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.
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 $b=a+\frac1{2a}+O(a^{-2})$.)
$$\sqrt{(b-a)^2+(b^2-a^2)^2}=1\implies \sqrt{(b-a)^2(1+(b+a)^2)}=1\\
\implies (b-a)^2(1+(b+a)^2)=1$$
$$\lim_{a\to\infty}a^3f(a)=\lim_{a\to\infty}a^3\int_a^{b}\left(\frac{b^2-a^2}{b-a}(x-a)-x^2\right){\rm d}x$$
These two relations can get you started.