Step 1: Recognise the shape inside the root.
The factor $(x-\alpha)(\beta-x)$ is a downward parabola that is zero at $x=\alpha$ and $x=\beta$ and positive between them, so its square root traces a semicircle.
Step 2: Complete the square.
$(x-\alpha)(\beta-x)=-x^2+(\alpha+\beta)x-\alpha\beta$. Setting $r=\dfrac{\beta-\alpha}{2}$ and centre $m=\dfrac{\alpha+\beta}{2}$, this equals $r^2-(x-m)^2$.
Step 3: Rewrite the integral.
So $I=\displaystyle\int_{\alpha}^{\beta}\sqrt{r^2-(x-m)^2}\,dx$. This is the area under the upper half of a circle of radius $r$.
Step 4: Use the known semicircle area.
Over the full width from $\alpha$ to $\beta$ (which is the diameter $2r$), the curve is a semicircle, and a semicircle of radius $r$ has area $\dfrac{\pi r^2}{2}$.
Step 5: Substitute $r$.
$I=\dfrac{\pi}{2}r^2=\dfrac{\pi}{2}\left(\dfrac{\beta-\alpha}{2}\right)^2=\dfrac{\pi}{2}\cdot\dfrac{(\beta-\alpha)^2}{4}$.
Step 6: Simplify.
$I=\dfrac{\pi}{8}(\beta-\alpha)^2$. \[ \boxed{\dfrac{\pi}{8}(\beta-\alpha)^2} \]