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人気のある微分積分 >

integral from 0 to 1 of 2^{2x+1}

解

\int_{0}^{1} 2^{2 x + 1} d x

解

\frac{3}{ln ( 2 )}

+1

十進法表記

4.32808 \dots

解答ステップ

\int_{0}^{1} 2^{2 x + 1} d x

u置換積分法を適用する

= \int_{1}^{3} 2^{u - 1} d u

簡素化

2^{u - 1} : 2^{u} \cdot 2^{- 1}

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

定数を除く:

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

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

指数の規則を適用する:

a^{- 1} = \frac{1}{a}

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

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{2} [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{3}

限度を計算する:

\frac{6}{ln ( 2 )}

= \frac{1}{2} \cdot \frac{6}{ln ( 2 )}

簡素化

= \frac{3}{ln ( 2 )}

グラフ

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