How to Solve: Applications of Integration (Area Under Curves, Volumes of Revolution, Differential Equations)

How to Solve: Applications of Integration (Area Under Curves, Volumes of Revolution, Differential Equations)

GCSE / A-Level Maths

Introduction

"Mastering integration applications doesn’t just get you marks—it unlocks real-world problems like calculating the exact fuel needed for a rocket launch, designing medical implants, or predicting population growth. On your A-Level exam, this topic is worth 15-20% of your calculus questions, and one slip in limits or rotation axes can cost you 5+ marks. Let’s make sure you never lose those marks again."

What You Need To Know First

Before diving in, you must already understand: 1. Basic integration rules (power rule, trigonometric integrals, substitution). 2. Definite integrals (how to evaluate them, including negative areas). 3. Graph sketching (identifying intercepts, symmetry, and regions between curves).

If any of these feel shaky, pause and review them first—this guide assumes you’re solid on them.

Key Vocabulary

Formulas To Know

1. Area Under a Curve (or Between Curves)

Formula: - Area under y = f(x) from x = a to x = b: [ A = \int_a^b f(x) \, dx \quad \text{(MEMORISE THIS)} ] - f(x) = function (must be above the x-axis in [a, b] for this to work). - a, b = lower and upper limits (left and right bounds).

• Area between two curves y = f(x) and y = g(x): [ A = \int_a^b \left[ f(x) - g(x) \right] dx \quad \text{(MEMORISE THIS)} ]
• f(x) = top function (must be above g(x) in [a, b]).
• g(x) = bottom function.
• a, b = points of intersection (solve f(x) = g(x) to find them).

Note: If the curves cross, split the integral at the intersection point.

2. Volumes of Revolution

Formula (Disk Method – rotate around x-axis): [ V = \pi \int_a^b \left[ f(x) \right]^2 dx \quad \text{(MEMORISE THIS)} ] - f(x) = function being rotated. - a, b = limits along the axis of rotation.

Formula (Washer Method – rotate around x-axis with a hole): [ V = \pi \int_a^b \left[ (f(x))^2 - (g(x))^2 \right] dx \quad \text{(MEMORISE THIS)} ] - f(x) = outer radius (farther from axis). - g(x) = inner radius (closer to axis).

Formula (Rotate around y-axis): - Rewrite the function as x = f(y) and integrate with respect to y: [ V = \pi \int_c^d \left[ f(y) \right]^2 dy ] - c, d = y-limits.

3. Differential Equations (Separation of Variables)

General Solution: [ \frac{dy}{dx} = f(x)g(y) \implies \int \frac{1}{g(y)} dy = \int f(x) dx \quad \text{(MEMORISE THIS)} ] - Solve for y to get the general solution (includes +C). - Use an initial condition to find C and get the particular solution.

Step-by-Step Method

A. Area Under a Curve / Between Curves

Steps: 1. Sketch the graph(s). Label intercepts and points of intersection. 2. Decide if it’s area under one curve or between two curves.
- One curve: Use ∫ f(x) dx.
- Two curves: Use ∫ (top – bottom) dx. 3. Find limits of integration (a and b).
- For one curve: Given in the question.
- For two curves: Solve f(x) = g(x) to find intersection points. 4. Set up the integral. Write the correct integrand (top – bottom if needed). 5. Integrate. Use power rule, trig rules, or substitution. 6. Evaluate at the limits. Subtract lower limit from upper limit. 7. Check the sign. Area is always positive—if your answer is negative, take the absolute value.

B. Volumes of Revolution

Steps: 1. Sketch the curve and the solid. Label the axis of rotation. 2. Decide: Disk or Washer?
- Disk: No hole (rotate one curve).
- Washer: Hole (rotate region between two curves). 3. Identify the radius (or radii).
- Disk: Radius = f(x) (distance from curve to axis).
- Washer: Outer radius = f(x), inner radius = g(x). 4. Find limits of integration (a and b).
- Usually given, or solve f(x) = 0/g(x) for intercepts. 5. Set up the integral.
- Disk: V = π ∫ [f(x)]² dx.
- Washer: V = π ∫ [(f(x))² – (g(x))²] dx. 6. Integrate. Expand if needed, then integrate term by term. 7. Evaluate at the limits. Subtract lower limit from upper limit. 8. Simplify. Factor out π if possible.

C. Differential Equations (Separation of Variables)

Steps: 1. Write the equation in the form dy/dx = f(x)g(y). 2. Separate variables: Move all y terms to one side, all x terms to the other.
- Example: dy/dx = xy → (1/y) dy = x dx. 3. Integrate both sides. Add +C to the x-side only. 4. Solve for y. Isolate y to get the general solution. 5. Use the initial condition to find C (if given). 6. Write the particular solution. Substitute C back into the general solution.

