Unit 8: Applications of Integration

Average Value of a Function (and what an integral is “counting”)

Integration is powerful because it turns a local description (how something changes at each input) into a global description (a total accumulated amount over an interval). In Unit 8, you use that idea to compute things like areas, volumes, work, and other accumulated quantities. A great first application that shows how integrals aggregate information is the average value of a function.

What average value means

The ordinary average from arithmetic is “add everything up and then divide.” For a function over a continuous interval, you can’t literally add infinitely many outputs one by one, so you approximate the “sum” using Riemann sums and then take a limit. That limit is an integral.

If you want the average height of a curve given by

$y=f(x)$

over an interval

$[a,b]$

you want “total height accumulated” divided by “total width.” The “total height accumulated” is captured by the integral, and the width is the interval length.

So the average value of $f$ on $[a,b]$ is

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

This is sometimes called the mean value of the function (not to be confused with the Mean Value Theorem for derivatives).

A quick numeric-style example of the “divide by the interval length” idea: on an interval from $0$ to $40$, the average value is

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

Why it matters

Average value shows up whenever a quantity varies continuously and you need a representative “typical” value.

In physics, if velocity is given by

$v(t)$

then average velocity on $[a,b]$ is

$\frac{1}{b-a}\int_a^b v(t)\,dt$

In geometry, when $f(x)\ge 0$ represents a height, the average value is the height of a rectangle with the same base and the same area as the region under $f$. This rectangle interpretation is important: the average value is exactly the constant height that would produce the same accumulated area.

How it works (conceptually)

Partition $[a,b]$ into many subintervals of equal width, sample heights, and take an average. If you start with a “plain” average of sampled heights,

$\frac{1}{n}\sum_{i=1}^n f(x_i)$

that treats each sample equally but doesn’t show the physical width. Multiply and divide by the subinterval width to connect it to accumulation:

$\frac{1}{n}\sum_{i=1}^n f(x_i)=\frac{1}{b-a}\sum_{i=1}^n f(x_i)\Delta x$

As the partition gets finer, the sum becomes an integral:

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

Mean Value Theorem for Integrals (the “there exists a point” fact)

If $f$ is continuous on $[a,b]$, then there exists some $c$ in $[a,b]$ such that

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

Interpretation: at least one point on the graph hits the average height.

Example 1: Compute average value

Find the average value of

$f(x)=x^2$

on

$[0,2]$

Compute the integral:

$\int_0^2 x^2\,dx=\left[\frac{x^3}{3}\right]_0^2=\frac{8}{3}$

Divide by interval length:

$f_{\text{avg}}=\frac{1}{2}\cdot\frac{8}{3}=\frac{4}{3}$

Example 2: Find a point where the function equals its average

Using the previous result, find $c$ in $[0,2]$ such that

$c^2=\frac{4}{3}$

Solve:

$c=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}$

This is a direct use of the Mean Value Theorem for Integrals.

Exam Focus

• Typical question patterns:
  • “Find the average value of $f$ on $[a,b]$.”
  • “Find a value $c$ such that $f(c)=f_{\text{avg}}$.”
  • “Interpret the meaning of $\int_a^b f(x)\,dx$ and $f_{\text{avg}}$ in context (include units).”
• Common mistakes:
  • Dividing by $b$ instead of $b-a$.
  • Using average value when the question is actually asking for average rate of change (slope) rather than average height.
  • Ignoring sign: if $f$ is negative on much of the interval, the average value will be negative.

Position, Velocity, and Acceleration (using integrals to “undo” derivatives)

Derivatives move you from position to velocity to acceleration. Integrals move you back the other way by accumulating change over time.

The core relationships

If position is $s(t)$, velocity is $v(t)$, and acceleration is $a(t)$, then

$v(t)=s'(t)$

$a(t)=v'(t)$

Integrating reverses these relationships in an accumulated way.

