The area between two curves - is my answer correct? 
Find the area between a parabola $y=x^2-2$ and an asymptote to $y=\sqrt{x^2+4x}+2x $ for $x \to -\infty$. As an answer, write the area multiplied by 6.

This is a very easy problem, but I made some careless mistake at first and my answer was incorrect. However, the answer listed for this problem is just 1.
First, we find the asymptote:
$$\lim_{x \to -\infty} \sqrt{x^2+4x}+2x=|x|+2x=x$$
Then we find the intersection points:
$$x^2-2=x$$
$$x_1=-1,~~~~~x_2=2$$
Now the area is:
$$S=\int^{2}_{-1}(x-x^2+2)dx=2-\frac{1}{2}-\frac{8}{3}-\frac{1}{3}+4+2=4.5$$
Now the final answer is $6S=4.5 \cdot 6=27$
Where is my mistake? How could anyone get 1 as a 'correct' answer?

We have $\sqrt{x^2+4x}+2x\sim x-2+\frac{2}{x}+ O(\frac{1}{x^2})$ as $x\rightarrow -\infty$. Thus the asymptote is $y=x-2$. The points of intersection are the solutions of $x^2-2=x-2,$ i.e. $x=0\,$ and $x=1$. And therefore the answer is
$$ 6 \Big|\int_0^1  (x^2-2 - (x-2)) dx\Big|= 6 \Big|\int_0^1  (x^2- x) dx\Big| = 
6\Big|[\frac{x^3}{3}-\frac{x^2}{2}]_0^1\Big|= 6\Big|-\frac{1}{2}\Big| = 1 $$
