Lesson 35: Applications of Integration — Area, Volume & Average Value

(a) Opening Question

A region R is bounded above by the line y = 4 and below by the parabola y = x².

Before you compute anything, answer three questions that will frame this entire lesson:

1. The two curves meet where x² = 4. What are the two x-values, and which curve is "on top" between them?
2. To find the area of R, you integrate (top − bottom). Write the integral — but don't evaluate it yet.
3. Now imagine spinning R around the x-axis. Each slice perpendicular to the axis sweeps out a flat washer. What is the outer radius of that washer? What is the inner radius?

Take two minutes. The whole of Unit 8 is contained in those three questions: find where curves meet, identify top-and-bottom (or outer-and-inner), and integrate the right expression. By the end of this lesson you'll set up the area, three different volumes, and the average value of x² over R — all from this same picture.

(We'll return to this region in part (b).)

(b) Core Concepts

This is the last content lesson, and it is almost entirely about setting up the integral correctly. The integration itself is Unit 6. The skill here is translating a picture into a definite integral. Move slowly through the setup; the antiderivative is the easy part.

1. Area between two curves

If f(x) ≥ g(x) for all x in [a, b], the area of the region between them is

A = ∫_a^b (top − bottom) dx = ∫_a^b [f(x) − g(x)] dx

The integrand is always (top − bottom), and because top ≥ bottom the result is automatically positive — you do not need absolute values, and you do not subtract the x-axis. Three setup rules:

• Find the intersection points first. Set f(x) = g(x) and solve; these are your limits a and b.
• If the curves switch order, split the integral at the crossing point and use (top − bottom) on each piece separately.
• If the region is easier to describe sideways (bounded left and right rather than top and bottom), integrate with respect to y:

A = ∫_c^d (right − left) dy = ∫_c^d [x_right(y) − x_left(y)] dy

Worked area example. Find the area between y = x and y = x².

Intersections: x = x² → x² − x = 0 → x = 0 and x = 1. On (0, 1), test x = ½: the line gives 0.5, the parabola gives 0.25, so the line is on top.

A = ∫_0^1 (x − x²) dx = [x²/2 − x³/3]_0^1 = 1/2 − 1/3 = 1/6

• Line y = x is a straight diagonal through (0,0) and (1,1)
• Parabola y = x² passes through (0,0) and (1,1), bowing below the line
• Curves intersect at (0, 0) and (1, 1)
• Shade the lens-shaped region between them on [0, 1]; the line is the upper boundary, the parabola the lower]

The area is 1/6. Notice the structure: intersections gave the limits, a test point gave the order, and (top − bottom) gave the integrand.

2. Volumes of revolution — disc and washer

Spin a region around a line and it sweeps out a solid. Slice perpendicular to the axis of rotation; each slice is a thin disc or washer.

Disc method (region touches the axis, no hole):

V = π ∫_a^b [R(x)]² dx

where R(x) is the distance from the axis to the curve.

Washer method (region has a gap between it and the axis, leaving a hole):

V = π ∫_a^b ([R_outer]² − [R_inner]²) dx

Critical: you square the radii first, then subtract. It is R_outer² − R_inner², never (R_outer − R_inner)². Those are different numbers.

The radius is always distance from the axis to the curve, so when you revolve about a line other than the axis, adjust every radius:

• About y = k (horizontal line): radius = |curve − k|.
• About x = h (vertical line): integrate in y, radius = |curve − h|.

Worked washer example. Revolve the region between y = √x (top) and y = x (bottom) about the x-axis.

On [0, 1], √x ≥ x. Revolving about the x-axis, the outer edge is the farther curve √x, the inner edge is x:

V = π ∫_0^1 ([√x]² − [x]²) dx = π ∫_0^1 (x − x²) dx = π [x²/2 − x³/3]_0^1 = π · (1/6) = π/6

The volume is π/6. The most common error here is dropping the π, so write it down the moment you set up the integral.

3. Volumes by known cross-sections

Now the cross-sections are not discs — they're squares, semicircles, triangles, or rectangles built on a base region. The principle is the same: slice, find the area of one slice as a function of x, then integrate.

V = ∫_a^b A(x) dx

where A(x) is the cross-sectional area. The base region gives you a length — usually s = top − bottom — and the shape tells you how to turn that length into an area:

There is no π unless the shape is round, and for a semicircle the diameter is s, so the radius is s/2 and the area is ½ · π · (s/2)² = (π/8) s² — don't use s itself as the radius.

Worked cross-section example. The base is the region between y = x and y = x² on [0, 1]. Cross-sections perpendicular to the x-axis are squares. Here s = x − x², so A(x) = (x − x²)²:

V = ∫_0^1 (x − x²)² dx = ∫_0^1 (x² − 2x³ + x⁴) dx = [x³/3 − x⁴/2 + x⁵/5]_0^1 = 1/3 − 1/2 + 1/5 = 1/30

The volume is 1/30. No π — the cross-sections are squares.

