Unit 8 Applications of Integration: Understanding Area Between Curves

Finding the Area Between Curves Expressed as Functions of $x$

What “area between curves” means (and why you need calculus)

When you hear “area between curves,” you’re being asked to find the area of a region in the coordinate plane whose boundary is formed by two graphs. In AP Calculus AB, the graphs are usually given as functions like $y=f(x)$ and $y=g(x)$.

If both curves were straight lines, you could sometimes use geometry. But most of the time the boundaries are curved, so the region doesn’t break nicely into triangles and rectangles. That’s where integration matters: a definite integral adds up infinitely many very thin rectangles to approximate area, and then takes the limit to get the exact result.

The key idea is:

• If you slice the region with vertical rectangles, each rectangle’s height is “top curve minus bottom curve,” and its thickness is $dx$.
• Adding all those rectangle areas from a left endpoint $a$ to a right endpoint $b$ gives the total area.

This is why the standard “function of $x$” area formula looks like an integral with respect to $x$.

Vertical slicing: the fundamental setup

Suppose over an interval from $x=a$ to $x=b$, one curve is always above the other. If $f(x)$ is the upper function and $g(x)$ is the lower function, then a typical vertical slice at position $x$ has:

• height $f(x)-g(x)$
• width $dx$

So the area is:

$A=\int_a^b \left(f(x)-g(x)\right)dx$

This formula is simple, but it only works as written when you have correctly identified:

1. the correct bounds $a$ and $b$, and
2. which function is on top for the entire interval.

If the “top” and “bottom” switch somewhere in the interval, you must split the integral.

Where do the bounds $a$ and $b$ come from?

Most AP problems define the region in one of these ways:

1. Intersection points of the curves

   • You solve $f(x)=g(x)$ to find the $x$-coordinates where the curves cross.
   • Those intersection $x$-values often become $a$ and $b$.

2. Given vertical lines

   • The problem might explicitly say “between $x=1$ and $x=4$.” Then $a=1$ and $b=4$.

3. A mix of both

   • One boundary could be an intersection, the other a given vertical line.

A common misconception is to assume the bounds are always the intersection points. Sometimes the region is cut off earlier by a vertical line, an axis, or another curve.

“Top minus bottom” is a relationship, not a label

Students often try to decide which formula to use before thinking about the picture. A better habit is:

1. Sketch both curves (even a rough sketch).
2. Pick a typical vertical slice.
3. Ask: at that $x$, which $y$-value is higher?

The “top” function is the one that gives larger $y$ for the same $x$.

If the top/bottom relationship changes, you can’t fix it by taking an absolute value blindly. On the AP exam, the expected method is usually to split at intersection points (or other switching points).

When you must split the integral

If $f(x)$ and $g(x)$ intersect at some value $x=c$ inside $[a,b]$, then the top curve on $[a,c]$ might be different from the top curve on $[c,b]$.

In that case, area becomes a sum:

$A=\int_a^c \left(\text{top}-\text{bottom}\right)dx+\int_c^b \left(\text{top}-\text{bottom}\right)dx$

This “split when the top changes” idea is one of the most tested skills in area-between-curves problems.

Notation reference (common equivalent forms)

Worked Example 1: Two functions with clean intersections

Find the area of the region bounded by $y=x$ and $y=x^2$.

Step 1: Understand the region

These two curves intersect where:

$x=x^2$

$x^2-x=0$

$x(x-1)=0$

So intersections occur at $x=0$ and $x=1$.

Step 2: Decide top minus bottom

On the interval $0<x<1$, compare $x$ and $x^2$. For example, at $x=\frac{1}{2}$:

$x=\frac{1}{2}$

$x^2=\frac{1}{4}$

So $y=x$ is above $y=x^2$ on $[0,1]$.

Step 3: Set up and compute the integral

$A=\int_0^1 \left(x-x^2\right)dx$

Antiderivative:

$\int \left(x-x^2\right)dx=\frac{x^2}{2}-\frac{x^3}{3}$

Evaluate from $0$ to $1$:

$A=\left(\frac{1}{2}-\frac{1}{3}\right)-\left(0-0\right)=\frac{1}{6}$

So the area is $\frac{1}{6}$ square units.

Worked Example 2: A “switching top” situation (splitting required)

Find the area between $y=\sin x$ and $y=\cos x$ from $x=0$ to $x=\pi$.

Step 1: Find where they intersect in the interval

Solve:

$\sin x=\cos x$

This happens when:

$\tan x=1$

In $[0,\pi]$, the solution is:

$x=\frac{\pi}{4}$

Step 2: Determine which is on top in each sub-interval

• At $x=0$: $\sin 0=0$ and $\cos 0=1$, so $\cos x$ is on top near $0$.
• At $x=\frac{\pi}{2}$: $\sin \frac{\pi}{2}=1$ and $\cos \frac{\pi}{2}=0$, so $\sin x$ is on top near $\frac{\pi}{2}$.

