Unit 8: Applications of Integration

Average Value of a Function

When you hear “average value,” it’s tempting to think it means “add up outputs and divide by how many.” That idea works for a finite list of numbers, but a function on an interval produces infinitely many values. The average value of a function on an interval is the continuous version of an average: you average all the function outputs across every input in the interval.

A helpful memory hook is the everyday average idea: “add everything up, then divide.” For functions, the definite integral is what “adds everything up,” and dividing by the interval length completes the average.

What it is (and why integrals appear)

Imagine the graph of f(x) on [a,b]. The integral

\int_a^b f(x)\,dx

measures the signed area between the curve and the x-axis. If you divide that total accumulated area by the width of the interval b-a, you get a “typical height” that would produce the same area if the function were constant. That constant height is exactly the average value.

Definition and formula

The average value of f on [a,b] is

f_{avg} = \frac{1}{b-a}\int_a^b f(x)\,dx

• a and b are the interval endpoints.
• f_{avg} is a single number (a height).

For instance, on the interval from 0 to 40, the average value setup is

f_{avg} = \frac{1}{40}\int_0^{40} f(x)\,dx

A useful interpretation is: there is some input c in [a,b] where the function actually equals its average value (this is the Mean Value Theorem for Integrals). You don’t usually have to find c in AP Calculus AB, but the idea helps you trust that the average value isn’t just an abstract construction.

How it works (step-by-step)

1. Identify the interval [a,b].
2. Compute the definite integral \int_a^b f(x)\,dx.
3. Divide by b-a.

Be careful: if f(x) is sometimes negative, the integral includes negative area. So the average value is an average of signed values, not an average of distances from the axis.

Example 1: Average value of a polynomial

Find the average value of f(x) = x^2 on [0,3].

Compute the integral:

\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9

Divide by the interval length 3-0=3:

f_{avg} = \frac{1}{3}\cdot 9 = 3

Interpretation: the constant function y=3 makes the same area over [0,3] as y=x^2 does.

Example 2: Average value with negative parts

Find the average value of f(x) = x-2 on [0,4].

\int_0^4 (x-2)\,dx = \left[\frac{x^2}{2} - 2x\right]_0^4 = \left(8 - 8\right) - 0 = 0

Then

f_{avg} = \frac{1}{4-0}\cdot 0 = 0

Even though the line is not always at 0, the positive area above the axis cancels the negative area below the axis.

Common misconceptions to avoid

• Forgetting the “signed” nature of area: if you want an “average height above the axis,” you would need an average of |f(x)|, which is not the standard average value.
• Dividing by the wrong quantity: you divide by b-a, not by b or a.
• Averaging endpoints: \frac{f(a)+f(b)}{2} is generally not the average value (it’s related to the trapezoidal rule approximation, not the exact average).

Exam Focus

Typical question patterns:

• “Find the average value of f(x) on [a,b].”
• “A function is defined by an integral; compute its average value on an interval.”
• “Use an integral to find a constant value that represents the average output.”

Common mistakes:

• Dropping parentheses and computing \frac{1}{b} - a instead of \frac{1}{b-a}.
• Treating average value as average rate of change.
• Ignoring negative values and incorrectly using geometric area.

Position, Velocity, and Acceleration

Similar to how you can go from position to velocity using a derivative, you can do the reverse using an integral. In motion problems, integrals connect the three main quantities by accumulating change over time.

The key accumulation relationships

Over a time interval, integrating velocity gives displacement (net change in position), and integrating acceleration gives change in velocity. If instead you integrate speed (the absolute value of velocity), you get total distance traveled.

A compact reference is:

Using the Fundamental Theorem of Calculus (FTC)

The FTC still applies: a definite integral of a rate of change gives a net change. For example, if acceleration is a(t), then

\int_a^b a(t)\,dt = v(b) - v(a)

This is often the fastest way to compute velocity at a time if you know an initial velocity.

Exam Focus

Typical question patterns:

• “Given v(t), find displacement on a time interval.”
• “Given v(t), find total distance traveled (watch for absolute value).”
• “Given a(t) and an initial velocity, find v(t) or compute v(b).”

Common mistakes:

• Forgetting that displacement can be negative.
• Using \int v(t)\,dt when the question asks for total distance (which requires \int |v(t)|\,dt).
• Treating \int_a^b a(t)\,dt as a velocity value by itself, instead of as v(b)-v(a).

Area Between Curves

One of the most important applications of integration is turning geometry into an integral you can compute. Area between curves is the classic example: you use integration to add up infinitely many thin rectangles whose heights are set by the distance between two graphs.

