1. **State the problem:** Evaluate the definite integral $$\int_0^1 \frac{1}{\sqrt{4 - x^2}} \, dx$$. 2. **Recall the formula:** The integral $$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C$$ where $a > 0$. 3. **Identify parameters:** Here, $a = 2$ since $4 = 2^2$. 4. **Apply the formula:** $$\int \frac{1}{\sqrt{4 - x^2}} \, dx = \arcsin\left(\frac{x}{2}\right) + C$$ 5. **Evaluate the definite integral:** $$\int_0^1 \frac{1}{\sqrt{4 - x^2}} \, dx = \left[ \arcsin\left(\frac{x}{2}\right) \right]_0^1 = \arcsin\left(\frac{1}{2}\right) - \arcsin(0)$$ 6. **Calculate values:** - $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$ (since $\sin(\pi/6) = 1/2$) - $\arcsin(0) = 0$ 7. **Final answer:** $$\int_0^1 \frac{1}{\sqrt{4 - x^2}} \, dx = \frac{\pi}{6}$$ This means the area under the curve from $x=0$ to $x=1$ is $\frac{\pi}{6}$.