Find that $\lim_{a\to \infty}a^3 f(a)=?$ I would appreciate if somebody could help me with the following problem:

Let $f(a)$ area of S, $A(a,a^2)$, $B(b,b^2)$ and $\overline{AB}=1$, given that:


Find that $$\lim_{a\to \infty}a^3 f(a)=?$$
I tried  but couldn't get it that way.
 A: It can be shown that the area above an upward-facing parabola, defined by $y=q(x)$, that is cut off by the secant line between $x=a$ and $x=b$ is equal to
$$
(b-a) \bigg( \frac13 q(a) - \frac23 q\bigg(\frac{a+b}2\bigg) + \frac13 q(b) \bigg).
$$
(Indeed, this calculation is what goes into Simpson's Rule!)
In this problem, $b\approx a+\frac1{2a}$ when $a$ is large (this makes the "rise" of the segment tend to $1$ and the "run" tend to $0$). Therefore we want to calculate
$$
\lim_{a\to\infty} a^3 \cdot \frac1{2a}\bigg( \frac13 a^2 - \frac23 \bigg( a+\frac1{4a} \bigg) ^2 + \frac13 \bigg( a+\frac1{2a} \bigg) ^2 \bigg) = \frac1{48}.
$$
(To make this calculation rigorous, one would use a statement like $b=a+\frac1{2a}+O(a^{-2})$.)
A: $${(a-b)^2+(a^2-b^2)^2}=1$$
Set $b-a=\cos \theta$ and $b^2-a^2=\sin \theta$
then $a+b=\tan\theta$.
So
$$a={1\over 2}\{\tan\theta-\cos \theta\}$$
but the area of $S$ is 
$$\frac{(a^2+b^2)(b-a)}{2}-\frac{b^3-a^3}{3}={(b-a)^3\over 6}$$
Therefore the desired limit is $$\lim_{a\to\infty}a^{3}f(a)={1\over 8}\times {1\over 6}\lim_{\theta\to{\pi\over 2}}\cos^3\theta\times  \big(\tan\theta-\cos \theta\big)^3={1\over 48}\lim_{\theta\to{\pi\over 2}}(\sin\theta-\cos^{2}\theta)^{3}=\color\red {{1\over 48}}$$
A: The area of S is $\frac12 (b-a) (b^2+a^2) - \frac13 (b^3-a^3) = \frac16 (b-a)^3 $.
As $a \to \infty$, the difference betwen $b$ and $a$ vanishes.  In this case,
 $$\left |\overline{AB}\right |^2 = 1 = (a-b)^2 \left [ 1+(a+b)^2 \right ] \approx (b-a)^2 (1+4 a^2) $$  
Thus, 
$$\lim_{a \to \infty} a^3 \frac16 (b-a)^3 = \frac16 \lim_{a \to \infty} a^3 (1+4 a^2)^{-3/2} = \frac1{48} $$
A: $$\sqrt{(b-a)^2+(b^2-a^2)^2}=1\implies \sqrt{(b-a)^2(1+(b+a)^2)}=1\\
\implies (b-a)^2(1+(b+a)^2)=1$$
$$\lim_{a\to\infty}a^3f(a)=\lim_{a\to\infty}a^3\int_a^{b}\left(\frac{b^2-a^2}{b-a}(x-a)-x^2\right){\rm d}x$$
These two relations can get you started.