4. Average value and the MVT for Integrals

The average value of f on [a, b] is

f_avg = (1/(b−a)) ∫_a^b f(x) dx

This is the height of the rectangle on [a, b] whose area equals the area under f. For example, the average value of f(x) = x² on [0, 3] is (1/3) ∫_0^3 x² dx = (1/3)(9) = 3.

The Mean Value Theorem for Integrals guarantees this average is actually attained: if f is continuous on [a, b], then there exists at least one c in (a, b) such that

f(c) = (1/(b−a)) ∫_a^b f(x) dx

For f(x) = x² on [0, 3], solving c² = 3 gives c = √3 ≈ 1.73, which lies in (0, 3). State the continuity hypothesis when you cite this theorem.

5. Accumulation / net change

If r(t) is a rate (the derivative of some quantity), then the integral of the rate is the net change in that quantity:

∫_a^b r(t) dt = (net change in the quantity from t = a to t = b)

Units carry through: if r(t) is in gallons per minute and t is in minutes, then ∫_a^b r(t) dt is in gallons. Always state units in your answer. If water flows into a tank at r(t) = 2t gal/min for 0 ≤ t ≤ 5, the total added is ∫_0^5 2t dt = 25 gallons.

Calculator note

On calculator-active parts, evaluate these integrals numerically with fnInt:

TI-84: MATH → 9:fnInt( → fnInt(√(X) − X², X, 0, 1) → 0.16667
Means ∫_0^1 (√x − x²) dx ≈ 1/6

To find intersection points numerically, graph both curves and use 2nd → TRACE → 5:intersect, or solve f(X) − g(X) = 0 with the Solver. Set up the integral by hand; let the calculator do only the arithmetic.

(c) Worked Examples

Example 1 — Area between two curves (finding intersections) · [NO CALC]

Problem. Find the area of the region bounded by y = 2x and y = x².

Strategy. Find where they meet, decide which is on top, integrate (top − bottom).

Solution. Set x² = 2x → x² − 2x = 0 → x(x − 2) = 0 → x = 0, 2. Test x = 1: line gives 2, parabola gives 1, so the line 2x is on top.

A = ∫_0^2 (2x − x²) dx = [x² − x³/3]_0^2 = (4 − 8/3) − 0 = 4/3

Answer: 4/3.

Example 2 — Washer volume about a horizontal axis · [NO CALC]

Problem. Let R be the region bounded by y = x² and y = 4. Find the volume of the solid formed by revolving R about the x-axis.

Strategy. The curves meet at x² = 4 → x = ±2. On [−2, 2] the top is y = 4, the bottom is y = x². Revolving about the x-axis, the line y = 4 (farther from the axis) gives the outer radius, the parabola gives the inner radius. This leaves a hole, so it's a washer.

Solution.

V = π ∫_{−2}^{2} (4² − (x²)²) dx = π ∫_{−2}^{2} (16 − x⁴) dx
  = π [16x − x⁵/5]_{−2}^{2}
  = π [(32 − 32/5) − (−32 + 32/5)]
  = π (64 − 64/5) = 256π/5

Answer: 256π/5. If you had mistakenly written (4 − x²)², you'd be squaring the difference instead of subtracting the squares — a guaranteed lost point.

Example 3 — Known cross-sections (semicircles) · [NO CALC]

Problem. The base of a solid is the region bounded by y = √x, the x-axis, and x = 4. Cross-sections perpendicular to the x-axis are semicircles with diameter in the base. Find the volume.

Strategy. The base height at x (from the x-axis up to y = √x) is s = √x. This is the diameter of the semicircle, so the radius is √x / 2 and A(x) = ½ π (√x/2)² = (π/8) x.

Solution.

V = ∫_0^4 (π/8) x dx = (π/8) [x²/2]_0^4 = (π/8)(8) = π

Answer: π. The diameter-vs-radius trap is the whole game here: the base length is the diameter, never the radius.

Example 4 — Average value · [NO CALC]

Problem. Find the average value of f(x) = sin x on [0, π], then find a value c guaranteed by the Mean Value Theorem for Integrals.

Strategy. Apply the formula; then solve f(c) = f_avg.

Solution.

f_avg = (1/(π − 0)) ∫_0^π sin x dx = (1/π) [−cos x]_0^π = (1/π)(1 − (−1)) = 2/π

Since sin x is continuous on [0, π], the MVT for Integrals guarantees a c in (0, π) with sin c = 2/π. Solving, c = arcsin(2/π) ≈ 0.690 (and by symmetry c = π − 0.690 ≈ 2.452 also works).

Answer: f_avg = 2/π ≈ 0.637.

(d) Common Mistakes

Reversing top and bottom. Writing ∫ (bottom − top) gives a negative area, and on a split region it can partially cancel. Fix: always test an interior point to confirm which curve is larger, and write the integrand as (top − bottom) so the result is positive by construction.

Forgetting the π in volumes of revolution. Students set up ∫ R² dx and forget that revolving a disc multiplies by π. Fix: write the π before the integral sign the instant you choose disc or washer — V = π ∫ ....

