Question

Интеграл.$\int\limits^2_0 \, \sqrt{4-x^2}dx$

Answer 1

Ответ:

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Пошаговое объяснение:

Answer 2

$\int_{0}^{2}{\sqrt{4-x^2}}dx=\begin{vmatrix}x=2\sin(u)\\ dx=2\cos(u)du\\a=0\,\,\,\,\,b=\dfrac{\pi}{2}\end{vmatrix}=2\int_{0}^{\pi/2}\cos(u)\sqrt{4-4\sin^2(u)}du=4\int_{0}^{\pi/2}\cos(u)\sqrt{\cos^{2} u}du=4\int_{0}^{\pi/2}\cos^2(u)du=4\int_{0}^{\pi/2}\dfrac{1+\cos(2u)}{2}du=2\int_{0}^{\pi/2}{1+\cos(2u)}du=2\left(u+\dfrac{sin(2u)}{2}\right)\Big|^{\pi/2}_0=\pi$