{\frac{9}{17}}} u^{\frac{1}{17}} d u

u置換積分法を適用する

= 2 e \frac{1}{17} \cdot \int_{(- 1)^{\frac{9}{17}}}^{0} \frac{17 e ^{v} v}{9} d v

定数を除く:

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

= 2 e \frac{1}{17} \cdot \frac{17}{9} \cdot \int_{(- 1)^{\frac{9}{17}}}^{0} e^{v} v d v

部分積分を適用する

= 2 e \frac{1}{17} \cdot \frac{17}{9} [e^{v} v - \int e^{v} d v]_{(- 1)^{\frac{9}{17}}}^{0}

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

= 2 e \frac{1}{17} \cdot \frac{17}{9} [e^{v} v - e^{v}]_{(- 1)^{\frac{9}{17}}}^{0}

簡素化

2 e \frac{1}{17} \cdot \frac{17}{9} [e^{v} v - e^{v}]_{(- 1)^{\frac{9}{17}}}^{0} : \frac{2 e}{9} [e^{v} v - e^{v}]_{- 1}^{0}

= \frac{2 e}{9} [e^{v} v - e^{v}]_{- 1}^{0}

限度を計算する:

- 1 + \frac{2}{e}

= \frac{2 e}{9} (- 1 + \frac{2}{e})

簡素化

= \frac{2 ( - e + 2 )}{9}

グラフ

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