Displacement, total distance, and the Fundamental Theorem of Calculus

Displacement on a time interval $[a,b]$ is the signed change in position:

$\text{Displacement}=\int_a^b v(t)\,dt$

Total distance traveled accumulates speed, so it uses absolute value:

$\text{Total distance}=\int_a^b |v(t)|\,dt$

Position can be recovered from velocity by adding the starting position:

$s(b)=s(a)+\int_a^b v(t)\,dt$

Similarly, velocity can be recovered from acceleration:

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

Using the Fundamental Theorem of Calculus, the last statement is often also written as

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

A compact summary (matching common AP language) is:

Exam Focus

• Typical question patterns:
  • “Given $v(t)$, find displacement or total distance traveled on $[a,b]$.”
  • “Given $a(t)$ and $v(a)$, find $v(b)$.”
  • “Use FTC to interpret a definite integral like $\int_a^b a(t)\,dt$.”
• Common mistakes:
  • Confusing displacement with total distance (forgetting the absolute value for distance).
  • Forgetting to add the initial value (like $s(a)$ or $v(a)$) when asked for the actual position or velocity, not just the change.
  • Losing the sign meaning: negative velocity contributes negative displacement.

Area Between Curves (turning geometry into integrals)

A definite integral naturally measures signed area: area above the $x$-axis counts positive and area below counts negative. But many geometry problems ask for area in the everyday sense, which is always nonnegative. To find the area between two curves, you accumulate the distance between them across an interval.

The main idea: “top minus bottom”

Suppose you have two functions

$y=f(x)$

and

$y=g(x)$

on an interval $[a,b]$. If $f(x)\ge g(x)$ on that interval, then the area between them is

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

This comes from slicing into thin vertical rectangles: each slice has width $dx$ and height “top minus bottom,” so

$dA=\left(f(x)-g(x)\right)dx$

Choosing bounds (start to intersection)

A common workflow is:

1. Find intersection points by solving $f(x)=g(x)$.
2. Use those intersection points as bounds.
3. Integrate “top minus bottom” on each interval where the identity of the top function stays the same.

Many problems are exactly “integrate the top function and subtract the bottom function” across the interval where the region is enclosed.

When curves cross: split the integral

If the curves cross, you must split the integral at the crossing points. Otherwise, signed area can cancel.

Integrating with respect to $y$ (right minus left)

Sometimes the region is easier to describe using horizontal slices. If curves are written as

$x=F(y)$

and

$x=G(y)$

and $F(y)\ge G(y)$ on $[c,d]$, then

$A=\int_c^d \left(F(y)-G(y)\right)\,dy$

Here you use “right minus left” because a horizontal slice measures width.

Example 1: Classic top-minus-bottom

Find the area between

$y=2x$

and

$y=x^2$

Step 1: Intersections.

$2x=x^2$

$x^2-2x=0$

$x(x-2)=0$

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

Step 2: Which is on top? On $[0,2]$, $2x\ge x^2$.

Step 3: Integrate top minus bottom.

$A=\int_0^2 (2x-x^2)\,dx$

$\int_0^2 (2x-x^2)\,dx=\left[x^2-\frac{x^3}{3}\right]_0^2=4-\frac{8}{3}=\frac{4}{3}$

Example 2: When you must split the interval

Find the area between

$y=x^3$

and

$y=x$

on

$[-1,1]$

Step 1: Intersections.

$x^3=x$

$x^3-x=0$

$x(x^2-1)=0$

So $x=-1,0,1$.

Step 2: Determine top/bottom on each subinterval. On $[-1,0]$, $x^3>x$. On $[0,1]$, $x>x^3$.

Step 3: Add two integrals.

