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:


I end up with the answer $$\frac{-32}{3}$$

Where am I making a mistake?

  • 2
    $\begingroup$ Why did you drop the absolute value in the second version of the integrand? $\endgroup$
    – DJohnM
    Commented Aug 27, 2013 at 23:55
  • $\begingroup$ How do you know you are making a mistake? Is there solution given? $\endgroup$
    – Stefan4024
    Commented Aug 27, 2013 at 23:57
  • $\begingroup$ @User58220 I honestly don't know how to properly deal with the absolute value. $\endgroup$
    – nullByteMe
    Commented Aug 27, 2013 at 23:59
  • 1
    $\begingroup$ 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 $2\int_0^4 (3x-(x^2-4))dx$. $\endgroup$
    – user84413
    Commented Aug 28, 2013 at 0:00

1 Answer 1


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.

  • $\begingroup$ Where did you get the $2$? $\endgroup$
    – nullByteMe
    Commented Aug 28, 2013 at 2:12
  • $\begingroup$ Ohhh I see... because it's symmetrical you're just getting one half of the integral and multiplying by 2. $\endgroup$
    – nullByteMe
    Commented Aug 28, 2013 at 2:18

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