Question

Find the value of \[ \int_0^a \sqrt{a^2 - x^2} \, dx. \]

Hint:

This integral represents the area of a quarter circle of radius \(a\).

Answer 1

This integral represents the area under the curve of a semicircle of radius \(a\). Using the formula: \[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + C. \] Evaluate from 0 to \(a\): \[ \int_0^a \sqrt{a^2 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} \right]_0^a. \] At \(x = a\), \[ \frac{a}{2} \times 0 + \frac{a^2}{2} \times \frac{\pi}{2} = \frac{a^2 \pi}{4}. \] At \(x = 0\), \[ 0 + \frac{a^2}{2} \times 0 = 0. \] Therefore, \[ \boxed{ \int_0^a \sqrt{a^2 - x^2} \, dx = \frac{a^2 \pi}{4}. } \]