Question

Integrala de la 0 la 1 din radical din 1-x^2

Prin parti
Nu cu cos

Urgeeent

Answer 1

$\displaystyle I =\int_{0}^1 \sqrt{1-x^2}\, dx = \int_{0}^1 \dfrac{1-x^2}{\sqrt{1-x^2}}\, dx = \\ \\ =\int_{0}^1\Big(\dfrac{1}{\sqrt{1-x^2}}-x\cdot \dfrac{x}{\sqrt{1-x^2}}\Big)\, dx = \\ \\ = \int_{0}^1 \dfrac{1}{\sqrt{1-x^2}}\, dx - \int_{0}^1x\cdot (-\sqrt{1-x^2}\,)' = \\ \\ = \int_{0}^1 \dfrac{1}{\sqrt{1-x^2}}\, dx + \int_{0}^1x\cdot (\sqrt{1-x^2}\,)' \\ \\ =\arcsin(x)\Big|_{0}^1 +x\sqrt{1-x^2}\Big|_{0}^1-\int_{0}^1 x'\sqrt{1-x^2}\, dx =$

$= \arcsin(1)-0+0-0-I\\ \\\\ I = \arcsin(1) - I\\ \\ 2I = \arcsin(1) \\ \\2I = \dfrac{\pi}{2}\\\\ \Rightarrow \boxed{I = \dfrac{\pi}{4}}$