# Definite Integral from 0 to a of Root of a Squared minus x Squared/Proof 1

## Theorem

$\ds \int_0^a \sqrt {a^2 - x^2} \rd x = \frac {\pi a^2} 4$

for $a > 0$.

## Proof

 $\ds \int_0^a \sqrt {a^2 - x^2} \rd x$ $=$ $\ds \intlimits {\frac {x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a} 0 a$ Primitive of $\sqrt {a^2 - x^2}$ $\ds$ $=$ $\ds \paren {\frac {a \sqrt {a^2 - a^2} } 2 + \frac {a^2} 2 \arcsin \frac a a} - \paren {\frac {0 \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac 0 a}$ $\ds$ $=$ $\ds \frac {a^2} 2 \arcsin 1 - \frac {a^2} 2 \arcsin 0$ removing vanishing terms $\ds$ $=$ $\ds \frac {a^2} 2 \arcsin 1$ Sine of Zero is Zero $\ds$ $=$ $\ds \frac \pi 2 \frac {a^2} 2$ Sine of Right Angle

Hence the result.

$\blacksquare$