AP Calculus AB: Applications of Integration
This is where integration becomes a tool for measuring the world: areas, volumes, distances, and accumulated change. This free, complete lesson covers area under a curve and using symmetry, area between curves, volumes with known cross sections, volumes of revolution using disks and washers, motion along a line (distance vs. displacement), and net change from a rate. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.
1. Area Under a Curve and Using Symmetry
The area bounded by y = f(x), the x-axis, and the vertical lines x = a and x = b (with f(x) ≥ 0) is ∫ab f(x) dx. When a region is symmetric about an axis, integrating over half the region and doubling the result is often simpler. Area can also be found with respect to y, integrating x = g(y) between horizontal bounds.
Worked Example: Find the area under f(x) = x² + 2 from x = 0 to x = 3.
∫03 (x² + 2) dx = [x³3 + 2x]03 = 9 + 6 = 15
1. Find the area bounded by f(x) = 4 − x² and the x-axis, using symmetry about the y-axis.
Explanation: The curve crosses the x-axis at x = ±2. By symmetry: 2∫02(4 − x²) dx = 2[4x − x³/3]02 = 2(16/3) = 32/3.
2. Find the area under f(x) = 3x² from x = 1 to x = 4.
Explanation: ∫143x² dx = [x³]14 = 64 − 1 = 63.
3. The region bounded by y = cos(2x) and the x-axis on [−π/4, π/4] is symmetric about the y-axis. If the area to the right of the y-axis is 12, what is the total area?
Explanation: By symmetry, the total area is twice the area of one half: 2 × 1/2 = 1.
4. Find the area bounded by x = y², the y-axis, and the lines y = 0 and y = 3.
Explanation: Integrating with respect to y: ∫03y² dy = [y³/3]03 = 27/3 = 9.
2. Area Between Curves
To find the area between y = f(x) and y = g(x), first find where they intersect, then integrate (top − bottom) between those intersection points. If which curve is on top changes, split the integral at the crossing point.
Worked Example: Find the area between y = 2x and y = x² (they intersect at x = 0 and x = 2).
∫02 (2x − x²) dx = [x² − x³3]02 = 4 − 83 = 43
1. Find the area between y = x + 2 and y = x² (they intersect at x = −1 and x = 2).
Explanation: The line lies above the parabola on this interval. ∫−12[(x + 2) − x²] dx = [x²/2 + 2x − x³/3]−12 = 10/3 − (−7/6) = 9/2.
2. Find the area between y = 6 − x² and y = x (they intersect at x = −3 and x = 2).
Explanation: ∫−32[(6 − x²) − x] dx = [6x − x³/3 − x²/2]−32 = 22/3 − (−27/2) = 125/6.
3. Two curves intersect such that f(x) ≥ g(x) on [a, c] and g(x) ≥ f(x) on [c, b]. Write the total area between them from a to b.
Explanation: Each piece must use (top − bottom) for that piece specifically, so the integral splits at the crossing point c with the order reversed on each side.
4. Why is it essential to determine which curve is on top before setting up an area-between-curves integral?
Explanation: Reversing the order (bottom − top) produces a negative value; the integrand must be top minus bottom to correctly represent a positive area.
3. Volumes with Known Cross Sections
When the area of a cross section perpendicular to an axis is known in terms of x, the volume is ∫ab A(x) dx, where A(x) is the cross-sectional area formula.
Worked Example: A solid has a base bounded by the circle x² + y² = 9, with cross sections perpendicular to the x-axis that are squares. Find the volume.
The square's side is 2y = 2√(9 − x²), so A(x) = 4(9 − x²).
V = 2∫034(9 − x²) dx = 8[9x − x³3]03 = 8(18) = 144
1. A solid has a base bounded by the circle x² + y² = 4, with square cross sections perpendicular to the x-axis. Find the volume.
Explanation: Side = 2√(4 − x²), A(x) = 4(4 − x²). V = 2∫024(4 − x²) dx = 8[4x − x³/3]02 = 8(16/3) = 128/3.
2. A solid's base is bounded by y = x² and y = 4, with semicircular cross sections whose diameter stretches from the parabola to the line. Find the radius as a function of x.