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

An antiderivative used in the computation is

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

First part:

$\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^0=\frac{1}{4}$

Second part:

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

Total area:

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

Example 3: Setup by subtracting one area from another

If the top curve is

$y=5x-x^2$

and the bottom curve is

$y=x$

over the interval from $0$ to $4$, one correct setup for the area between them is

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

Equivalently, you can combine into one integral of “top minus bottom,” but the key idea is the same: subtract the accumulated area under the bottom function from the accumulated area under the top function over the same bounds.

Exam Focus

• Typical question patterns:
  • “Find the area of the region enclosed by $y=f(x)$ and $y=g(x)$.”
  • “Set up (not necessarily evaluate) an integral for the area between curves; justify bounds.”
  • “Decide whether to integrate with respect to $x$ or $y$ and explain why.”
• Common mistakes:
  • Forgetting to split at intersection points (leading to cancellation).
  • Using $\int (\text{bottom}-\text{top})$ and getting a negative answer for area.
  • Mixing up “top minus bottom” (for $dx$) with “right minus left” (for $dy$).

Volume by Cross Sections (slicing solids)

Many volume problems look complicated in 3D but become straightforward when you remember that you can build a solid by stacking many thin slices, each with a measurable cross-sectional area.

The slicing principle

If a solid extends from $x=a$ to $x=b$ and a cross section perpendicular to the $x$-axis at position $x$ has area $A(x)$, then

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

A thin slice has thickness $dx$ and volume approximately $A(x)dx$, so integrating adds all slices.

You can also integrate with respect to $y$:

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

Rectangular cross-sections as a special case

If the cross section is a rectangle at position $x$ with length $L(x)$ and width $W(x)$, then

$A(x)=L(x)W(x)$

and the volume becomes

$V=\int_a^b L(x)W(x)\,dx$

This matches the idea “cross-sectional area times thickness,” where the thickness is represented by $dx$.

Why it matters

This method is the foundation beneath disk/washer and shell methods. It also handles solids that are not formed by rotation but instead have cross sections like squares, semicircles, or equilateral triangles.

Common cross-section shapes (and their areas)

In AP-style problems, a “base region” in the plane is given, and you’re told what the cross sections look like.

• Square with side $s$:

$A=s^2$

• Rectangle with sides $s$ and $ks$:

$A=ks^2$

• Semicircle with diameter $d$:

$A=\frac{\pi}{8}d^2$

• Equilateral triangle with side $s$:

$A=\frac{\sqrt{3}}{4}s^2$

The key skill is expressing the side length or diameter as a distance between curves (as a function of $x$ or $y$).

Example 1: Squares on a base region

The base is the region under

$y=\sqrt{x}$

above the $x$-axis from $x=0$ to $x=4$. Cross sections perpendicular to the $x$-axis are squares.

At a given $x$, the side length is

$s(x)=\sqrt{x}$

So the cross-sectional area is

$A(x)=s(x)^2=x$

Volume:

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

Example 2: Semicircles with diameter between two curves

The base is the region between

$y=\sin x$

and

$y=\cos x$

on

$\left[0,\frac{\pi}{4}\right]$

Cross sections perpendicular to the $x$-axis are semicircles. The diameter is the vertical distance:

$d(x)=\cos x-\sin x$

So the area is

$A(x)=\frac{\pi}{8}(\cos x-\sin x)^2$

Volume setup:

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

Exam Focus

• Typical question patterns:
  • “A solid has base region $R$; cross sections perpendicular to the $x$-axis are squares/semicircles/triangles. Find the volume.”
  • “Write (but do not evaluate) an integral for the volume given a base bounded by curves.”
  • “Choose whether to integrate with respect to $x$ or $y$ based on how the cross-section dimension is described.”
• Common mistakes:
  • Using the function value itself as the side length when the side is actually a distance between two curves.
  • Forgetting that a semicircle area uses diameter-to-radius conversion.
  • Using wrong bounds because you didn’t find where the boundary curves intersect.

Volumes of Revolution: Disk and Washer Methods

A very common way to create a 3D solid is to revolve a 2D region around a line (an axis of rotation). The integral comes from slicing that solid into thin circular pieces.

