05-09 - Applications of Integration
Phase: 5 | Subject: 05-09 Prerequisites: 05-02-the-definite-integral.md, 05-05-integration-by-parts.md Next subject: 05-10-parametric-equations-and-polar-coordinates.md
Learning Objectives
By the end of this subject, you will be able to:
- Calculate area between two curves
- Calculate volume by disks and washers
- Calculate volume by cylindrical shells
- Calculate arc length
- Calculate surface area of revolution
- Calculate average value of a function
Core Content
Area Between Two Curves
$Area = ∫[a to b] [top(x) - bottom(x)] dx $
where the curves intersect at x = a and x = b.
Example: Area between y = x² and y = x from x = 0 to x = 1. Top: y = x, Bottom: y = x² Area = ∫[0 to 1] (x - x²)dx = [x²/2 - x³/3][0 to 1] = 1/2 - 1/3 = 1/6
Volume by Disks
When rotating y = f(x) around the x-axis:
$V = π ∫[a to b] [f(x)]² dx $
Example: Rotate y = √x from x = 0 to x = 4 around x-axis. V = π ∫[0 to 4] x dx = π[x²/2][0 to 4] = π(8) = 8π
Volume by Washers
When rotating the region between two curves around an axis:
$V = π ∫[a to b] ([outer]² - [inner]²) dx $
Volume by Shells
When rotating around the y-axis:
$V = 2π ∫[a to b] x·f(x) dx $
When to use shells: When the region is easier to describe as x = f(y) or when rotating around y-axis with y = f(x).
Arc Length
$L = ∫[a to b] √(1 + (dy/dx)²) dx $
Example: Length of y = x² from x = 0 to x = 1. dy/dx = 2x. L = ∫[0 to 1] √(1 + 4x²)dx. This requires trig substitution — result involves ln and √5.
Surface Area of Revolution
Around x-axis:
$S = 2π ∫[a to b] y·√(1 + (dy/dx)²) dx $
Average Value
$f_avg = (1/(b-a)) ∫[a to b] f(x)dx $
Mean Value Theorem for Integrals: There exists c in [a, b] such that f(c) = f_avg.
Key Terms
- 05 09 Applications Of Integration
- Arc Length
- Area Between Two Curves
- Average Value
- Correct: A)
- Correct: B)
- Correct: C)
- Example 1: Area between curves
- Example 2: Volume by washers
- Example 3: Average value
- Surface Area of Revolution
- Volume by Disks
Worked Examples
Example 1: Area between curves
y = x² and y = 2x. Intersection: x² = 2x → x = 0, 2. Top: y = 2x, Bottom: y = x². Area = ∫[0 to 2] (2x - x²)dx = [x² - x³/3][0 to 2] = 4 - 8/3 = 4/3
Example 2: Volume by washers
Rotate region between y = x and y = x² from x = 0 to 1 around x-axis. Outer: y = x, Inner: y = x². V = π ∫[0 to 1] (x² - x⁴)dx = π[x³/3 - x⁵/5][0 to 1] = π(1/3 - 1/5) = 2π/15
Example 3: Average value
Average of f(x) = x² on [0, 2]. f_avg = (1/2) ∫[0 to 2] x²dx = (1/2)[x³/3][0 to 2] = (1/2)(8/3) = 4/3
Practice Problems
Problem 1: Area between y = x and y = x² from 0 to 1
Answer
∫[0 to 1] (x - x²)dx = 1/2 - 1/3 = 1/6Problem 2: Volume of rotating y = √x, 0 ≤ x ≤ 4, around x-axis
Answer
π ∫[0 to 4] x dx = π[x²/2][0 to 4] = 8πProblem 3: Average value of f(x) = sin(x) on [0, π]
Answer
(1/π) ∫[0 to π] sin(x)dx = (1/π)[-cos(x)][0 to π] = (1/π)(1 + 1) = 2/πProblem 4: Arc length of y = (2/3)x^(3/2) from x = 0 to x = 3
Answer
dy/dx = x^(1/2). L = ∫[0 to 3] √(1 + x)dx = ∫[0 to 3] (1+x)^(1/2)dx. u = 1+x: ∫[1 to 4] u^(1/2)du = (2/3)[u^(3/2)][1 to 4] = (2/3)(8 − 1) = 14/3.Problem 5: Volume: rotate region bounded by y = x², x = 0, x = 2, y = 0 around y-axis using shells
Answer
V = 2π ∫[0 to 2] x·x² dx = 2π ∫[0 to 2] x³ dx = 2π[x⁴/4][0 to 2] = 2π(4) = 8π.Summary
Key takeaways:
- Area between curves: ∫(top - bottom)
- Disk method: V = π∫y²dx (rotation around x-axis)
- Washer: V = π∫(outer² - inner²)
- Shell: V = 2π∫x·y dx (rotation around y-axis)
- Arc length: ∫√(1 + (dy/dx)²) dx
- Average value: (1/(b-a)) ∫[a to b] f(x)dx
Pitfalls
- Choosing the wrong curve as top/bottom. Area between curves is ∫(top − bottom). If the curves cross within the interval, you must split at the intersection point(s). Assuming one curve is always above the other without checking is a common error.
