Why Applications of Integration Are Different
In a basic integration problem, you are usually given an expression and asked to find an antiderivative. Application problems are different. The difficult part is often deciding what quantity should be integrated, what the limits should be, and whether the integral should represent a total, a difference, an area, or an accumulation.
On the AP Calculus exam, several different-looking problems can actually use the same underlying idea:
The integral adds up infinitely many small contributions. Your job is to identify what each small contribution represents before calculating it.
The Major Applications You Should Recognize
1. Accumulation from a Rate
If represents a rate of change, then the amount accumulated between and is:
If you know the initial amount, you can find the final amount by adding the net change to the initial value:
2. Area Between Curves
When integrating with respect to , the area between two curves is the integral of top minus bottom:
When integrating with respect to , use right minus left instead:
3. Volume with Known Cross-Sections
If the cross-sectional area perpendicular to the -axis is , then the volume is:
The most important step is determining the cross-sectional area. For example, a square cross-section with side length has area , while a semicircular cross-section with diameter has area .
4. Volumes of Revolution
When a region is revolved around an axis, the resulting volume can often be found using washers or disks:
If there is a hole in the middle, use the washer formula:
The key question is always: what are the outer and inner radii?
5. Average Value of a Function
The average value of a continuous function on is:
Notice that this is not simply the average of the two endpoint values. The integral accounts for the function's behavior across the entire interval.
6. Position, Velocity, and Acceleration
Integration connects velocity and position, and acceleration and velocity. If velocity is known, displacement is:
The final position is the initial position plus displacement:
Be careful: displacement is not the same as total distance traveled. Distance requires you to account for whether the particle is moving forward or backward.
A Recognition Strategy for Mixed Problems
When you encounter a mixed integration problem, do not immediately reach for an antiderivative. First identify what the problem is asking you to accumulate.
- Identify the quantity being accumulated. Is it area, volume, distance, displacement, total change, or an average value?
- Identify the variable of integration. Are you adding horizontal slices with respect to , or vertical slices with respect to ?
- Determine the bounds. Use the interval or intersection points given in the problem.
- Build the integrand. This is usually the most important step.
- Check units. A rate integrated over time should produce an amount; an area integrated over distance should produce volume.
- Then calculate. Once the setup is correct, the integration itself is often the easy part.
Worked Examples
Water enters a tank at a rate of gallons per hour, where . The tank initially contains 120 gallons. How much water is in the tank at ?
Find the area enclosed by and on the interval where the curves intersect.
The base of a solid is the region under and above the -axis from to . Cross-sections perpendicular to the -axis are squares. Find the volume.
A particle moves along the -axis with velocity for . Find the displacement and total distance traveled.
Find the average value of on .
The Most Important AP Exam Distinctions
Displacement can be positive, negative, or zero because positive and negative velocity can cancel. Total distance cannot be negative and requires integrating the magnitude of velocity. If velocity changes sign, split the integral at the zeros of velocity.
A definite integral gives signed net area. If a function crosses the axis, a single integral may contain cancellation. For geometric area, split the interval at the zeros and make each contribution positive.
When integrating with respect to , use top minus bottom. When integrating with respect to , use right minus left. Always identify the geometry before writing the integral.
A rate such as gallons per hour, people per hour, or dollars per hour is not itself the accumulated amount. Integrating the rate over time produces the total change. If an initial amount is provided, add the accumulated change to that initial value.
In a cross-section problem, the function describing the base is usually not itself the volume integrand. First determine the cross-sectional area, then integrate that area across the relevant interval.
Mixed Practice Problems
A tank contains 80 gallons of water at time . Water enters at a rate of gallons per hour. How much water is in the tank at ?
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Find the area enclosed by and .
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The base of a solid is the region under and above the -axis from to . Cross-sections perpendicular to the -axis are semicircles. Find the volume.
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A particle has velocity on . Find its displacement.
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A particle moves with velocity for . Find the total distance traveled.
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Find the average value of on .
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Evaluate . Explain why the result should not automatically be interpreted as geometric area.
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The rate at which a population changes is , where is measured in years. The population at is 2,400. Estimate the population at .
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A Final Checklist
Before submitting an AP Calculus application-of-integration answer, ask:
- What quantity am I accumulating?
- What does the integrand represent?
- Are my bounds correct?
- Am I using the correct geometric difference?
- Does the function change sign?
- Do I need displacement or total distance?
- Do I need an initial value?
- Do my units make sense?
The biggest improvement you can make on mixed integration problems is to stop thinking of the integral as merely an antiderivative. Think of it as an accumulation machine. Once you identify what is being accumulated, the correct integral usually becomes much easier to construct.
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