The core geometric idea

If you revolve a region around an axis and take a cross section perpendicular to that axis, the slice is usually a circle (disk) or a ring (washer).

• A disk occurs when the region touches the axis, so there is no hole.
• A washer occurs when there is a gap between the region and the axis, creating a hole.

Many rotation problems you see in practice produce circular cross sections, so disk/washer methods are extremely common. Still, you should not assume “disk always works best” without checking whether the algebra is reasonable.

Disk method (no hole)

If each slice has radius $R(x)$, then the cross-sectional area is

$A(x)=\pi R(x)^2$

and the volume is

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

If you instead slice horizontally, you use $dy$:

$V=\pi\int_c^d R(y)^2\,dy$

In all cases, the radius is a distance to the axis of rotation.

Washer method (hole in the middle)

If a slice is a washer with outer radius $R$ and inner radius $r$, then

$A=\pi(R^2-r^2)$

and

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

A common way to remember the structure is “outer disk minus inner disk.” Written as separate integrals, it looks like

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

Rotating around lines other than the coordinate axes

If you rotate around a horizontal line $y=k$ and your curve is $y=f(x)$, a radius is often

$R(x)=|f(x)-k|$

If you rotate around a vertical line $x=h$ and your curve is written as $x=g(y)$, then

$R(y)=|g(y)-h|$

In practice, you typically pick outer and inner radii so the distances are automatically nonnegative and you can avoid absolute value.

Example 1: Disk method about the $x$-axis

Rotate the region under

$y=x$

from $x=0$ to $x=1$ about the $x$-axis.

Radius:

$R(x)=x$

Volume:

$V=\pi\int_0^1 x^2\,dx=\pi\left[\frac{x^3}{3}\right]_0^1=\frac{\pi}{3}$

Example 2: Washer method between two curves

Rotate the region between

$y=x$

and

$y=x^2$

on $[0,1]$ about the $x$-axis. On $[0,1]$, $x\ge x^2$, so outer radius is $x$ and inner radius is $x^2$.

Setup:

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

Evaluate:

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

A practical decision: $dx$ or $dy$?

Disk/washer uses slices perpendicular to the axis of rotation.

• If the axis is horizontal, perpendicular slices are vertical, which suggests integrating with $dx$.
• If the axis is vertical, perpendicular slices are horizontal, which suggests integrating with $dy$.

If that orientation forces you to rewrite curves in an inconvenient form (for example, solving for $x$ in terms of $y$), that’s a strong hint the shell method might be cleaner.

Exam Focus

• Typical question patterns:
  • “Find the volume of the solid formed by revolving a region about the $x$-axis or $y$-axis.”
  • “Revolve around a shifted line like $y=1$; set up the washer integral.”
  • “Decide whether to use $dx$ or $dy$ and justify using ‘perpendicular slices.’”
• Common mistakes:
  • Squaring errors: forgetting that radii get squared inside $R^2-r^2$.
  • Mixing up radii with function values when the axis is shifted (you need distance to the axis, not just $y$).
  • Using a single integral even though the outer/inner curve changes (requires splitting).

Volumes of Revolution: Shell Method

The shell method is another way to compute volumes of revolution. Instead of slicing perpendicular to the axis (disks/washers), you slice parallel to the axis and get thin cylindrical shells.

What a shell is

Revolving a thin vertical strip at position $x$ about the $y$-axis creates a hollow cylinder with:

• radius = distance from the strip to the axis
• height = vertical length of the strip (top minus bottom)
• thickness = $dx$

A thin shell’s volume is approximately circumference times height times thickness:

$dV=2\pi(\text{radius})(\text{height})\,dx$

So the volume is

$V=2\pi\int_a^b (\text{radius})(\text{height})\,dx$

If you use horizontal shells (common when rotating about a horizontal axis), then

$V=2\pi\int_c^d (\text{radius})(\text{height})\,dy$

Why it matters (when shells are better)

Shells are often best when the axis is vertical but the functions are naturally written as $y=f(x)$ and $y=g(x)$, because washers would require rewriting everything in terms of $x$ as a function of $y$.

