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人気のある微分積分 >

integral from 0 to 1 of (x^2+5)e^{2x}

解

\int_{0}^{1} (x^{2} + 5) e^{2 x} d x

解

\frac{11 e ^{2} - 11}{4}

+1

十進法表記

17.56990 \dots

解答ステップ

\int_{0}^{1} (x^{2} + 5) e^{2 x} d x

部分積分を適用する

= [\frac{1}{2} e^{2 x} (x^{2} + 5) - \int e^{2 x} x d x]_{0}^{1}

\int e^{2 x} x d x = \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

= [\frac{1}{2} e^{2 x} (x^{2} + 5) - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})]_{0}^{1}

簡素化

[\frac{1}{2} e^{2 x} (x^{2} + 5) - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})]_{0}^{1} : [\frac{1}{4} (2 e^{2 x} x^{2} - 2 e^{2 x} x + 11 e^{2 x})]_{0}^{1}

= [\frac{1}{4} (2 e^{2 x} x^{2} - 2 e^{2 x} x + 11 e^{2 x})]_{0}^{1}

限度を計算する:

\frac{11 e ^{2}}{4} - \frac{11}{4}

= \frac{11 e ^{2}}{4} - \frac{11}{4}

簡素化

= \frac{11 e ^{2} - 11}{4}

グラフ

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