Worked Examples

Example 1 – Basic: Area Under a Curve

Question: Find the area bounded by y = x², the x-axis, and the lines x = 1 and x = 3.

Solution: 1. Sketch: y = x² is a parabola opening upwards. Shade the region from x=1 to x=3. 2. One curve, no intersections: Use ∫ f(x) dx. 3. Limits: a = 1, b = 3. 4. Set up integral:
[
A = \int_1^3 x^2 \, dx
] 5. Integrate:
[
\int x^2 \, dx = \frac{x^3}{3} + C
] 6. Evaluate:
[
\left. \frac{x^3}{3} \right|_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}
] 7. Check sign: Positive → area = 26/3.

What we did and why: - We used the basic area formula because there was only one curve and the x-axis. - The limits were given, so we didn’t need to find intersections. - Always check the sign—if the curve were below the x-axis, the integral would be negative, but area is positive.

Example 2 – Medium: Area Between Two Curves

Question: Find the area enclosed by y = x² and y = 2x.

Solution: 1. Sketch: y = x² (parabola) and y = 2x (straight line). They intersect at two points. 2. Two curves: Use ∫ (top – bottom) dx. 3. Find limits: Solve x² = 2x → x² – 2x = 0 → x(x – 2) = 0 → x = 0 or x = 2. 4. Set up integral:
- Top function: y = 2x (above y = x² in [0, 2]).
- Bottom function: y = x².
[
A = \int_0^2 (2x - x^2) \, dx
] 5. Integrate:
[
\int (2x - x^2) \, dx = x^2 - \frac{x^3}{3} + C
] 6. Evaluate:
[
\left. \left( x^2 - \frac{x^3}{3} \right) \right|_0^2 = \left(4 - \frac{8}{3}\right) - (0) = \frac{4}{3}
] 7. Check sign: Positive → area = 4/3.

What we did and why: - We found the intersection points to set the limits. - We confirmed which curve was on top by testing a point (e.g., x=1: 2(1) = 2 > 1² = 1). - If the curves crossed, we’d split the integral at the intersection.

Example 3 – Exam-Style: Volume of Revolution

Question: The region R is bounded by y = √x, the x-axis, and the line x = 4. Find the volume of the solid formed when R is rotated 360° about the x-axis.

Solution: 1. Sketch: y = √x is the upper half of a sideways parabola. Region R is from x=0 to x=4. 2. Disk method: Rotating one curve around the x-axis. 3. Radius: f(x) = √x (distance from curve to x-axis). 4. Limits: a = 0, b = 4. 5. Set up integral:
[
V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx
] 6. Integrate:
[
\pi \int x \, dx = \pi \left( \frac{x^2}{2} \right) + C
] 7. Evaluate:
[
\pi \left. \frac{x^2}{2} \right|_0^4 = \pi \left( \frac{16}{2} - 0 \right) = 8\pi
] 8. Simplify: Volume = 8π.

What we did and why: - We used the disk method because there was no hole (only one curve). - Squaring √x gave us x, simplifying the integral. - Always include π in the volume formula—missing it is a common error.

Example 4 – Differential Equation (Exam-Style)

Question: Solve the differential equation dy/dx = 3x²y, given that y = 2 when x = 0.

Solution: 1. Separate variables:
[
\frac{dy}{dx} = 3x^2y \implies \frac{1}{y} dy = 3x^2 dx
] 2. Integrate both sides:
[
\int \frac{1}{y} dy = \int 3x^2 dx \implies \ln|y| = x^3 + C
] 3. Solve for y:
[
y = e^{x^3 + C} = e^C \cdot e^{x^3} = Ae^{x^3} \quad \text{(where } A = e^C\text{)}
] 4. Use initial condition (y=2 when x=0):
[
2 = Ae^{0} \implies A = 2
] 5. Particular solution:
[
y = 2e^{x^3}
]

What we did and why: - We separated variables to isolate y and x. - The constant C was combined into A for simplicity. - The initial condition let us find the exact value of A.

Common Mistakes

Exam Traps

1-Minute Recap

"Listen up—this is your 60-second cheat sheet for integration applications. For area under a curve, it’s just ∫ f(x) dx between limits. For area between curves, it’s ∫ (top – bottom) dx—find where they cross to set the limits. Volumes? Disk method: π ∫ [f(x)]² dx. Washer method: π ∫ (outer² – inner²) dx. Rotating around the y-axis? Rewrite as x = f(y) and integrate with respect to y. Differential equations? Separate variables, integrate both sides, add +C, then use the initial condition to kill C. Common traps: forgetting π in volumes, mixing up top/bottom curves, and losing the +C. Sketch every graph, label everything, and check your signs. You’ve got this—now go smash that exam!