Example 1: Rotating under a curve about the $y$-axis

Rotate the region under

$y=x^2$

from $x=0$ to $x=1$ about the $y$-axis.

Radius:

$r(x)=x$

Height:

$h(x)=x^2$

Volume:

$V=2\pi\int_0^1 x\cdot x^2\,dx=2\pi\int_0^1 x^3\,dx=2\pi\left[\frac{x^4}{4}\right]_0^1=\frac{\pi}{2}$

Example 2: Rotating around a shifted vertical line

Rotate the region bounded by

$y=x$

$y=0$

and

$x=2$

about the line

$x=3$

Using vertical shells:

$r(x)=3-x$

$h(x)=x$

Bounds are $0$ to $2$. Setup:

$V=2\pi\int_0^2 (3-x)x\,dx$

Evaluate:

$V=2\pi\int_0^2 (3x-x^2)\,dx=2\pi\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^2=\frac{20\pi}{3}$

Disk/washer vs shell: a quick conceptual comparison

Disk/washer: slices are perpendicular to the axis, volume comes from stacking circular areas.

Shell: slices are parallel to the axis, volume comes from wrapping rectangles into cylindrical shells.

Neither method is “more correct.” The goal is to choose the method that makes the integral simplest.

Exam Focus

• Typical question patterns:
  • “Compute the volume of revolution using the shell method (often explicitly requested).”
  • “Set up an integral for volume about a vertical line $x=h$ or horizontal line $y=k$.”
  • “Choose between washer and shell and explain which is more efficient.”
• Common mistakes:
  • Using radius $x$ automatically, even when rotating about $x=h$ (must use distance).
  • Using height incorrectly (should be top minus bottom or right minus left, depending on orientation).
  • Mixing $dx$ and $dy$ with the wrong slice orientation.

Work Done by a Variable Force (integration as accumulated effort)

In physics, work measures energy transferred when a force moves an object through a distance.

What work is

If a constant force $F$ pushes an object through a distance $d$ in the force direction, then

$W=Fd$

If the force changes with position, you break the motion into tiny segments where the force is approximately constant and then add the contributions.

If force is given by $F(x)$ along the direction of motion, then the work from $x=a$ to $x=b$ is

$W=\int_a^b F(x)\,dx$

Units (a frequent interpretation point)

If force is in newtons and distance is in meters, then work is in newton-meters, also called joules. Tracking units helps you catch setup errors (for example, accidentally multiplying by an extra distance factor).

Hooke’s Law (springs)

A common variable-force model is a spring. Hooke’s Law says the force needed to stretch or compress a spring is proportional to displacement from its natural length:

$F(x)=kx$

Work to stretch from $x=a$ to $x=b$ is

$W=\int_a^b kx\,dx$

Example 1: Work to stretch a spring

A spring has constant $k=12$. Find the work required to stretch it from $x=0$ to $x=0.5$.

$W=\int_0^{0.5} 12x\,dx=12\left[\frac{x^2}{2}\right]_0^{0.5}=1.5$

A common misconception is trying to do “average force times distance” without doing the integral. That approach only works if you correctly compute the average force, which itself comes from the integral/average-value idea.

Pumping and lifting (work as force times small distance)

Many work setups use

$dW=(\text{force to lift slice})(\text{distance lifted})$

Often the “force to lift slice” is the weight of the slice, which is (weight density) times (volume of slice).

Example 2: Pumping water (setup-focused)

A tank is full of water to height $H$. At height $y$ above the bottom, cross-sectional area is $A(y)$. You pump all water out the top at height $H$.

