Question

помогите найти определенный интеграл
$\int\limits^5_0 { \sqrt{25-x^2} } \, dx$

Answer 1

$\displaystyle \int\limits^5_0 { \sqrt{25-x^2} } \, dx=x\sqrt{25-x^2}|^5_0+\int\limits^5_0\frac{x^2dx}{\sqrt{25-x^2}}=x\sqrt{25-x^2}|^5_0-\\-\int\limits^5_0\frac{(25-x^2-25)dx}{\sqrt{25-x^2}}=x\sqrt{25-x^2}|^5_0-\int\limits^5_0\sqrt{25-x^2}dx+\\+25\int\frac{dx}{\sqrt{25-x^2}}\\\\\\ \int\limits^5_0 { \sqrt{25-x^2} } \, dx=\frac{x}{2}\sqrt{25-x^2}|^5_0+\frac{25}{2}arcsin\frac{x}{5}|^5_0\\\\\int\limits^5_0 { \sqrt{25-x^2} } \, dx=\frac{25\pi}{4}\approx19,635$

$u=\sqrt{25-x^2} ;du=-\frac{xdx}{\sqrt{25-x^2}}\\dv=dx;v=x$