6. Determine the positive value of $a$ such that the parabola $y=x^{2}+1$ bisects the area of the rectangle with vertices $(0,0),(a, 0),\left(0, a^{2}+1\right)$, and $\left(a, a^{2}+1\right)$.
Solution: $\sqrt{3}$ The area of the rectangle is $a^{3}+a$. The portion under the parabola has area $\int_{0}^{a} x^{2}+1 d x=a^{3} / 3+a$. Thus we wish to solve the equation $a^{3}+a=2\left(a^{3} / 3+a\right)$; dividing by $a$ and rearranging gives $a^{2} / 3=1$, so $a=\sqrt{3}$.