# Complicated Absolute Value of Integral Problem

I'm not sure where to start with this problem, it's not like the previous examples of absolute value integration I've seen, and I feel like the steps are different? When I plugged it into symbolap, it couldnt even give me an answer. Any help would be appreciated.

Find the value of the following integral and use any means necessary to set this up without the absolute values to that you calculate the answer. $$\int_{-4}^4 \left| \frac{1}{2} x^2 - 3 - \tan^{-1}(x) \right| dx$$

• That's a weird problem, to be sure. The usual technique for solving integrals of the form $\int |f(x)| dx$ is to split the region of integration into regions where $f(x)$ is either positive or negative. But in this case, $f(x)$ switches sign when $\arctan(x) = \frac{1}{2}x^2 - 3$, which I don't think can be solved exactly (i.e., the values of $x$ for which this holds are transcendental numbers.) – Michael Seifert May 23 '17 at 19:50

Think of this integral as an area. Normally, the area of the curve under the x-axis would be treated as negative area, but now the absolute value makes all of the area positive. What you need to do in this case is find the interval(s) of $$\frac{1}{2}x^2-3-\tan^{-1}x$$ within $[-4,4]$ that is below the x-axis, separate that area from the rest of the integral, and make it positive. I don't want to do this one for you, but here is a similar example: $$\int_{-3}^{3} \mid x^2-1 \mid dx$$ First of all, notice that the zeroes of $x^2-1$ come at $-1$ and $1$. It is negative between $-1$ and $1$, and positive otherwise. Since we are integrating the absolute value, we need to separate this interval and make it positive. We can break this integral up into $$\int_{-3}^{-1} (x^2-1) dx - \int_{-1}^{1} (x^2-1) dx+ \int_{1}^{3} (x^2-1) dx$$ Now we can evaluate each of this separately to get $$\frac{20}{3} + \frac{4}{3}+ \frac{20}{3}$$ $$\frac{44}{3}$$ You should be able to do your problem similarly. You may have a little trouble finding the zeroes of your function, but you can always approximate them for an approximate answer.