Net Signed Area
\[ \int_a^b f(x)\,dx \]Area below the \(x\)-axis counts negatively.
Calculus Guide
Applications of Integrals
Learn how definite integrals are used to find area, volume, average value, accumulated change, and quantities related to motion.
Putting Integrals to Work
What Can We Do with an Integral?
Definite integrals measure accumulated quantities.
Once we can evaluate \(\int_a^b f(x)\,dx\), we can use integrals to calculate area, volume, displacement, total change, average value, and many other physical and geometric quantities.
Main Idea
\[
\int_a^b f(x)\,dx
\]
A definite integral adds together infinitely many small contributions over an interval.
Big Idea
Integral = Accumulation
In many applications, we can think of the integral as
\[ \text{Total} = \int \text{small contribution}. \]The main challenge is deciding what quantity each small contribution represents.
Geometric Interpretation
Area Under a Curve
If \(f(x)\geq 0\) on \([a,b]\), then the area between the graph of \(f\), the \(x\)-axis, and the vertical lines \(x=a\) and \(x=b\) is
\[
\boxed{
A
=
\int_a^b
f(x)\,dx
}
\]
Example
Find the Area Under \(y=x^2\)
Find the area from \(x=0\) to \(x=2\).
Visualize the Region
The shaded region represents the area under \(y=x^2\) from \(x=0\) to \(x=2\).
Find an antiderivative:
\[ \int x^2\,dx = \frac{x^3}{3}. \]Evaluate:
\[ A = \left[ \frac{x^3}{3} \right]_0^2 \] \[ = \frac{8}{3}. \]
\[
\boxed{
A=\frac83
}
\]
Signed Area
Net Area vs. Total Area
Total Geometric Area
\[ \int_a^b |f(x)|\,dx \]Every region contributes positively.
Positive and Negative Area
Regions above the \(x\)-axis contribute positively to the definite integral, while regions below the \(x\)-axis contribute negatively.
Positive contribution
Negative contribution
Common Issue
Crossing the \(x\)-Axis
If the function changes sign, split the integral at each zero when the problem asks for total area.
Two Boundaries
Area Between Two Curves
Vertical Slices
\[ \boxed{ A = \int_a^b \left[ \text{top} - \text{bottom} \right] dx } \]Example
Find the Area Between \(y=2x\) and \(y=x^2\)
First find the intersections:
\[ 2x=x^2. \] \[ x(x-2)=0. \] \[ x=0,\;2. \]Visualize Top Minus Bottom
Between the intersection points, \(y=2x\) is the top function and \(y=x^2\) is the bottom function.
\[
\text{Area}
=
\int_0^2
\left(
\underbrace{2x}_{\text{top}}
-
\underbrace{x^2}_{\text{bottom}}
\right)
dx
\]
On this interval, \(2x\) lies above \(x^2\).
\[ A = \int_0^2 \left( 2x-x^2 \right) dx. \] \[ = \left[ x^2 - \frac{x^3}{3} \right]_0^2. \] \[ = 4-\frac83. \]
\[
\boxed{
A=\frac43
}
\]
Choose the Easier Direction
Vertical vs. Horizontal Slices
Integrate with Respect to \(x\)
\[ A = \int \left( \text{top} - \text{bottom} \right) dx \]Use vertical slices.
Integrate with Respect to \(y\)
\[ A = \int \left( \text{right} - \text{left} \right) dy \]Use horizontal slices.
Strategy
Choose the direction that gives the simplest integral and requires the fewest pieces.
Three-Dimensional Accumulation
Volume with Integrals
A volume can be approximated by thin slices and then accumulated with an integral.
\[
\boxed{
V
=
\int_a^b
A(x)\,dx
}
\]
where \(A(x)\) is the cross-sectional area.
Solids of Revolution
Disk Method
If a region is rotated around an axis and there is no hole in the middle, each cross-section is a disk.
\[
\boxed{
V
=
\pi
\int_a^b
[R(x)]^2\,dx
}
\]
Example
Rotate \(y=x\) About the \(x\)-Axis
Use the region from \(x=0\) to \(x=2\).
