Trouble finding the area of the region $y=|3x|, y=x^2-4$ I'm supposed to find the area of this region:
$$y=|3x|, y=x^2-4$$
So I first tried to find points of intersection:
$$|3x|=x^2-4$$
$$3|x|=x^2-4$$
$$|x|=\frac{x^2-4}{3}$$
And from here I found that $x=4, x=-4$.  So I set these as the bounds of my integral and set it up this way:
$$\int_{-4}^{4}|3x|-(x^2-4)dx$$
$$\int_{-4}^{4}(3x-x^2+4)dx$$
And when I anti-differentiated it I got:
$$\frac{3x^2}{2}-\frac{x^3}{3}+4x$$
I end up with the answer $$\frac{-32}{3}$$
Where am I making a mistake?

Let's get rid of that pesky $|3x|$ by observing that $|3x|=|3(-x)|$ and, luckily for us, $x^2-4=(-x)^2-4$. That means that both functions are symmetric about the $y$ axis (so when graphed the part for negative $x$ will have the same shape as the part for positive $x$), so
$$
\int_{-4}^4 |3x|-(x^2-4) dx = 2\int_{0}^4 |3x|-(x^2-4)dx
$$
and for $x\ge 0$ we have $|3x|=3x$, so our integral becomes
$$
2\int_{0}^4 3x-(x^2-4)dx
$$
which I have every confidence you can evaluate.
Edit. Not to hog the credit, I noticed that user84413 also came up with the solution while I was typing mine in.
