In the problem statement we were given two limits on \(x\). However, as seen in the sketch of the graph above the curves intersect in this region and the upper/lower functions differ depending on what range of \(x\)’s we are looking for.

Therefore we’ll need to find the intersection points. However, we should never just assume that our graph is accurate or that we will be able to read it accurately enough to guess the coordinates from the graph. This is especially true when the intersection points of the two curves do not occur on an axis (as they don’t in this case).

So, to determine the intersection points correctly we’ll need to find them directly. The intersection points are where the two curves intersect and so all we need to do is set the two equations equal and solve. Doing this gives,

\[6 - x - 2{x^2} = 4x + 3\hspace{0.25in} \to \hspace{0.25in}2\,{x^2} + 5x - 3 = \left( {2x - 1} \right)\left( {x + 3} \right) = 0\hspace{0.25in} \to \hspace{0.25in}x = - 3,\,\,\,\,x = \frac{1}{2}\]

Note that you may well have found the intersection points in the first step to help with the graph if you were graphing by hand which is not a bad idea with faced with graphing this kind of region.

So, from the graph then it looks like we’ll need three integrals since there are three ranges of \(x\) (\( - 4 \le x \le - 3\), \( - 3 \le x \le \frac{1}{2}\) and \(\frac{1}{2} \le x \le 2\)) for which the upper/lower functions are different.

Show Step 3

At this point there isn’t much to do other than step up the integrals and evaluate them.

We are assuming that you are comfortable with basic integration techniques so we’ll not be including any discussion of the actual integration process here and we will be skipping some of the intermediate steps.

The area is,

\[\begin{align*}A & = \int_{{ - 4}}^{{ - 3}}{{4x + 3 - \left( {6 - x - 2{x^2}} \right)\,dx}} + \int_{{ - 3}}^{{\frac{1}{2}}}{{6 - x - 2{x^2} - \left( {4x + 3} \right)\,dx}} + \int_{{\frac{1}{2}}}^{2}{{4x + 3 - \left( {6 - x - 2{x^2}} \right)\,dx}}\\ & = \int_{{ - 4}}^{{ - 3}}{{2{x^2} + 5x - 3\,dx}} + \int_{{ - 3}}^{{\frac{1}{2}}}{{3 - 5x - 2{x^2}\,dx}} + \int_{{\frac{1}{2}}}^{2}{{2{x^2} + 5x - 3\,dx}}\\ & = \left. {\left( {\frac{2}{3}{x^3} + \frac{5}{2}{x^2} - 3x} \right)} \right|_{ - 4}^{ - 3} + \left. {\left( {3x - \frac{5}{2}{x^2} - \frac{2}{3}{x^3}} \right)} \right|_{ - 3}^{\frac{1}{2}} + \left. {\left( {\frac{2}{3}{x^3} + \frac{5}{2}{x^2} - 3x} \right)} \right|_{\frac{1}{2}}^2\\ & = \frac{{25}}{6} + \frac{{343}}{{24}} + \frac{{81}}{8} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{343}}{{12}}}}\end{align*}\]