Minimum of $\frac{{(b-a)} ^{3}}{6}$

Question

I need help with this problem. Any insight is appreciated. Thanks in advance.

A parabola $y = x^2$ intersects a line at the points $(a,a^2)$ and $(b,b^2)$ $(a < b)$. The area $S$ between the line and the parabola is given by $\frac{{(b-a)} ^{3}}{6}$. Find an expression for the minimum value $m$ of $S$ and the corresponding value of $b$.

Answer 1

I think I found the answer. Since $a$ = $\frac{-2b^2-1}{2b}$, $S$ = $\frac{(\frac{4b^2+1}{2b})^3}{6}$. Therefore, $\frac{d}{db}$[$\frac{(\frac{4b^2+1}{2b})^3}{6}$] = 0, and after derivation, $b$ comes out to be either $\frac{1}{2}$ or -$\frac{1}{2}$. Then, using the second derivative test, $b$=-$\frac{1}{2}$ is the the value with negative concavity. Replacing $b$ in $S =$ $\frac{(\frac{4b^2+1}{2b})^3}{6}$, the minimum value turns out to be $-\frac{4}{3}$ and the value of $b$ is $-\frac{1}{2}$. Is that correct?

Answer 2

To me, there is something which is bizarre in the problem.

The equation of the line passing through points $(a,a^2)$ and $(b,b^2)$ is $$y = -a b + (a+b)\,x$$ so the area between the line and the curve is $$S=\int_ a^b(-ab+ (a+b)\,x-x^2)\,dx=\frac{(b-a)^3}{6} $$ which is the given result and an identity.

What else could we do without any further information ?