Question

\[\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx\]

Answer 1

\[Let\ I = \int_0^1 \sqrt{x\left( 1 - x \right)} d x . Then, \]
\[I = \int_0^1 \sqrt{\frac{1}{4} - \left( x - \frac{1}{2} \right)^2} dx\]
\[ \Rightarrow I = \frac{1}{2} \int_0^1 \sqrt{1 - \frac{\left( x - \frac{1}{2} \right)^2}{\frac{1}{4}}} dx\]
\[ \Rightarrow I = \frac{1}{2} \int_0^1 \sqrt{1 - \left( \frac{x - \frac{1}{2}}{\frac{1}{2}} \right)^2} dx\]
\[Let \left( \frac{x - \frac{1}{2}}{\frac{1}{2}} \right) = \sin u\]
\[ \Rightarrow 2 dx = \cos u du\]
\[ \therefore I = \frac{1}{4} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sqrt{1 - \sin^2 u} \cos u du\]
\[ \Rightarrow I = \frac{1}{4} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos^2 u du\]
\[ \Rightarrow I = \frac{1}{4} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( \frac{\cos 2u + 1}{2} \right) du\]
\[ \Rightarrow I = \frac{1}{8} \left[ \frac{\sin 2u}{2} + u \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{1}{8}\left[ \frac{\pi}{2} + \frac{\pi}{2} \right]\]
\[ \Rightarrow I = \frac{\pi}{8}\]