Finding the Area Between Two Curves (with respect to x)

The Basic Idea

When you find the area under a single curve, you use a definite integral. When the region lies between two curves, the idea is almost the same: at every value of \(x\), you measure the vertical distance between the curves and then add those distances across the interval.

If \(f(x)\) is the curve on top and \(g(x)\) is the curve on the bottom, the vertical height of a thin slice is:

Therefore, the area between the curves from \(x=a\) to \(x=b\) is:

The most important phrase to remember is:

That is the entire method. Most errors come from identifying the wrong functions, using the wrong interval, or forgetting that the curves may switch which one is on top.

Step 1: Find the Intersection Points

Before you can find the area, you usually need to know where the two curves meet. Set the functions equal to each other:

The solutions give the \(x\)-coordinates of the intersection points. These values become the limits of integration when the region is bounded by the two curves.

For example, suppose:

Set them equal:

So the curves intersect at \(x=0\) and \(x=2\). The bounded region between them therefore runs from \(x=0\) to \(x=2\).

Step 2: Determine Which Curve Is on Top

This is the step that causes many AP Calculus mistakes. You cannot simply assume that the first function listed is the top function.

Pick an \(x\)-value inside the interval and compare the two functions. For the interval from \(0\) to \(2\), a convenient test value is \(x=1\).

Since \(f(1)>g(1)\), the function \(f(x)\) is above \(g(x)\) on this interval.

Therefore:

Step 3: Write the Area Integral

Once you know the intersection points and which curve is on top, write:

For the example above:

Notice the order: top minus bottom. This makes the integrand nonnegative throughout the region.

Step 4: Evaluate the Integral

Now integrate normally. For our example:

The final answer is \( \dfrac{4}{3} \) square units.

Worked Examples

The Geometric Meaning

It helps to visualize what the integral is doing. Imagine slicing the region into extremely thin vertical rectangles.

Each rectangle has:

• Width: \(dx\)
• Height: top curve minus bottom curve
• Area: \((\text{top}-\text{bottom})\,dx\)

Adding infinitely many of these rectangles gives the exact area:

This is why the problem is an application of the definite integral: integration adds up infinitely many small pieces of area.

What "with Respect to x" Means

When a problem says to find area with respect to \(x\), it means that your slices are vertical. You are integrating with \(dx\).

For vertical slices, the general setup is:

By contrast, if you were finding area with respect to \(y\), you would use horizontal slices and subtract the left function from the right function. That is a different setup.

When You Need to Split the Integral

Sometimes the same curve is not on top throughout the entire interval. When that happens, you must split the integral at the \(x\)-value where the curves switch positions.

For example, suppose the top function changes at \(x=c\). Then:

Do not try to force a single integral to work when the "top minus bottom" expression changes. Splitting the integral is the correct mathematical approach.

Common Mistakes

A Fast AP Calculus Checklist

1. Find the intersection points by setting the curves equal, unless the interval is already given.
2. Identify the interval over which the region is being measured.
3. Determine the top curve by comparing the functions at a test point.
4. Write top minus bottom.
5. Use \(dx\) because the slices are vertical.
6. Split the integral if the curves switch positions.
7. Evaluate and report the area in square units.

Practice Problems