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.

$\blacksquare$