Let \(A\) be the area under the curve \(y=\sqrt{ 1- x^2 }\) for \(0\leq x\leq1\).

By definition: $$ A = \int_0^1 { \sqrt{1-x^2} }\, dx$$ We can compute this integral by using the substitution \(x = sin(u)\): $$ A = \int_0^1 { \sqrt{cos^2(u)} }\, \frac{dx}{du} du$$ Now at \(x=0\), \(u=0\) and at \(x=1\), \(u=\pi/2\). So: $$ A = \int_{u=0}^{\pi/2} { cos(u) \times cos(u) }\, du$$ Remembering that \(cos^2(u) = \frac12 \big[ cos(2u)+1\big]\): $$ A = \frac12 \int_{u=0}^{\pi/2} { cos(2u) + 1 }\, du$$ $$ = \frac12 \bigg[ \frac12 sin(2u) + u \bigg]_{0}^{\pi/2} $$ $$ = \frac\pi4 $$

So the area of a unit circle is \(4 \times A = \pi\).