Disk Method Visual
A vertical slice rotated around the \(x\)-axis forms a disk with radius \(R(x)=x\).
The radius is
\[ R(x)=x. \]Therefore,
\[ V = \pi \int_0^2x^2\,dx. \] \[ = \pi \left[ \frac{x^3}{3} \right]_0^2. \]
\[
\boxed{
V=\frac{8\pi}{3}
}
\]
A Hole in the Middle
Washer Method
\[
\boxed{
V
=
\pi
\int_a^b
\left[
R(x)^2
-
r(x)^2
\right]
dx
}
\]
Washer Method Visual
The outer curve determines \(R(x)\), while the inner curve determines \(r(x)\).
Outer Radius
\[ R(x) \]Distance from the axis to the outer curve.
Inner Radius
\[ r(x) \]Distance from the axis to the inner curve.
Remember
\[ \boxed{ \text{Washer} = \text{Outer Disk} - \text{Inner Disk} } \]Cylindrical Shells
Shell Method
Instead of slicing perpendicular to the axis of rotation, shells use slices parallel to the axis.
\[
\boxed{
V
=
2\pi
\int_a^b
(\text{radius})
(\text{height})
\,dx
}
\]
Shell Method Visual
A vertical strip rotated around the \(y\)-axis creates a cylindrical shell. Its distance from the axis is the radius and its vertical length is the height.
Radius
Distance from the slice to the axis of rotation.
Height
Length of the region being rotated.
Example
Rotate the region under \(y=x\), from \(x=0\) to \(x=2\), around the \(y\)-axis.
Radius:
\[ r=x. \]Height:
\[ h=x. \]Therefore,
\[ V = 2\pi \int_0^2x^2\,dx. \] \[ = 2\pi \left[ \frac{x^3}{3} \right]_0^2. \]
\[
\boxed{
V=\frac{16\pi}{3}
}
\]
Choosing a Method
Disks, Washers, or Shells?
Disks / Washers
Slices Perpendicular to the Axis
Often convenient when the radius is easy to express.
Shells
Slices Parallel to the Axis
Often convenient when a washer setup would require solving for the other variable or splitting the region.
Best Method
There is not always one mandatory method. Choose the setup that produces the simpler integral.
Average Height of a Function
Average Value of a Function
\[
\boxed{
f_{\text{avg}}
=
\frac{1}{b-a}
\int_a^b f(x)\,dx
}
\]
Example
Find the Average Value of \(f(x)=x^2\)
Use the interval \([0,2]\).
\[ f_{\text{avg}} = \frac{1}{2-0} \int_0^2x^2\,dx. \] \[ = \frac12 \left( \frac83 \right). \]
\[
\boxed{
f_{\text{avg}}
=
\frac43
}
\]
Accumulated Rate of Change
The Net Change Theorem
\[
\boxed{
\int_a^b F'(x)\,dx
=
F(b)-F(a)
}
\]
Interpretation
Integral of a Rate = Net Change
If \(r(t)\) represents the rate at which a quantity changes, then \(\int_a^b r(t)\,dt\) represents the net change in that quantity.
Position from Velocity
Motion Applications
Displacement
\[ \boxed{ \int_a^b v(t)\,dt } \]Net change in position.
Total Distance
\[ \boxed{ \int_a^b |v(t)|\,dt } \]Total amount traveled.
Important
Displacement Is Not Always Distance
If velocity changes sign, split the interval at the times when \(v(t)=0\) before calculating total distance.
Building Totals Over Time
Accumulation Functions
\[
\boxed{
A(x)
=
\int_a^x
f(t)\,dt
}
\]
\(A(x)\) represents the amount accumulated from the starting point \(a\) to the current location \(x\).
Fundamental Theorem Connection
\[ \boxed{ A'(x)=f(x) } \]The rate of change of the accumulated amount is the original integrand.