What it is

The integral gives the area below a function (as signed area). Therefore, if you want the area between two curves, you can subtract one accumulated area from the other, as long as you set it up as a nonnegative distance.

The main formula (vertical slices)

If f(x) is above g(x) on [a,b], then

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

The integrand f(x)-g(x) is the vertical distance between the curves, so the area comes out nonnegative.

How to choose bounds correctly

The limits a and b often come from intersection points. If the region is “enclosed,” the curves cross at the endpoints of the region, so you solve

f(x) = g(x)

to find the boundary x-values. A common habit is to start “where the functions start” (often at 0) and go to an intersection, but on exams you should rely on the actual region description: the correct bounds come from endpoints given in the problem or from where boundaries meet.

When “top minus bottom” changes

Sometimes the two curves cross inside the interval, meaning the “top” function changes. In that case, you must split the integral at the intersection point(s) so that each piece uses the correct top-minus-bottom order. If you don’t split, your integral will compute a signed area (with cancellations), which is not the actual geometric area.

Example 1: Region between two functions

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

Step 1: Find intersection points.

x = x^2

x^2 - x = 0

x(x-1)=0

So x=0 and x=1.

Step 2: Determine which is on top on [0,1].

For inputs strictly between 0 and 1, x > x^2, so the top is y=x and the bottom is y=x^2.

Step 3: Set up and compute the integral.

A = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}

Example 2: Area where curves cross inside the interval

Find the area between y = x^3 and y = x on [-1,1].

Step 1: Find intersections.

x^3 = x

x^3 - x = 0

x(x^2 - 1)=0

So x=-1, 0, 1.

Step 2: Decide top and bottom on each subinterval.

On [-1,0], test the point

x = -\frac{1}{2}

Then

x^3 = -\frac{1}{8}

so x^3 > x there.

On [0,1], test the point

x = \frac{1}{2}

Then

x^3 = \frac{1}{8}

so x > x^3 there.

Step 3: Split the area integral.

A = \int_{-1}^0 (x^3 - x)\,dx + \int_0^1 (x - x^3)\,dx

Compute each:

\int (x^3 - x)\,dx = \frac{x^4}{4} - \frac{x^2}{2}

So

\int_{-1}^0 (x^3 - x)\,dx = \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^0 = 0 - \left(\frac{1}{4} - \frac{1}{2}\right) = \frac{1}{4}

Similarly,

\int_0^1 (x - x^3)\,dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

Total area:

A = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

Example 3: Subtracting two “area under the curve” integrals

Because the integral gives accumulated area below a graph, another perfectly valid setup is “area under the top curve minus area under the bottom curve” over the same bounds. For example, if the top function is y=5x-x^2 and the bottom function is y=x on [0,4], you can write the area between them as

\int_0^4 (5x-x^2)\,dx - \int_0^4 x\,dx

This is equivalent to a single integral of the difference, but it can be a useful way to organize your thinking.

Horizontal slices (solving for x)

Sometimes a region is easier to describe using horizontal distances (right minus left) instead of vertical distances. That requires rewriting curves as x in terms of y.

If the region runs from y=c to y=d and the right boundary is x=R(y) and left boundary is x=L(y), then

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

In AP Calculus AB, vertical slices are most common, but you should recognize that either orientation is possible depending on how the region is given.

Common misconceptions to avoid

• Integrating the wrong difference: you must use top minus bottom (or right minus left). If you do bottom minus top, you’ll get a negative value.
• Not checking which function is on top: don’t assume based on the function name; test a point.
• Forgetting to split at intersections: if curves switch order, one integral will “cancel” parts of the area.

Exam Focus

Typical question patterns:

• “Find the area of the region enclosed by f(x) and g(x).”
• “Set up (and sometimes evaluate) an integral representing area between curves.”
• “A region is shown/defined; write an expression for its area.”

Common mistakes:

• Using intersection y-values as bounds when integrating with respect to x.
• Missing an intersection point and using incorrect limits.
• Treating signed area as geometric area.

Volumes with Cross Sections (Slicing)

Area between curves uses rectangles; volume problems use 3D “slices.” The main idea is consistent: if you can describe the cross-sectional area of a solid at position x (or y), then integrating those areas adds up the entire volume.

What it is

A cross section is the shape you get when you “slice” a solid with a plane. If you take many thin slices perpendicular to an axis, each slice has a small thickness and some area. Volume is approximately

\text{volume} \approx \sum A(x_i)\,\Delta x

Taking the limit as the slices become infinitely thin produces an integral.

The slicing formula

If cross sections perpendicular to the x-axis have area A(x) for a \le x \le b, then

V = \int_a^b A(x)\,dx

Similarly, for slices perpendicular to the y-axis,

V = \int_c^d A(y)\,dy

The entire challenge is converting geometry into the function A(x) or A(y).

