# 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?

• Why did you drop the absolute value in the second version of the integrand? Commented Aug 27, 2013 at 23:55
• How do you know you are making a mistake? Is there solution given? Commented Aug 27, 2013 at 23:57
• @User58220 I honestly don't know how to properly deal with the absolute value. Commented Aug 27, 2013 at 23:59
• Since you have absolute values in the integral, you want to split up the integral, in this case at 0 since that's where $|3x|$ changes sign. By symmetry, you can find $2\int_0^4 (3x-(x^2-4))dx$. Commented Aug 28, 2013 at 0:00

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.
• Where did you get the $2$? Commented Aug 28, 2013 at 2:12