Definite Integral from 0 to a of Root of a Squared minus x Squared

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Theorem

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

for $a > 0$.


Proof 1

\(\ds \int_0^a \sqrt {a^2 - x^2} \rd x\) \(=\) \(\ds \intlimits {\frac {x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a} 0 a\) 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.

$\blacksquare$


Proof 2

\(\ds \int_0^a \sqrt {a^2 - x^2} \rd x\) \(=\) \(\ds \frac {a^{1 + \frac 2 2} } 2 \frac {\map \Gamma {\frac 1 2} \map \Gamma {1 + \frac 1 2} } {\map \Gamma {\frac 1 2 + \frac 1 2 + 1} }\) Definite Integral from 0 to a of $x^m \paren {a^n - x^n}^p$
\(\ds \) \(=\) \(\ds \frac {a^2} 2 \frac {\map \Gamma {\frac 3 2} \map \Gamma {\frac 1 2} } {\map \Gamma 2}\)
\(\ds \) \(=\) \(\ds \frac {a^2} {2 \times 1!} \paren {\frac 1 2} \paren {\map \Gamma {\frac 1 2} }^2\) Gamma Difference Equation, Gamma Function Extends Factorial
\(\ds \) \(=\) \(\ds \frac {\pi a^2} 4\) Gamma Function of One Half

$\blacksquare$


Sources

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