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人気のある微分積分 >

integral from 0 to pi/2 of 5e^xcos^2(x)

解

\int_{0}^{\frac{π}{2}} 5 e^{x} cos^{2} (x) d x

解

2 e^{\frac{π}{2}} - 3

+1

十進法表記

6.62095 \dots

解答ステップ

\int_{0}^{\frac{π}{2}} 5 e^{x} cos^{2} (x) d x

定数を除く:

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

= 5 \cdot \int_{0}^{\frac{π}{2}} e^{x} cos^{2} (x) d x

部分積分を適用する

= 5 [e^{x} cos^{2} (x) - \int - e^{x} sin (2 x) d x]_{0}^{\frac{π}{2}}

\int - e^{x} sin (2 x) d x = \frac{2}{5} e^{x} cos (2 x) - \frac{1}{5} e^{x} sin (2 x)

= 5 [e^{x} cos^{2} (x) - (\frac{2}{5} e^{x} cos (2 x) - \frac{1}{5} e^{x} sin (2 x))]_{0}^{\frac{π}{2}}

簡素化

5 [e^{x} cos^{2} (x) - (\frac{2}{5} e^{x} cos (2 x) - \frac{1}{5} e^{x} sin (2 x))]_{0}^{\frac{π}{2}} : 5 [\frac{1}{5} (5 e^{x} cos^{2} (x) - 2 e^{x} cos (2 x) + e^{x} sin (2 x))]_{0}^{\frac{π}{2}}

= 5 [\frac{1}{5} (5 e^{x} cos^{2} (x) - 2 e^{x} cos (2 x) + e^{x} sin (2 x))]_{0}^{\frac{π}{2}}

限度を計算する:

\frac{2 e ^{\frac{π}{2}}}{5} - \frac{3}{5}

= 5 (\frac{2 e ^{\frac{π}{2}}}{5} - \frac{3}{5})

簡素化

= 2 e^{\frac{π}{2}} - 3

グラフ

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