Section 6.2 : Area Between Curves
9. Determine the area of the region bounded by \(\displaystyle y = \frac{1}{{x + 2}}\), \(y = {\left( {x + 2} \right)^2}\), \(x = - \frac{3}{2}\), \(x = 1\).
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Let’s start off with getting a sketch of the region we want to find the area of.
We are assuming that, at this point, you are capable of graphing most of the basic functions that we’re dealing with in these problems and so we won’t be showing any of the graphing work here.
Here is a sketch of the bounded region we want to find the area of.
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 point. 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 point of the two curves does not occur on an axis (as they don’t in this case).
So, to determine the intersection points correctly we’ll need to find it directly. The intersection point is where the two curves intersect and so all we need to do is set the two equations equal and solve. Doing this gives,
\[\frac{1}{{x + 2}} = {\left( {x + 2} \right)^2}\hspace{0.25in} \to \hspace{0.25in}{\left( {x + 2} \right)^3} = 1\hspace{0.25in}\, \to \hspace{0.25in}\,x + 2 = \sqrt[3]{1} = 1\hspace{0.25in}\, \to \hspace{0.25in}x = - 1\]So, from the graph then it looks like we’ll need two integrals since there are two ranges of \(x\) (\( - \frac{3}{2} \le x \le - 1\) and \( - 1 \le x \le 1\)) for which the upper/lower functions are different.
Show Step 3At 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,
\[\begin{align*}A & = \int_{{ - \,\,\frac{3}{2}}}^{{ - 1}}{{\frac{1}{{x + 2}} - {{\left( {x + 2} \right)}^2}\,dx}} + \int_{{ - 1}}^{1}{{{{\left( {x + 2} \right)}^2} - \frac{1}{{x + 2}}\,dx}}\\ & = \left. {\left( {\ln \left| {x + 2} \right| - \frac{1}{3}{{\left( {x + 2} \right)}^3}} \right)} \right|_{ - \,\frac{3}{2}}^{ - 1} + \left. {\left( {\frac{1}{3}{{\left( {x + 2} \right)}^3} - \ln \left| {x + 2} \right|} \right)} \right|_{ - 1}^1\\ & = \left[ { - \frac{7}{{24}} - \ln \left( {\frac{1}{2}} \right)} \right] + \left[ {\frac{{26}}{3} - \ln \left( 3 \right)} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{67}}{8} - \ln \left( {\frac{1}{2}} \right) - \ln \left( 3 \right) = 7.9695}}\end{align*}\]