Question

Вычислить определенный интеграл

Answer 1

$\int _0^{\sqrt3}\frac{x\, dx}{\sqrt{4-x^2}}=-\frac{1}{2}\cdot \int _0^{\sqrt3}\frac{-2x\, dx}{\sqrt{4-x^2}}=-\frac{1}{2}\int _0^{\sqrt3}\frac{d(4-x^2)}{\sqrt{4-x^2}}=[\int \frac{dt}{\sqrt{t}}=2\sqrt{t}+C]=\\\\=-\frac{1}{2}\cdot 2\sqrt{4-x^2}|_0^{\sqrt3}=-(\sqrt{4-3}-\sqrt{4-0})=-(1-2)=1$