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📐 Concept diagram

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:

  1. Calculate area between two curves
  2. Calculate volume by disks and washers
  3. Calculate volume by cylindrical shells
  4. Calculate arc length
  5. Calculate surface area of revolution
  6. 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

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 = 1/6

Problem 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] = 8π

Problem 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/π

Problem 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:


Pitfalls


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