Explanation: The diameter equals the vertical distance between the curves, 4 − x², so the radius is half that.
3. For a solid whose cross sections perpendicular to the x-axis are equilateral triangles with side length s(x), what is the area formula for each cross section?
Explanation: This is the standard area formula for an equilateral triangle with side length s.
4. Set up (but do not evaluate) the integral for the volume of a solid whose base is bounded by y = √x, the x-axis, and x = 4, with square cross sections perpendicular to the x-axis.
Explanation: The square's side is y = √x, so A(x) = (√x)² = x.
4. Volumes of Revolution: Disks
When a region bounded by a curve and an axis is revolved about that axis, each cross section is a disk. V = π∫ab r² dx (or dy), where r is the distance from the axis to the curve.
Worked Example: Find the volume when the region bounded by y = x², x = 2, y = 0 is rotated about the x-axis.
V = π∫02(x²)² dx = π[x⁵5]02 = 32π5
1. Find the volume when the region bounded by y = √x, x = 4, y = 0 is rotated about the x-axis.
Explanation: V = π∫04(√x)² dx = π∫04x dx = π[x²/2]04 = 8π.
2. Find the volume when the region bounded by y = x², y = 0, x = 3 is rotated about the y-axis (integrate in y, using radius x = √y, for y from 0 to 9).
Explanation: V = π∫09y dy = π[y²/2]09 = 81π/2.
3. A region bounded by y = x³, x = 0, y = 8 is rotated about the y-axis. Using radius x = y1/3, set up the integral for the volume.
Explanation: V = π∫(y1/3)² dy = π∫y2/3 dy.
4. Using disks, find the volume of a sphere of radius 3 generated by rotating a semicircle of radius 3 about the x-axis.
Explanation: V = (4/3)πr³ = (4/3)π(27) = 36π, matching the standard sphere volume formula.
5. Volumes of Revolution: Washers
When the revolved region is bounded away from the axis (leaving a hole), each cross section is a washer: V = π∫ab(R² − r²) dx, where R is the outer radius and r is the inner radius.
Worked Example: Find the volume when the region between y = x² and y = 4 is rotated about the x-axis (they intersect at x = ±2).
V = 2π∫02(16 − x⁴) dx = 2π[16x − x⁵5]02 = 2π(1285) = 256π5
1. Find the volume when the region between y = x and y = x² (on [0, 1]) is rotated about the x-axis.
Explanation: Outer radius R = x, inner radius r = x². V = π∫01(x² − x⁴) dx = π(1/3 − 1/5) = 2π/15.
2. Find the volume when the region between y = x² and y = 9 is rotated about the x-axis (they intersect at x = ±3).
Explanation: R = 9, r = x². V = 2π∫03(81 − x⁴) dx = 2π[81x − x⁵/5]03 = 2π(972/5) = 1944π/5.
3. In the washer method, if R is the outer radius and r is the inner radius, what is the formula for the volume element ΔV?
Explanation: A washer's volume is the outer disk's volume minus the inner disk's volume: πR²Δx − πr²Δx = π(R² − r²)Δx.
4. The region bounded by y = x² and y = 2x (intersecting at x = 0 and x = 2, with y = 2x above y = x²) is rotated about the x-axis. Which radius corresponds to y = 2x?
Explanation: Since y = 2x lies farther from the x-axis than y = x² on this interval, it forms the outer radius R.
6. Motion Along a Line: Distance vs. Displacement
Displacement (net change in position) is ∫abv(t) dt. Total distance traveled is ∫ab|v(t)| dt, found by splitting the interval wherever v(t) changes sign and summing the absolute value of each piece.
Worked Example: A particle has v(t) = t² − 4t + 3 on [0, 3]. Find the displacement and total distance traveled.
Displacement: ∫03(t² − 4t + 3) dt = [t³/3 − 2t² + 3t]03 = 0.
v(t) = (t − 1)(t − 3): positive for t < 1, negative for 1 < t < 3.
Distance = ∫01v dt − ∫13v dt = 43 − (−43) = 83
1. A particle has v(t) = 3t² − 6t on [0, 3]. Find the displacement from t = 0 to t = 3.
Explanation: ∫03(3t² − 6t) dt = [t³ − 3t²]03 = 27 − 27 = 0.
