1/√(a-x)の積分

1/√(a-x)の積分

不定積分

$$\begin{array}{rcl}\int \frac{1}{\sqrt{a-x}}\mathrm{d}x& =& \int (a-x{)}^{-\frac{1}{2}}\mathrm{d}x\\ & =& \int u^{-\frac{1}{2}}(-1)\mathrm{d}u\cdots u=a-x,\frac{\mathrm{d}u}{\mathrm{d}x}=-1,\mathrm{d}x=-\mathrm{d}u\\ & =& -\int u^{-\frac{1}{2}}\mathrm{d}u\cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& -\frac{1}{-\frac{1}{2}+1}{u}^{-\frac{1}{2}+1}\cdots \int x^a\mathrm{d}x=\frac{1}{a+1}{x}^{a+1}+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& -\frac{1}{\frac{1}{2}}{u}^{\frac{1}{2}}\\ & =& -2u^{\frac{1}{2}}\\ & =& -2(a-x{)}^{\frac{1}{2}}\cdots u=a-x\\ & =& -2(a-x{)}^{\frac{1}{2}}+C\cdots C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\\ & =& -2\sqrt{a-x}+C\end{array}$$

定積分

$$\begin{array}{rcl}{\int }_0^a\frac{1}{\sqrt{a-x}}\mathrm{d}x& =& {[-2\sqrt{a-x}]}_0^a\\ & =& (-2\sqrt{a-a})-(-2\sqrt{a-0})\\ & =& 0-(-2\sqrt{a})\\ & =& 2\sqrt{a}\\ & =& 2a^{\frac{1}{2}}\end{array}$$