A slice has volume

$dV=A(y)dy$

Its weight (force) is

$dF=\rho A(y)dy$

It must be lifted distance $H-y$, so

$dW=\rho A(y)(H-y)dy$

Total work:

$W=\int_0^H \rho A(y)(H-y)\,dy$

Even when you’re not asked to evaluate, setting up the factor $H-y$ correctly is a major exam skill.

Exam Focus

• Typical question patterns:
  • “A force $F(x)$ acts along a line. Find the work from $x=a$ to $x=b$.”
  • “Spring work using $F(x)=kx$; sometimes from a nonzero starting stretch.”
  • “Write an integral for work required to pump/lift a substance (interpret each factor).”
• Common mistakes:
  • Mixing up the variable: using total distance where the slice-by-slice distance (like $H-y$) is needed.
  • Forgetting that Hooke’s Law uses displacement from natural length, not the object’s position in space.
  • Dropping units and losing track of what each factor represents (force vs distance vs volume).

Setting Up Integrals from Context (a unifying skill for Unit 8)

Unit 8 is less about memorizing separate formulas and more about learning one transferable skill: translating a description into an integral that accumulates tiny contributions.

The universal template

Most application problems follow this structure:

1. Identify the quantity you want (area, volume, work, average value).
2. Choose a slicing/accumulation variable.
3. Describe a small piece of the quantity, such as $dA$, $dV$, or $dW$.
4. Add up all pieces using an integral over correct bounds.

A strong mental model is:

Total = sum of (rate per unit) times (small amount of unit)

For example:

• Area = (height) times (small width)
• Volume = (cross-sectional area) times (small thickness)
• Work = (force) times (small distance)

Choosing bounds correctly

Bounds are not decoration; they tell calculus what region or motion you are accumulating over.

Common ways bounds are determined:

• Given directly
• Intersection points (solve $f(x)=g(x)$)
• Physical constraints (height from $y=0$ to $y=H$)

A frequent AP error is writing a correct-looking integrand with incorrect bounds.

Choosing $dx$ vs $dy$ (and justifying it)

Think in slices:

• Vertical slices mean integrating with $dx$.
• Horizontal slices mean integrating with $dy$.

Then match slice direction to the geometry:

• Area between curves: vertical slices use “top minus bottom,” horizontal slices use “right minus left.”
• Volume by disks/washers: slices must be perpendicular to axis of rotation.
• Volume by shells: slices must be parallel to axis of rotation.

Interpreting results (including sign)

• Areas and volumes are nonnegative by definition, so your setup should produce a nonnegative result.
• Work can be positive or negative depending on force direction, but problems are often framed as “work required,” which is typically positive.
• Average value has the same units as the original function output.

Example: From words to an integral (without overcomplicating)

A region has upper boundary

$y=f(x)$

and lower boundary

$y=g(x)$

for $a\le x\le b$. Cross sections perpendicular to the $x$-axis are equilateral triangles.

At position $x$, the side length is the vertical distance:

$s(x)=f(x)-g(x)$

Area of an equilateral triangle:

$A(x)=\frac{\sqrt{3}}{4}s(x)^2=\frac{\sqrt{3}}{4}(f(x)-g(x))^2$

So the volume is

$V=\int_a^b \frac{\sqrt{3}}{4}(f(x)-g(x))^2\,dx$

This highlights a common Unit 8 pattern: a distance between curves becomes an input to a geometry formula for cross-sectional area.

Exam Focus

• Typical question patterns:
  • “Write an integral expression for [area/volume/work] and label what each part represents.”
  • “Explain why the integrand has the form it does (distance, radius, cross-sectional area).”
  • “Decide on method (washer vs shell) and defend the choice based on algebraic simplicity.”
• Common mistakes:
  • Treating a distance as a single value instead of a function of position.
  • Forgetting to square when a formula involves radius squared or side squared.
  • Not checking which function is larger, leading to negative ‘area’ or negative cross-sectional dimensions.