integral from 0 to h of 1/(2sqrt(hx))

解

\int_{0}^{h} \frac{1}{2 h x} d x

1

解答ステップ

\int_{0}^{h} \frac{1}{2 h x} d x

定数を除く:

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

= \frac{1}{2} \cdot \int_{0}^{h} \frac{1}{h x} d x

\frac{1}{h x} = \frac{1}{h} \frac{1}{x}

= \frac{1}{2} \cdot \int_{0}^{h} \frac{1}{h} \cdot \frac{1}{x} d x

定数を除く:

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

= \frac{1}{2} \cdot \frac{1}{h} \cdot \int_{0}^{h} \frac{1}{x} d x

べき乗則を適用する

= \frac{1}{2} \cdot \frac{1}{h} [2 x^{\frac{1}{2}}]_{0}^{h}

簡素化

\frac{1}{2} \cdot \frac{1}{h} [2 x^{\frac{1}{2}}]_{0}^{h} : \frac{[ 2 x ] _{0}^{h}}{2 h}

= \frac{[ 2 x ] _{0}^{h}}{2 h}

限度を計算する:

2 h

= \frac{2 h}{2 h}

簡素化

= 1