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人気のある微分積分 >

integral from 0 to 1 of 2/((5+3x)^{1/2)}

解

\int_{0}^{1} \frac{2}{( 5 + 3 x ) ^{\frac{1}{2}}} d x

解

\frac{2 ( 4 \cdot 2 ^{\frac{1}{2}} - 2 \cdot 5 ^{\frac{1}{2}} )}{3}

+1

十進法表記

0.78981 \dots

解答ステップ

\int_{0}^{1} \frac{2}{( 5 + 3 x ) ^{\frac{1}{2}}} d x

定数を除く:

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

= 2 \cdot \int_{0}^{1} \frac{1}{( 5 + 3 x ) ^{\frac{1}{2}}} d x

部分積分を適用する

= 2 [\frac{x}{( 5 + 3 x ) ^{\frac{1}{2}}} - \int - \frac{3 x}{2 ( 5 + 3 x ) ^{\frac{3}{2}}} d x]_{0}^{1}

\int - \frac{3 x}{2 ( 5 + 3 x ) ^{\frac{3}{2}}} d x = - \frac{3 x + 10}{3 ( 3 x + 5 ) ^{\frac{1}{2}}}

= 2 [\frac{x}{( 5 + 3 x ) ^{\frac{1}{2}}} - (- \frac{3 x + 10}{3 ( 3 x + 5 ) ^{\frac{1}{2}}})]_{0}^{1}

簡素化

2 [\frac{x}{( 5 + 3 x ) ^{\frac{1}{2}}} - (- \frac{3 x + 10}{3 ( 3 x + 5 ) ^{\frac{1}{2}}})]_{0}^{1} : 2 [\frac{2}{3} (3 x + 5)^{\frac{1}{2}}]_{0}^{1}

= 2 [\frac{2}{3} (3 x + 5)^{\frac{1}{2}}]_{0}^{1}

限度を計算する:

\frac{4 \cdot 2 ^{\frac{1}{2}} - 2 \cdot 5 ^{\frac{1}{2}}}{3}

= 2 \cdot \frac{4 \cdot 2 ^{\frac{1}{2}} - 2 \cdot 5 ^{\frac{1}{2}}}{3}

簡素化

= \frac{2 ( 4 \cdot 2 ^{\frac{1}{2}} - 2 \cdot 5 ^{\frac{1}{2}} )}{3}

グラフ

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