Differentiate then integrate?

Let's start by solving the first parametric equation for $t$ in terms of $x$ : $t = (x+1)^{1/3}$ , and substitution into the second parametric equation gives: $y = (x+1)^{2/3} - 1.$ (i) First and second derivatives of (i) yield:

$y' = \frac{2}{3}(x+1)^{-1/3}, y'' = -\frac{2}{9}(x+1)^{-4/3}$

and the quotient: $\frac{y''}{(y')^4} = \frac{-\frac{2}{9}(x+1)^{-4/3}}{( \frac{2}{3}(x+1)^{-1/3})^4} = -\frac{9}{8} = -\frac{3^2}{2^3}.$

thus $a = 3, b = 2$ with the line $y = 3x+2$ (ii). Now curves (i) and (ii) intersect at:

$(x+1)^{2/3} - 1 = 3x+2 \Rightarrow 0 = (x+1)^{2/3} - (3x + 3) \Rightarrow 0 = (x+1)^{2/3} \cdot (1 - 3(x+1)^{1/3}) \Rightarrow x = -1, -\frac{26}{27}$ .

The required area between (i) and (ii) (with (i) $\ge$ (ii) for all $x \in [-1, -\frac{26}{27}])$ computes to:

$\int_{-1}^{-\frac{26}{27}} [(x+1)^{2/3} - 1] - (3x+2) dx = \frac{3}{5}(x+1)^{5/3} - \frac{3}{2}x^2 - 3x|_{-1}^{-\frac{26}{27}} \approx \boxed{4.1 \times 10^{-4}}.$