A common “first model” is rectangular cross sections: if a slice has length and width, then the cross-sectional area is length times width, so the volume setup looks like

V = \int (\text{length})(\text{width})\,dx

where the integration bounds match the extent of the solid along the direction you slice.

Building A(x): common cross-sectional shapes

In many AP problems, you’re given a base region in the xy-plane and told that cross sections perpendicular to an axis are squares, rectangles, semicircles, circles, or equilateral triangles. The base region tells you a “width” of a slice.

• With vertical slices, the slice width is typically

\text{width}(x) = \text{top}(x) - \text{bottom}(x)

• With horizontal slices, the slice width is typically

\text{width}(y) = \text{right}(y) - \text{left}(y)

Then you plug that width into the area formula for the cross-sectional shape.

Notation reference (width-to-area)

A big theme: many cross sections lead to A(x) being a constant multiple of w(x)^2.

Example 1: Squares on a base region

The base region is bounded by y=\sqrt{x} and y=0 from x=0 to x=4. Cross sections perpendicular to the x-axis are squares. Find the volume.

Step 1: Identify the slice width.

With vertical slices, the segment in the base goes from bottom y=0 to top y=\sqrt{x}, so

w(x) = \sqrt{x} - 0 = \sqrt{x}

Step 2: Convert width to cross-sectional area.

Square area:

A(x) = w(x)^2 = (\sqrt{x})^2 = x

Step 3: Integrate area over the interval.

V = \int_0^4 x\,dx = \left[\frac{x^2}{2}\right]_0^4 = 8

So the volume is 8 cubic units.

Example 2: Semicircles on a base region

The base region is bounded by y=4-x^2 and y=0. Cross sections perpendicular to the x-axis are semicircles. Find the volume.

Step 1: Find the interval for x.

Intersections with y=0:

4 - x^2 = 0

x^2 = 4

So x=-2 to x=2.

Step 2: Width of the diameter.

Top minus bottom:

w(x) = (4-x^2) - 0 = 4-x^2

Step 3: Area of a semicircle with diameter w.

A(x) = \frac{\pi}{8}w(x)^2 = \frac{\pi}{8}(4-x^2)^2

Step 4: Integrate.

V = \int_{-2}^2 \frac{\pi}{8}(4-x^2)^2\,dx

You can expand or use symmetry. Since \left(4-x^2\right)^2 is even,

V = 2\int_0^2 \frac{\pi}{8}(4-x^2)^2\,dx = \frac{\pi}{4}\int_0^2 (4-x^2)^2\,dx

Expand:

(4-x^2)^2 = 16 - 8x^2 + x^4

Integrate:

V = \frac{\pi}{4}\left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_0^2

Evaluate at 2:

V = \frac{\pi}{4}\left(32 - \frac{64}{3} + \frac{32}{5}\right)

Combine:

32 = \frac{480}{15}

\frac{64}{3} = \frac{320}{15}

\frac{32}{5} = \frac{96}{15}

So inside parentheses:

\frac{480}{15} - \frac{320}{15} + \frac{96}{15} = \frac{256}{15}

Thus

V = \frac{\pi}{4}\cdot \frac{256}{15} = \frac{64\pi}{15}

Common misconceptions to avoid

• Using the wrong “width”: the width comes from the base region in the plane; it’s not automatically x or y.
• Mixing up radius and diameter: many problems give diameter; the radius is half of it.
• Integrating with the wrong variable: if slices are perpendicular to the x-axis, integrate with respect to x.

Exam Focus

Typical question patterns:

• “A solid has a base region bounded by … Cross sections perpendicular to the x-axis are squares/semicircles/etc. Find the volume.”
• “Set up (and possibly evaluate) the integral for the volume.”
• “Given a diagram of a base region, express A(x) and write V.”

Common mistakes:

• Forgetting to square the width when the cross section area depends on w^2.
• Using the wrong bounds (especially when the base is defined by intersections).
• Applying the wrong area formula (for example, using circle instead of semicircle).

Volumes of Revolution (Disk and Washer Methods)

Another major application is building a 3D solid by spinning a 2D region around a line (an axis of rotation). If you rotate a 2D region, you create a 3D object, and integration adds up the volumes of many thin rotated slices.

In many common setups, those rotated slices form disks (solid circles) or washers (circles with holes). Because the area of a circle is \pi r^2, disk and washer problems often reduce to integrating a circle-based cross-sectional area.