So the top switches at $x=\frac{\pi}{4}$.

Step 3: Split and integrate

Area equals:

$A=\int_0^{\pi/4} \left(\cos x-\sin x\right)dx+\int_{\pi/4}^{\pi} \left(\sin x-\cos x\right)dx$

Compute the first integral:

$\int \left(\cos x-\sin x\right)dx=\sin x+\cos x$

So:

$\int_0^{\pi/4} \left(\cos x-\sin x\right)dx=\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)-\left(\sin 0+\cos 0\right)$

Use $\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}$:

$=\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)-\left(0+1\right)=\sqrt{2}-1$

Compute the second integral:

$\int \left(\sin x-\cos x\right)dx=-\cos x-\sin x$

So:

$\int_{\pi/4}^{\pi} \left(\sin x-\cos x\right)dx=\left(-\cos \pi-\sin \pi\right)-\left(-\cos \frac{\pi}{4}-\sin \frac{\pi}{4}\right)$

Evaluate values:

$-\cos \pi-\sin \pi=-(-1)-0=1$

And:

$-\cos \frac{\pi}{4}-\sin \frac{\pi}{4}=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}$

So the second integral is:

$1-(-\sqrt{2})=1+\sqrt{2}$

Total area:

$A=(\sqrt{2}-1)+(1+\sqrt{2})=2\sqrt{2}$

This example highlights the big idea: area requires “top minus bottom,” and when “top” changes, you split.

Why this topic connects to other integration ideas

Area between curves is one of the first places where you must combine multiple calculus skills:

• solving equations to find intersections
• understanding graphs and sign changes
• interpreting a definite integral as accumulated quantity

It also previews later applications (some in AB, more in BC): volumes by slicing use the same “slice area times thickness” logic.

Exam Focus

• Typical question patterns:
  • “Find the area of the region bounded by $y=f(x)$ and $y=g(x)$.” (You must find intersection points.)
  • “Find the area between the curves from $x=a$ to $x=b$.” (Bounds are given; you still must decide top minus bottom.)
  • “Set up (but do not evaluate) an integral for the area …” (Setup is the main skill.)
• Common mistakes:
  • Using the wrong bounds because you didn’t solve $f(x)=g(x)$ or you chose the wrong intersection points.
  • Forgetting to split when the curves cross inside the interval, leading to cancellation (negative area).
  • Reversing top and bottom (you get a negative number, which signals your setup is wrong for an area question).

Finding the Area Between Curves Expressed as Functions of $y$

Why change the variable of integration?

Sometimes a region is much easier to describe using horizontal slices instead of vertical slices. When you slice horizontally:

• each slice has a length measured left-to-right, which is “right curve minus left curve,” and
• each slice has a thickness of $dy$.

This matters because not every curve is easiest to work with as $y=f(x)$. Some graphs fail the vertical line test (like circles), or a region might be bounded by curves that are naturally given as $x$ in terms of $y$.

A good rule of thumb is:

• Use $dx$ when vertical slices give a single “top minus bottom” expression over a clean interval.
• Use $dy$ when horizontal slices give a single “right minus left” expression over a clean interval.

Horizontal slicing: “right minus left”

Suppose the region is bounded by two curves written as:

$x=R(y)$

$x=L(y)$

where $R(y)$ is the right boundary (larger $x$-value) and $L(y)$ is the left boundary (smaller $x$-value) for each $y$ in an interval $[c,d]$.

A typical horizontal slice at height $y$ has:

• length $R(y)-L(y)$
• thickness $dy$

So the area is:

$A=\int_c^d \left(R(y)-L(y)\right)dy$

Notice the parallel structure to the $dx$ formula:

• $dx$ uses top minus bottom
• $dy$ uses right minus left

A helpful memory aid is: T-B with $dx$, R-L with $dy$.

How to find the $y$-bounds $c$ and $d$

The bounds for a $dy$ integral are $y$-values, so you typically find them by locating the lowest and highest points of the region in the vertical direction.

Common ways they appear:

• The region includes intersection points; you solve for the corresponding $y$ values.
• The problem states “from $y=c$ to $y=d$.”
• The region is bounded by a curve and the axis $y=0$ (or another horizontal line).

A common mistake is to keep using $x$-bounds out of habit. If you integrate with respect to $y$, the limits must be $y$-values.

Converting from $y=f(x)$ to $x=g(y)$ (and when that’s tricky)

To use $dy$, you often need boundaries in the form $x=\text{(something in }y\text{)}$. That usually requires solving equations for $x$.

For example, if:

$y=x^2$

then solving for $x$ gives:

$x=\pm \sqrt{y}$

This immediately raises an important point: some curves become two separate functions of $y$ (a left branch and a right branch). In area problems with $dy$, that can be a feature, not a bug, because horizontal slices naturally see a left boundary and a right boundary.

Worked Example 1: Same region, different variable (seeing the symmetry)

Find the area bounded by $y=x^2$ and $y=2x$ using integration with respect to $y$.