- Confusing the disk and shell methods. Disk/washer methods integrate perpendicular to the axis of rotation (cross-sectional area). The shell method integrates parallel to the axis (cylindrical shells). Using the wrong method leads to integrals that are impossible to set up or have the wrong variable of integration.
- Forgetting to square in the disk formula. V = π∫[f(x)]²dx. The square comes from the area of a circular cross-section: πr². Writing π∫f(x)dx instead misses the square and gives a completely wrong (and dimensionally inconsistent) result.
- Using the wrong radius in the shell method. When rotating around the y-axis using shells, the radius is x (the horizontal distance from the axis of rotation to the shell). Using f(x) or y as the radius is incorrect.
- Mishandling the arc length formula under the radical. L = ∫√(1 + (dy/dx)²)dx. Students sometimes write √((dy/dx)²) = |dy/dx| (dropping the 1) or misinterpret (1 + dy/dx)². The expression under the radical is 1 plus the square of the derivative, not the square of (1 + dy/dx).
Quiz
Q1: Area between y = x and y = x² from x = 0 to 1:
A) 1/6 B) 1/3 C) 1/2 D) 1/12
Answer and Explanations
**Correct: A)** - If you chose A: ∫[0 to 1] (x - x²)dx = [x²/2 - x³/3][0 to 1] = 1/2 - 1/3 = 1/6. Correct! - If you chose B: You may have computed 1/2 - 1/6 = 1/3. Check: x³/3 at x=1 is 1/3, not 1/6. - If you chose C: That's just ∫[0 to 1] x dx = 1/2, ignoring x². - If you chose D: You divided by 2 again somewhere.Q2: Volume by rotating y = x² from 0 to 2 around x-axis:
A) 8π/3 B) 16π/3 C) 32π/5 D) 8π
Answer and Explanations
**Correct: C)** - If you chose C: V = π ∫[0 to 2] x⁴dx = π[x⁵/5][0 to 2] = 32π/5. Correct! - If you chose A: That's ∫[0 to 2] x²dx = 8/3. Missing the π and the square in the disk method. - If you chose B: You may have computed 2 × 8π/3 = 16π/3. That's not a standard formula. - If you chose D: That's for y = √x from 0 to 4, not y = x² from 0 to 2.Q3: Average value of f(x) = x² on [0, 2]:
A) 4/3 B) 2 C) 4 D) 8/3
Answer and Explanations
**Correct: A)** - If you chose A: (1/2) ∫[0 to 2] x²dx = (1/2)(8/3) = 4/3. Correct! - If you chose B: That's f(2) = 4? No, f(2) = 4. 2 is the midpoint value or the interval length. - If you chose C: That's f(2) = 4, the value at the endpoint. - If you chose D: You computed ∫[0 to 2] x²dx = 8/3 but forgot to divide by (b-a) = 2.Q4: Volume by shells for rotating y = x², 0 ≤ x ≤ 1, around y-axis:
A) π/2 B) π/3 C) π/4 D) π/5
Answer and Explanations
**Correct: A)** - If you chose A: V = 2π ∫[0 to 1] x·x²dx = 2π ∫[0 to 1] x³dx = 2π[x⁴/4][0 to 1] = 2π(1/4) = π/2. Correct! - If you chose B: You may have used disk method around x-axis incorrectly, or forgot the 2 in 2π. - If you chose C: That's π/4, half of π/2. You may have used shells but with wrong formula. - If you chose D: That's π/5, which doesn't appear in standard calculations here.Q5: The average value of f(x) on [a, b] is:
A) (f(a) + f(b))/2 B) (1/(b-a)) ∫[a to b] f(x)dx C) ∫[a to b] f(x)dx D) f((a+b)/2)
Answer and Explanations
**Correct: B)** - If you chose B: Average value = (1/(b-a)) times the integral. Correct! - If you chose A: That's the average of the ENDPOINTS, not the average value over the interval. - If you chose C: That's the definite integral itself, not the average. - If you chose D): That's the value at the midpoint, which equals the average only for linear functions.Next Steps
Next up: 05-10-parametric-equations-and-polar-coordinates.md