What it means to “revolve” a region

If you rotate a region around a horizontal or vertical line, every point sweeps out a circle. Thin slices of the region sweep out thin disks or washers. Adding up the volumes of those disks/washers produces the total volume.

A key skill is translating “distance from the axis” into a radius.

Disk method (no hole)

Use the disk method when the slices rotate into solid disks. If you rotate around a horizontal line and use vertical slices, each slice forms a disk with radius R(x). Then

V = \pi\int_a^b \left(R(x)\right)^2\,dx

Washer method (hole in the middle)

Use the washer method when rotation creates a hole. Each slice becomes a washer with outer radius R(x) and inner radius r(x).

V = \pi\int_a^b \left(\left(R(x)\right)^2 - \left(r(x)\right)^2\right)\,dx

You may also see the washer idea written as “outer disk minus inner disk,” which matches the structure

\int \pi\left(R(x)\right)^2\,dx - \int \pi\left(r(x)\right)^2\,dx

When the bounds are the same, this is equivalent to a single integral of \pi\left(R(x)^2 - r(x)^2\right).

How to decide radii: “distance to axis” thinking

This is where many mistakes happen, so it’s worth slowing down.

• If you rotate around the x-axis (the line y=0), then a radius is typically a y-value: R(x)=f(x) if the region touches the axis.
• If you rotate around a horizontal line y=k, then radius is a vertical distance: |f(x)-k|.
• If you rotate around the y-axis (the line x=0) using horizontal slices, radii are often x-values.
• If you rotate around a vertical line x=h, then radius is a horizontal distance: |x-h|.

On AP problems, absolute values are often avoidable if you set things up so distances are naturally positive on the interval, but the “distance” interpretation is always correct.

Choosing dx versus dy

• Vertical slices (integrate dx) are natural when curves are given as y=f(x) and you rotate around a horizontal axis.
• Horizontal slices (integrate dy) are natural when curves are given as x=g(y) and you rotate around a vertical axis.

You can do either in many cases, but one setup is usually much simpler.

Example 1: Disk method about the x-axis

Find the volume when the region under y=\sqrt{x} from x=0 to x=4 is revolved about the x-axis.

Step 1: Identify radius.

Rotating around y=0, the radius is the function value:

R(x) = \sqrt{x}

Step 2: Use disk formula.

V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx

Compute:

V = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi

Example 2: Washer method about the x-axis

Find the volume of the solid formed by revolving the region between y=x and y=x^2 on [0,1] about the x-axis.

Step 1: Outer and inner radii.

On [0,1], x is above x^2. Rotating around the x-axis:

R(x) = x

r(x) = x^2

Step 2: Washer integral.

V = \pi\int_0^1 \left((x)^2 - (x^2)^2\right)\,dx = \pi\int_0^1 (x^2 - x^4)\,dx

Compute:

V = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}

Example 3: Rotation about a horizontal line y=k

Revolve the region under y=x^2 from x=0 to x=2 about the line y=5. Find the volume.

Step 1: Determine whether it’s disks or washers.

The region is between y=0 and y=x^2. When rotating about y=5 (above the region), you get washers because there is a hole: the region does not touch the axis of rotation.

Step 2: Find outer and inner radii as distances to y=5.

Outer radius goes from y=5 down to y=0:

R(x) = 5 - 0 = 5

Inner radius goes from y=5 down to y=x^2:

r(x) = 5 - x^2

Step 3: Set up and compute.

V = \pi\int_0^2 \left(5^2 - (5-x^2)^2\right)\,dx

Simplify inside:

(5-x^2)^2 = 25 - 10x^2 + x^4

So

25 - (25 - 10x^2 + x^4) = 10x^2 - x^4

Thus

V = \pi\int_0^2 (10x^2 - x^4)\,dx = \pi\left[\frac{10x^3}{3} - \frac{x^5}{5}\right]_0^2

Evaluate at 2:

V = \pi\left(\frac{80}{3} - \frac{32}{5}\right)

Common denominator 15:

\frac{80}{3} = \frac{400}{15}

\frac{32}{5} = \frac{96}{15}

So

V = \pi\cdot \frac{304}{15} = \frac{304\pi}{15}

A note on setup vs evaluation

Some AP questions emphasize setting up the correct integral more than computing it, especially when the integral is messy or requires a calculator. Even then, your setup must include correct bounds, correct radii expressions, correct squaring and subtraction for washers, and the factor of \pi.

Common misconceptions to avoid

• Using function values as radii when rotating about y=k: radii are distances to the line, not just f(x).
• Swapping inner and outer radius: always use R^2 - r^2, where R is the larger distance.
• Forgetting to square radii: disk/washer volumes depend on area, and area depends on radius squared.

