7. Problem: A parabola is inscribed in equilateral triangle $A B C$ of side length 1 in the sense that $A C$ and $B C$ are tangent to the parabola at $A$ and $B$, respectively. Find the area between $A B$ and the parabola.
Solution: Suppose $A=(0,0), B=(1,0)$, and $C=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. Then the parabola in question goes through $(0,0)$ and $(1,0)$ and has tangents with slopes of $\sqrt{3}$ and $-\sqrt{3}$, respectively, at these points. Suppose the parabola has equation $y=a x^{2}+b x+c$. Then $\frac{d y}{d x}=2 a x+b$.

At point $(0,0), \frac{d y}{d x}=b$. Also the slope at $(0,0)$, as we determined earlier, is $\sqrt{3}$. Hence $b=\sqrt{3}$. Similarly, at point $(1,0), \frac{d y}{d x}=2 a+b$. The slope at $(1,0)$, as we determined earlier, is $-\sqrt{3}$. Then $a=-\sqrt{3}$.

Since the parabola goes through $(0,0), c=0$. Hence the equation of the parabola is $y=-\sqrt{3} x^{2}+\sqrt{3} x$. The desired area is simply the area under the parabolic curve in the interval $[0,1]$.
Hence $\int_{0}^{1}\left(-\sqrt{3} x^{2}+\sqrt{3} x\right) d x=\frac{\sqrt{3}}{6}$.