Step 1: Visualize and find intersection points

Solve:

$x^2=2x$

$x^2-2x=0$

$x(x-2)=0$

So intersections occur at $x=0$ and $x=2$.

Corresponding $y$-values (plug into either equation):

• at $x=0$, $y=0$
• at $x=2$, $y=4$

So $y$ runs from $0$ to $4$.

Step 2: Express boundaries as $x$ in terms of $y$

From $y=x^2$:

$x=\sqrt{y}$

(We use $x=\sqrt{y}$ as the right branch in the first quadrant, which matches the region here.)

From $y=2x$:

$x=\frac{y}{2}$

Now decide which is right and which is left for $0\le y\le 4$.
Take a test value, say $y=1$:

$x=\sqrt{1}=1$

$x=\frac{1}{2}$

So $x=\sqrt{y}$ is to the right, and $x=\frac{y}{2}$ is to the left.

Step 3: Set up and evaluate “right minus left”

$A=\int_0^4 \left(\sqrt{y}-\frac{y}{2}\right)dy$

Compute:

$\int \sqrt{y}dy=\int y^{1/2}dy=\frac{2}{3}y^{3/2}$

$\int \frac{y}{2}dy=\frac{1}{2}\cdot \frac{y^2}{2}=\frac{y^2}{4}$

So:

$A=\left(\frac{2}{3}y^{3/2}-\frac{y^2}{4}\right)_0^4$

Evaluate at $y=4$:

$\frac{2}{3}\cdot 4^{3/2}-\frac{16}{4}=\frac{2}{3}\cdot 8-4=\frac{16}{3}-4=\frac{4}{3}$

At $y=0$ the expression is $0$, so the area is:

$A=\frac{4}{3}$

This is the same region you could also do with $dx$; using $dy$ worked cleanly because the left and right boundaries were simple.

Worked Example 2: A circle cap where $dy$ is often cleaner

Find the area of the region inside the circle $x^2+y^2=4$ to the right of the $y$-axis.

Step 1: Describe the region

“Inside” the circle means all points with $x^2+y^2\le 4$ (a radius-2 circle centered at the origin). “To the right of the $y$-axis” means $x\ge 0$. So this is the right half of the circle.

This region is a classic case where writing the boundary as $y=f(x)$ would require splitting into upper and lower semicircles. Horizontal slices avoid that.

Step 2: Use horizontal slices and solve for $x$

From the circle equation:

$x^2=4-y^2$

Since we only want the right half (where $x\ge 0$):

$x=\sqrt{4-y^2}$

For a given $y$, the slice runs from the $y$-axis (left boundary) to the circle (right boundary):

$L(y)=0$

$R(y)=\sqrt{4-y^2}$

The lowest and highest $y$ values in the circle are $-2$ and $2$, so $y\in[-2,2]$.

Step 3: Set up the area integral

$A=\int_{-2}^{2} \left(\sqrt{4-y^2}-0\right)dy$

You are not expected in AP Calculus AB to evaluate this integral by hand using advanced techniques (it’s a semicircle area integral). But you can interpret it geometrically: it represents the area of a semicircle of radius $2$, so:

$A=\frac{1}{2}\pi (2^2)=2\pi$

This example shows a deeper point: integrals can encode geometric areas even when the antiderivative is not elementary in the tools you’ve learned.

Choosing between $dx$ and $dy$: what the exam expects you to notice

On many AP-style questions, both methods are possible, but one produces a single clean integral while the other requires splitting.

For example:

• If a region is bounded left-right by curves that are easy to express as $x$ in terms of $y$, $dy$ is often simpler.
• If the region is bounded top-bottom by curves that are easy to express as $y$ in terms of $x$, $dx$ is often simpler.

A frequent error is to pick $dy$ but forget to rewrite everything in terms of $y$, leading to an integral like $\int (f(x))dy$, which doesn’t make sense.

What can go wrong with “right minus left”

The same issue you saw with top/bottom switching can happen here: the right boundary might change as $y$ changes. When that happens, you must split the integral at the $y$-value where the switch occurs.

The best prevention is always the same:

• draw a quick sketch
• draw a representative horizontal slice
• verify which curve is left and which is right over the interval

Exam Focus

• Typical question patterns:
  • “Find the area of the region bounded by $x=R(y)$ and $x=L(y)$ from $y=c$ to $y=d$.”
  • “Find the area of the region enclosed by … using integration with respect to $y$.” (Often signals that $dx$ would be messy.)
  • “Set up an integral (or integrals) to find the area …” where the main challenge is expressing boundaries as functions of $y$.
• Common mistakes:
  • Using $x$-limits when integrating with respect to $y$ (limits must match the variable).
  • Not solving for $x$ correctly (missing the correct branch like $x=\sqrt{4-y^2}$ vs $x=-\sqrt{4-y^2}$).
  • Forgetting that the left/right boundary can change, so a single integral may not cover the whole region without splitting.