Exam Focus

Typical question patterns:

• “Find the volume of the solid formed by revolving the region bounded by … about the x-axis (or y-axis, or y=k).”
• “Set up an integral for volume using disks/washers.”
• “Given a shaded region, write the volume integral and evaluate.”

Common mistakes:

• Using the wrong method orientation (trying to use dx when dy would avoid solving for x in terms of y).
• Incorrect bounds from intersection points.
• Using “top minus bottom” language without translating it into distances to the axis.

Strategy for Modeling: Turning Geometry and Context into Integrals

Unit 8 problems often feel hard not because the calculus is advanced, but because the translation step is subtle. You’re given a picture, a region description, or a physical context, and your job is to produce the correct integral.

A reliable modeling workflow

When you’re not sure what to do, use a consistent process:

1. Sketch (even a rough sketch). Mark intersections, axes, and the axis of rotation if there is one.
2. Decide the slicing direction. Ask: are cross sections perpendicular to the x-axis or y-axis? Are you rotating around a horizontal or vertical line?
3. Write the key geometric quantity:
   • For area: a distance (top minus bottom, right minus left).
   • For volume by cross sections: a width that becomes an area formula.
   • For volume by revolution: radii measured as distances to the axis.
4. Choose bounds from intersections or given endpoints.
5. Build the integral with the correct differential dx or dy.
6. Only then evaluate (or leave it as a setup if appropriate).

Example: A “setup-first” volume problem

The region is bounded by y=\sin(x) and y=0 on [0,\pi]. Revolve about the x-axis.

Axis is horizontal and the region is given as y in terms of x, so vertical slices with dx are natural. The radius is

R(x)=\sin(x)

So

V = \pi\int_0^{\pi} (\sin(x))^2\,dx

Even if you don’t evaluate immediately, this setup is the essential calculus result.

Recognizing when you must split an integral

Splitting happens any time the rule that defines your integrand changes:

• For area: the top curve changes.
• For volume by revolution: the outer/inner radius expressions change.
• For cross sections: the width formula changes.

A quick way to detect this is to locate intersection points and consider whether the region description is uniform across the entire interval.

Common misconceptions to avoid

• Trying to memorize templates instead of reasoning from slices: templates fail when the axis changes.
• Not drawing a picture: many errors disappear once you see which curve is top or how far the axis is.
• Confusing the region with the solid: the region is 2D; the solid is 3D. Your integrand must represent an area (for volume) or a distance (for area).

Exam Focus

Typical question patterns:

• “Set up an integral expression (no evaluation required) for area/volume.”
• “A region is described in words; write the integral that models the quantity.”
• “Use a graph to determine bounds and expressions.”

Common mistakes:

• Missing a required split when a boundary changes.
• Using dx when the expressions are naturally in y (or vice versa).
• Writing an integrand with the wrong units (for volume, the integrand must be an area).

Connecting the Big Ideas (Why Unit 8 Looks Like One Theme)

At first glance, average value, area between curves, and volume might look like separate tricks. They’re really the same idea with different interpretations: a definite integral accumulates a quantity built from a geometric or physical meaning.

One unifying viewpoint: “sum of thin slices”

Average value: total accumulated output divided by width.

f_{avg} = \frac{1}{b-a}\int_a^b f(x)\,dx

Area between curves: accumulate many thin rectangles whose heights are distances.

A = \int_a^b (\text{top} - \text{bottom})\,dx

Volume: accumulate many thin slabs whose volumes are cross-sectional area times thickness.

V = \int_a^b A(x)\,dx

Volume by revolution: a special case where cross-sectional area comes from circles or washers.

V = \pi\int_a^b (R(x))^2\,dx

V = \pi\int_a^b ((R(x))^2 - (r(x))^2)\,dx

The formulas are different on the surface, but the reasoning is the same: create an expression for a slice, then integrate.

How to sanity-check your work

A quick “reasonableness check” can catch mistakes:

• Area should be nonnegative. If you got a negative answer, you likely reversed top and bottom or forgot to split.
• Volume should be nonnegative and have cubic units. If your integrand looks like a length, you probably forgot to square a radius or use an area formula.
• Bounds should match the variable of integration. If you integrate dx, bounds are x-values.

Exam Focus

Typical question patterns:

• “Explain the meaning of a definite integral in context” (often tied to geometry in this unit).
• “Choose an appropriate integral model and justify components (bounds, integrand).”
• “Compute an accumulated quantity and interpret it.”

Common mistakes:

• Treating the integral as a purely algebraic procedure rather than an accumulation of slices.
• Not checking whether the integrand represents the intended geometric quantity.
• Losing track of what the variable represents when switching between x and y descriptions.