1. **Problem Statement:** Calculate the definite integral $$\int_\alpha^\beta \sqrt{(x-\alpha)(\beta - x)}\, dx$$ where $\alpha < \beta$. 2. **Formula and Approach:** This integral represents the area under the curve of the function $f(x) = \sqrt{(x-\alpha)(\beta - x)}$ from $x=\alpha$ to $x=\beta$. 3. **Substitution:** Let $x = \alpha + t(\beta - \alpha)$, so when $x=\alpha$, $t=0$ and when $x=\beta$, $t=1$. 4. **Rewrite the integral:** $$ \int_\alpha^\beta \sqrt{(x-\alpha)(\beta - x)}\, dx = \int_0^1 \sqrt{t(1-t)} (\beta - \alpha) dt $$ 5. **Simplify:** $$ = (\beta - \alpha)^{3/2} \int_0^1 \sqrt{t(1-t)} dt $$ 6. **Evaluate the integral:** The integral $\int_0^1 \sqrt{t(1-t)} dt$ is a Beta function integral: $$ \int_0^1 t^{1/2} (1-t)^{1/2} dt = B\left(\frac{3}{2}, \frac{3}{2}\right) = \frac{\Gamma(\frac{3}{2}) \Gamma(\frac{3}{2})}{\Gamma(3)} $$ 7. **Use Gamma function values:** $$ \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}, \quad \Gamma(3) = 2! $$ 8. **Calculate Beta function:** $$ B\left(\frac{3}{2}, \frac{3}{2}\right) = \frac{\left(\frac{\sqrt{\pi}}{2}\right)^2}{2} = \frac{\pi}{4} \cdot \frac{1}{2} = \frac{\pi}{8} $$ 9. **Final answer:** $$ \int_\alpha^\beta \sqrt{(x-\alpha)(\beta - x)} dx = (\beta - \alpha)^{3/2} \cdot \frac{\pi}{8} $$ This gives the exact area under the curve between $\alpha$ and $\beta$ for the given function.