Rate to Quantity
Finding a Total Quantity from a Rate
Example
Water enters a tank at a rate
\[ r(t)=4t+2 \]liters per minute.
How much water enters from \(t=0\) to \(t=3\)?
\[ \text{Amount} = \int_0^3 (4t+2)\,dt. \] \[ = \left[ 2t^2+2t \right]_0^3. \] \[ = 18+6. \]
\[
\boxed{
24
\text{ liters}
}
\]
Dimensional Meaning
Units in Integral Applications
Integration multiplies the units of the integrand by the units of the variable of integration.
Rate
Integrate over time
Total quantity
For example, \(\text{miles/hour}\times \text{hours}=\text{miles}\).
Watch Out
Common Mistakes
1
Using Bottom Minus Top
For vertical area slices, use top minus bottom.
2
Using Left Minus Right
For horizontal area slices, use right minus left.
3
Forgetting to Find Intersections
Intersections often determine the correct bounds.
4
Confusing Net Area and Total Area
Negative regions subtract from a definite integral.
5
Forgetting to Square Radii
Disk and washer formulas use \(R^2\) and \(r^2\).
6
Mixing Up Radius and Height in Shells
Shell volume uses \(2\pi(\text{radius})(\text{height})\).
7
Forgetting the \(1/(b-a)\) in Average Value
The average is the integral divided by the interval length.
8
Confusing Displacement and Distance
Distance requires the absolute value of velocity.
9
Ignoring Units
Units often reveal whether the setup represents area, volume, distance, or another quantity.
10
Choosing a Harder Slice Direction
Check whether integrating with respect to the other variable produces a cleaner setup.
Problem-Solving Roadmap
Applications of Integrals Strategy
1
Identify
What Quantity Are You Finding?
Area, volume, average value, displacement, total distance, or accumulated change?
2
Sketch
Draw the Region When Geometry Is Involved
3
Choose
Decide on the Slice Direction
Use \(dx\) or \(dy\), and choose disks, washers, or shells if finding volume.
4
Bounds
Find the Correct Interval
5
Build
Write the Integral Before Calculating
6
Interpret
State the Answer with Units
Keep This Handy
Applications Quick Reference
Area Under Curve
\[ A = \int_a^b f(x)\,dx \]Area Between Curves
\[ A = \int_a^b (\text{top}-\text{bottom}) \,dx \]Disk Method
\[ V = \pi \int_a^b R^2\,dx \]Washer Method
\[ V = \pi \int_a^b (R^2-r^2)\,dx \]Shell Method
\[ V = 2\pi \int (\text{radius}) (\text{height}) \,dx \]Average Value
\[ f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x)\,dx \]Net Change
\[ \text{Net Change} = \int_a^b \text{rate}\,dt \]Displacement
\[ \int_a^b v(t)\,dt \]Total Distance
\[ \int_a^b |v(t)|\,dt \]Accumulation
\[ A(x) = \int_a^x f(t)\,dt \]Before You Practice
Applications of Integrals Checklist
- Understand a definite integral as accumulated change or signed area.
- Distinguish net area from total geometric area.
- Find intersection points before setting up area-between-curves integrals.
- Use top minus bottom with vertical slices.
- Use right minus left with horizontal slices.
- Recognize volume as an integral of cross-sectional area.
- Use the disk method when there is no inner radius.
- Use the washer method when the solid has a hole.
- Identify outer and inner radii correctly.
- Use shells as \(2\pi(\text{radius})(\text{height})\).
- Choose the volume method that produces the simplest setup.
- Calculate average value by dividing the integral by the interval length.
- Interpret the integral of a rate as net change.
- Distinguish displacement from total distance.
- Split intervals when velocity changes sign for total distance.
- Understand accumulation functions and their connection to the Fundamental Theorem.
- Check the units of your final answer.
Your Turn
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Practice area, volume, average value, net change, accumulation, displacement, and total distance problems.
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