Applications of Integration

Unit Overview

This capstone unit puts the definite integral to work. You will compute average value, use integrals for motion (displacement and total distance), find areas between curves, and calculate volumes of solids — both by cross sections and by rotation (disk/washer, with shells as an extension). These are the classic "set up and evaluate an integral" problems that dominate AP free-response.

The skill is twofold: set up the correct integral (the hardest part) and evaluate it using Unit 6 techniques.

Big picture: Slice the region/solid into thin pieces, express each piece's measure as an integrand, and integrate to total it.

Learning Objectives

• Compute the average value of a function and apply the Mean Value Theorem for Integrals.
• Use integrals for rectilinear motion: displacement ∫v and total distance ∫|v|; recover position from velocity and velocity from acceleration.
• Find area between curves, integrating with respect to x or y.
• Find volumes with known cross sections (squares, semicircles, triangles).
• Find volumes of revolution by the disk and washer methods (and shell method).

Prerequisite Concepts

• Definite integrals and the FTC (Unit 6); u-substitution.
• Geometry: areas of circles/triangles, volume intuition.
• Finding intersection points of curves.

Key Definitions & Formulas

Average value of f on [a, b]: f_avg = (1/(b − a)) ∫ₐᵇ f(x) dx.

Motion:

• Displacement on [a, b]: ∫ₐᵇ v(t) dt (net; can be negative).
• Total distance on [a, b]: ∫ₐᵇ |v(t)| dt (split at v = 0).
• Position: s(b) = s(a) + ∫ₐᵇ v(t) dt; velocity: v(t) = v(a) + ∫ₐᵗ a(τ) dτ.

Area between curves (top T, bottom B): ∫ₐᵇ [T(x) − B(x)] dx; or with horizontal strips ∫_c^d [R(y) − L(y)] dy (right − left).

Volume by cross sections with area A(x): V = ∫ₐᵇ A(x) dx. Common cross sections of "size" s:

• Square: A = s²
• Semicircle (diameter s): A = (π/8) s²
• Equilateral triangle (side s): A = (√3/4) s²

Volume of revolution about the x-axis:

• Disk: V = π ∫ₐᵇ [f(x)]² dx
• Washer: V = π ∫ₐᵇ ([R(x)]² − [r(x)]²) dx

Shell method (about the y-axis): V = 2π ∫ₐᵇ x·f(x) dx.

Important Theorem

Mean Value Theorem for Integrals: If f is continuous on [a, b], there exists c in [a, b] with f(c) = f_avg — the function attains its average value at some point.

Concept Explanations

Average value (8.1). The average value is the constant height of a rectangle on [a, b] with the same area as under f. It's an integral divided by length — not the average of the endpoints.

Motion via integrals (8.2). Integration reverses Unit 4's differentiation of motion. Displacement is the signed integral of velocity; total distance requires |v|, so find where v = 0 and integrate each piece with the correct sign. To recover position, add the accumulated displacement to the starting position.

Area between curves (8.3–8.4). Integrate (top − bottom) with respect to x, or (right − left) with respect to y. Choose the variable that avoids splitting the region: use dy (horizontal strips) when boundaries are naturally x = (function of y), as with y² = x.

Cross-section volumes (8.5, 8.7). The base region gives the size of each slice; the cross-section shape gives its area formula A(x). Integrate A(x) along the axis. For squares, A = side²; for semicircles, A = (π/8)(diameter)², etc.

Solids of revolution (8.6, 8.8). Rotating a region sweeps a solid. Disks (solid slice) use radius = distance to the axis; washers subtract an inner radius when there's a gap. The shell method integrates cylindrical shells 2π·(radius)(height) — handy for rotation about the y-axis when the curve is y = f(x).

Worked Examples

Example 1 (Average value). f(x) = x² on [0, 3]: f_avg = (1/3)∫₀³ x² dx = (1/3)(9) = 3.

Example 2 (Total distance). v(t) = t − 2 on [0, 3]: v = 0 at t = 2. Distance = ∫₀² (2 − t) dt + ∫₂³ (t − 2) dt = 2 + 0.5 = 2.5.

Example 3 (Area between curves). Between y = x² and y = 4: intersect at x = ±2, ∫₋₂² (4 − x²) dx = 32/3. (from the Vidaara NCERT/JEE bank)

Example 4 (Area with dy). Bounded by y² = x and y = x − 2: intersect at y = −1, 2. ∫₋₁² [(y + 2) − y²] dy = 10/3. (from the Vidaara NCERT/JEE bank)

Example 5 (Disk). Rotate y = x³ on [0, 1] about the x-axis: V = π∫₀¹ x⁶ dx = π/7. (from the Vidaara NCERT/JEE bank)

Example 6 (Sphere via disks). Rotate y = √(a² − x²) on [−a, a]: V = π∫₋ₐᵃ (a² − x²) dx = (4/3)πa³. (from the Vidaara NCERT/JEE bank)

Example 7 (Washer). Between y = x (outer) and y = x² (inner) on [0, 1] about the x-axis: V = π∫₀¹ (x² − x⁴) dx = 2π/15.

Example 8 (Cross sections). Base under y = 4 − x², square cross sections of side (4 − x²): V = ∫₋₂² (4 − x²)² dx.

Graphical Interpretation

• Area between curves is the region trapped between two graphs; a thin vertical strip has height (top − bottom).
• A disk/washer is what a vertical strip sweeps out when rotated — a circular slab of radius = distance to the axis.
• A shell is what a strip parallel to the axis sweeps — a thin cylinder of radius x and height f(x).

Common Mistakes

• Average of endpoints instead of the integral formula for average value.
• Using v instead of |v| for total distance (missing the split at v = 0).
• Top − bottom reversed, producing a negative area.
• Forgetting to square the radius in disk/washer (π∫R², not π∫R), or subtracting radii instead of their squares in the washer (R² − r², not (R − r)²).
• Wrong cross-section area formula (e.g., using r² instead of the semicircle's (π/8)s²).
• Mismatching the variable of integration with the strip orientation.

AP Exam Tips

• These are free-response staples — practice writing the integral set-up cleanly; a correct unevaluated integral earns significant credit.
• Always find intersection points first to get the limits of integration.
• For distance vs. displacement, state where v = 0 and split accordingly.
• Label the radius/height clearly for volume problems; for washers, identify outer vs. inner radius.
• On calculator-active questions, set up the integral by hand, then evaluate numerically.

Unit Summary

• Average value: (1/(b − a))∫ₐᵇ f.
• Motion: displacement ∫v; total distance ∫|v| (split at v = 0); position = s(a) + ∫v.
• Area: ∫(top − bottom) dx or ∫(right − left) dy.
• Volumes: cross sections ∫A(x) dx; disk π∫R²; washer π∫(R² − r²); shell 2π∫x·f(x).
• The universal method: slice → integrand → integrate.

This completes AP Calculus AB. Together, Units 1–8 build from limits to derivatives to integrals and their applications — the full arc of single-variable calculus tested on the AP exam.