Writing (R_outer − R_inner)² instead of R_outer² − R_inner². These are not equal. The washer's area is π(R_o² − R_i²), the difference of two circle areas. Fix: compute each squared radius separately, then subtract.

Using the wrong radius when revolving about y = k. Revolving about y = −1 or y = 8 instead of the x-axis, students keep using the bare curve as the radius. Fix: the radius is the distance from the axis to the curve — |curve − k|. Recompute both radii from the new axis, and be careful which curve becomes outer when the axis is above or below the region.

Mixing up the variable of integration / dropping the 1/(b−a). Integrating a "right − left" region in x (or omitting the 1/(b−a) factor in average value, which just gives the integral, not the average) are both setup errors. Fix: if the region is bounded left/right, integrate dy; and for average value, divide by the length of the interval every time.

(e) Practice Problems

— Interpretation. A pump removes oil from a tank at a rate of r(t) = 100 e^(−0.1t) liters per hour for 0 ≤ t ≤ 8. (a) Write an integral for the total volume of oil removed over the 8 hours, with units. (b) Evaluate it. (c) Explain in one sentence what the value of ∫_0^8 r(t) dt represents in the context of this problem.

— Justification. Let f(x) = x² on [0, 3]. (a) Find the average value of f on [0, 3]. (b) State whether the Mean Value Theorem for Integrals guarantees a value c in (0, 3) with f(c) equal to that average, justifying your answer by checking the hypothesis. (c) Find such a c.

(f) AP Exam Focus — Full Free-Response Question

> FRQ (9 points total). Calculator NOT permitted.

>

> Let R be the region in the first quadrant bounded by the graphs of y = 4x and y = x³.

>

> (a) Find the area of R. (2 points)

>

> (b) Find the volume of the solid generated when R is revolved about the x-axis. (3 points)

>

> (c) The region R is the base of a solid. For this solid, cross-sections perpendicular to the x-axis are squares. Find the volume of the solid. (2 points)

>

> (d) Find the average vertical distance between the two curves over the interval that bounds R, and explain what this value represents. (2 points)

Model Solution

Setup (do this once, use it for every part). The curves meet where 4x = x³ → x³ − 4x = 0 → x(x² − 4) = 0 → x = −2, 0, 2. In the first quadrant the relevant interval is [0, 2]. Test x = 1: 4x = 4, x³ = 1, so y = 4x is the top, y = x³ is the bottom on [0, 2].

Part (a) — Area.

A = ∫_0^2 (4x − x³) dx = [2x² − x⁴/4]_0^2 = (8 − 4) − 0 = 4

Area = 4.

Part (b) — Volume of revolution about the x-axis (washer). Revolving about the x-axis, the outer radius is the top curve 4x, the inner radius is the bottom curve x³:

V = π ∫_0^2 ([4x]² − [x³]²) dx = π ∫_0^2 (16x² − x⁶) dx
  = π [16x³/3 − x⁷/7]_0^2
  = π (128/3 − 128/7)
  = π · (896 − 384)/21 = 512π/21

Volume = 512π/21.

Part (c) — Square cross-sections. The side of each square is the vertical distance s = 4x − x³, so A(x) = (4x − x³)²:

V = ∫_0^2 (4x − x³)² dx = ∫_0^2 (16x² − 8x⁴ + x⁶) dx
  = [16x³/3 − 8x⁵/5 + x⁷/7]_0^2
  = 128/3 − 256/5 + 128/7

Over a common denominator of 105: = (4480 − 5376 + 1920)/105 = 1024/105.

Volume = 1024/105.

Part (d) — Average vertical distance. The vertical distance between the curves is 4x − x³, and the average value over [0, 2] is:

avg = (1/(2 − 0)) ∫_0^2 (4x − x³) dx = (1/2)(4) = 2

The average vertical distance is 2. This means that if the gap between the curves were constant across [0, 2], it would equal 2 — equivalently, a rectangle of width 2 and height 2 has the same area (4) as the region R.

Scoring Commentary — Where Students Lose Points

• Part (a), 2 pts: 1 point for the correct integrand (top − bottom) with correct limits 0 and 2; 1 point for the answer 4. Students who use x = −2 as a limit (ignoring "first quadrant") lose the answer point.
• Part (b), 3 pts: 1 point for the constant π and the washer form R_o² − R_i²; 1 point for the correct radii (4x)² and (x³)²; 1 point for the answer 512π/21. Writing (4x − x³)² instead of (4x)² − (x³)² loses the first two points. Dropping the π loses 1 point even if everything else is right.
• Part (c), 2 pts: 1 point for A(x) = (4x − x³)² (no π — the cross-sections are squares); 1 point for 1024/105. A common loss is carrying a π into a non-revolution cross-section.
• Part (d), 2 pts: 1 point for the average-value setup (1/2)∫_0^2 (4x − x³) dx (the 1/(b−a) factor is required); 1 point for the answer 2 with a correct interpretation. Omitting the 1/(b−a) gives 4 and loses both points; giving 2 with no interpretation loses the second point.