Integration: Negative areas

This chapter explores negative areas. It covers understanding that areas below the x axis are negative, calculating areas under a curve, some or all of which may be under the x axis. Before attempting this chapter you must have prior knowledge of integration to find the area under a curve.

Areas below the x axis

Let us recap on finding the are below the curve as we explored in a previous chapter. Suppose we wanted to find the area under the curve between x=1 and x=3.
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This simply means we have to integrate between x=1 and x=3 that is;
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Now we substitute in the x values in turn and then subtract.
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The area result above is a negative. We can sketch the curve to see what it looks like. It is a quadratic graph so we have;
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We know that it also crosses the x axis at;
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It looks like the following.
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The area we’re asked was between x=1 and x=3. It has been shown on the graph below.
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Above we can see that the region is below the x axis. This is why the area comes out to be negative. Note that we say that the area is;
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Areas which straddle the x axis

You may come across a problem which has a region containing a part below the axis and part above the axis. It is therefore important to do a sketch og the graph to check the nature of the integral. Below is an example.
Suppose we had to find the area between the curve and the x axis for the following between x=-1 and x=2.
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First we do a sktech to check the region. We have;
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We know ir crosses the x axis at;
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Below is the curve and the need region.
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We can see that the region has a positive part A and a negative part B. In this case we must treat each region separately otherwise the negative area will be cancel out the positive area.
We first find the first area A. This is between x=-1 and x=0, we have;
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You must be very careful;
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Now we can find the area for part B. This is between x=0 and x=2, so we have;
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So the total area must be;
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You might have realised by now that had we not taken each part in turn then the two areas would have cancelled each other out completely and would have resulted in an area of zero.

Example

Here is another example to work with;
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First let us factorise it first;
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Below is the graph with the region shown.
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First let us first find the area for part A. This is between x=-2 and x=0, so we have;
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Now we can find the area for part B. This is between x=0 and x=1, that is;
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So we can conclude that the total area is;
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Exam question

Sketch the following graph and find the area of the region enclosed by the curve and the x axis between x=1 and x=3
One feature we can regonise about the graph straight away is that it is a reciprocal graph. We can see this easily if we split the equation into two fractions.
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It is simply a graph of…
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…but shifted down by 1 unit. Belowis the graph of 1/x²…
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…and below is the graph of y=1/x² – 1
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We need to find out where the graph crosses the x axis. To do that we solve for;
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We now know that it crosses the x axis at x=1. The region that we need to find has been shown on the graph below. It should be negative.
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We integrate to find the area;
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We can rewrite it as shown below.
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Now we can integrate as usual.
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We conclude that the area is;
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