2. Using the same velocity v(t) = 3t² − 6t = 3t(t − 2), find the total distance traveled from t = 0 to t = 3.
Explanation: v < 0 on (0, 2) and v > 0 for t > 2. ∫02v dt = −4, ∫23v dt = 4. Distance = |−4| + |4| = 8.
3. If v(t) ≥ 0 for all t on [a, b], how does the total distance traveled compare to the displacement?
Explanation: When velocity never goes negative, the particle never reverses direction, so distance and displacement coincide exactly.
4. Why does total distance traveled require ∫|v(t)| dt rather than just ∫v(t) dt?
Explanation: Displacement can be small or zero even if the particle traveled a long path back and forth; taking the absolute value before integrating prevents that cancellation.
7. Net Change: Definite Integral of a Rate
If f(t) = F'(t), then ∫abf(t) dt = F(b) − F(a) is the net change (or total accumulation) in the quantity F over [a, b]. This principle applies to virtually any rate: population growth, spreading rumors, marginal cost, water flow, and more.
Worked Example: A chemical decomposes at a rate of R(t) = 5e−0.5t grams per minute. Find the amount decomposed during the first 4 minutes.
∫045e−0.5t dt = [−10e−0.5t]04 = 10(1 − e−2) ≈ 8.65 g
1. A population grows at a rate of G(t) = 50e0.1t organisms per day. Find the increase in population from t = 0 to t = 10.
Explanation: ∫01050e0.1t dt = [500e0.1t]010 = 500e − 500 = 500(e − 1) ≈ 859.1.
2. The marginal cost of production is C'(x) = 2 + 0.01x dollars per unit. Find the change in cost when production increases from 100 to 200 units.
Explanation: ∫100200(2 + 0.01x) dx = [2x + 0.005x²]100200 = 600 − 250 = 350.
3. If f(t) represents the rate (in people per week) at which a rumor spreads, what does ∫25f(t) dt represent?
Explanation: The definite integral of a rate over an interval gives the net accumulation during that specific interval, here, the new people reached between weeks 2 and 5.
4. Water flows into a tank at a rate of R(t) = 20 − 2t gallons per minute for 0 ≤ t ≤ 10. Find the total amount of water that flows in during this time.
Explanation: ∫010(20 − 2t) dt = [20t − t²]010 = 200 − 100 = 100.
Common Mistakes to Avoid
- Forgetting to find intersection points before setting up an area-between-curves integral. The bounds of integration come directly from where the curves meet, always solve for these first.
- Mixing up the radius in a cross-section problem. For a square cross section, the volume element uses (side)², not the side length itself, don't forget to square the expression for the side.
- Confusing disks and washers. Use disks when the region touches the axis of revolution (no hole); use washers when there's a gap between the region and the axis (a hole in the middle).
- Forgetting the absolute value when finding total distance traveled. ∫v(t) dt gives displacement, not distance, always check for sign changes in v(t) and split the integral accordingly when distance is asked for.
- Not connecting net change problems back to the Fundamental Theorem of Calculus. Any "rate of ___" problem asking for total accumulated change over an interval is simply ∫ab(rate) dt, recognizing this pattern quickly saves time.
Frequently Asked Questions
How do I know whether to integrate with respect to x or y?
Look at how the boundaries are most naturally described. If the region is bounded by horizontal lines and functions given as y in terms of x, integrate in x. If it's bounded by vertical lines and functions more naturally solved as x in terms of y, integrate in y, this often makes the setup significantly simpler.
Do I need to know the shell method for the AP exam?
No. The AP Calculus AB and BC exams test volumes of revolution using disks and washers, along with cross sections. The shell method is a valid technique and can be a helpful problem-solving tool, but it isn't required knowledge for the exam.
What's the difference between distance and displacement, and why does it matter?
Displacement is where a particle ends up relative to where it started (net change in position). Distance is how much ground it actually covered, including any back-and-forth motion. A particle can return exactly to its starting point (zero displacement) after traveling a large total distance.
Where can I practice more problems like these?
The Applications of Integration quizzes in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.