BrightChamps Logo
Login

Summarize this article:

Live Math Learners Count Icon103 Learners

Last updated on September 17, 2025

Area of Parabola

Professor Greenline Explaining Math Concepts

Area is the space inside the boundaries of a two-dimensional shape or surface. There are different methods for finding the area under curves, such as parabolas, which are widely used in calculus and physics. In this section, we will explore the area under a parabola.

Area of Parabola for US Students
Professor Greenline from BrightChamps

What is the Area Under a Parabola?

A parabola is a U-shaped curve that can open upwards or downwards. It is the graph of a quadratic function of the form y = ax² + bx + c.

 

The area under a parabola between two points on the x-axis can be calculated using definite integration. This area represents the space enclosed between the parabola and the x-axis.

Professor Greenline from BrightChamps

Formula for the Area Under a Parabola

To find the area under a parabola from x = a to x = b, we use definite integration. The formula is: Area = ∫[a to b] (ax² + bx + c) dx Let’s see how the formula is derived.

 

Derivation of the formula: 1. Start with the quadratic function y = ax² + bx + c. 2. Integrate the function with respect to x: ∫(ax² + bx + c) dx = (a/3)x³ + (b/2)x² + cx + C, where C is the constant of integration. 3. Evaluate the definite integral from x = a to x = b: Area = [(a/3)b³ + (b/2)b² + cb] - [(a/3)a³ + (b/2)a² + ca]

Professor Greenline from BrightChamps

How to Find the Area Under a Parabola?

We can find the area under a parabola using definite integration. Here’s a step-by-step method:

 

1. Identify the quadratic function y = ax² + bx + c.

 

2. Determine the interval [a, b] over which you want to find the area.

 

3. Set up the integral: ∫[a to b] (ax² + bx + c) dx.

 

4. Integrate the function and evaluate the definite integral over [a, b].

 

For example, if y = 2x² + 3x + 1 and you want to find the area from x = 0 to x = 2: Area = ∫[0 to 2] (2x² + 3x + 1) dx

Professor Greenline from BrightChamps

Units of the Area Under a Parabola

The area under a parabola is measured in square units. The unit depends on the measurement system used for the x and y axes:

 

In the metric system, it might be square meters (m²), square centimeters (cm²), etc.

 

In the imperial system, it could be square inches (in²), square feet (ft²), etc.

Professor Greenline from BrightChamps

Special Cases or Variations for the Area Under a Parabola

Depending on the form of the quadratic function and the interval, the area calculation can vary. Here are some special cases:

 

Case 1: Parabola symmetric about the y-axis If the parabola is symmetric about the y-axis, such as y = ax², integrate over [-b, b].

 

Case 2: Parabola intersecting x-axis If the parabola intersects the x-axis at points x = a and x = b, find the area between these points.

 

Case 3: Applications in physics In physics, finding the area under a velocity-time graph (a parabola) can represent the displacement of an object.

Professor Greenline from BrightChamps

Tips and Tricks for Calculating the Area Under a Parabola

To ensure accurate results while calculating the area under a parabola, consider these tips: 

 

  • Verify the limits of integration to ensure they correspond to the correct interval. 
     
  • Use symmetry properties of the parabola to simplify calculations if applicable. 
     
  • Double-check the integration process, especially the substitution of limits.
Max Pointing Out Common Math Mistakes

Common Mistakes and How to Avoid Them in Calculating Area Under a Parabola

Errors can occur when calculating the area under a parabola. Let’s examine some common mistakes.

Mistake 1

Red Cross Icon Indicating Mistakes to Avoid in This Math Topic

Incorrect integration of the quadratic function

Green Checkmark Icon Indicating Correct Solutions in This Math Topic

Ensure the integration of the quadratic function is performed correctly.

 

Missteps in the integration process can lead to incorrect area calculations.

Mistake 2

Red Cross Icon Indicating Mistakes to Avoid in This Math Topic

Using wrong limits of integration

Green Checkmark Icon Indicating Correct Solutions in This Math Topic

Check that the limits of integration correspond to the correct interval on the x-axis.

 

Incorrect limits can result in an inaccurate area.

Mistake 3

Red Cross Icon Indicating Mistakes to Avoid in This Math Topic

Confusing the parabola with other curves

Green Checkmark Icon Indicating Correct Solutions in This Math Topic

Ensure you are working with a quadratic function and not another type of curve, such as a circle or cubic function.

Mistake 4

Red Cross Icon Indicating Mistakes to Avoid in This Math Topic

Forgetting to evaluate definite integral

Green Checkmark Icon Indicating Correct Solutions in This Math Topic

After integrating, remember to evaluate the definite integral over the specified interval to find the exact area.

Mistake 5

Red Cross Icon Indicating Mistakes to Avoid in This Math Topic

Omitting units from the final answer

Green Checkmark Icon Indicating Correct Solutions in This Math Topic

Always include the correct units in your final answer to convey the area accurately.

arrow-right
Max from BrightChamps Saying "Hey"
Hey!

Area Under a Parabola Examples

Ray, the Character from BrightChamps Explaining Math Concepts
Max, the Girl Character from BrightChamps

Problem 1

Calculate the area under the parabola y = x² + 2x + 3 from x = 1 to x = 4.

Ray, the Boy Character from BrightChamps Saying "Let’s Begin"
Okay, lets begin

We will find the area as 39 square units.

Explanation

The quadratic function is y = x² + 2x + 3, and the interval is [1, 4].

Area = ∫[1 to 4] (x² + 2x + 3) dx = [(1/3)x³ + x² + 3x] (from 1 to 4) = [(1/3)(4)³ + (4)² + 3(4)] - [(1/3)(1)³ + (1)² + 3(1)] = [64/3 + 16 + 12] - [1/3 + 1 + 3] = [92/3 + 28] - [4.333] = 39 square units

Max from BrightChamps Praising Clear Math Explanations
Well explained 👍
Max, the Girl Character from BrightChamps

Problem 2

Find the area under the parabola y = 3x² - 2x + 1 from x = 0 to x = 3.

Ray, the Boy Character from BrightChamps Saying "Let’s Begin"
Okay, lets begin

We will find the area as 24.5 square units.

Explanation

The quadratic function is y = 3x² - 2x + 1, and the interval is [0, 3].

Area = ∫[0 to 3] (3x² - 2x + 1) dx = [(x³) - x² + x] (from 0 to 3) = [(3³) - 3² + 3] - [0] = [27 - 9 + 3] = 21 square units

Max from BrightChamps Praising Clear Math Explanations
Well explained 👍
Max, the Girl Character from BrightChamps

Problem 3

The area under the parabola y = 4x² - x + 2 from x = -1 to x = 2 is given as 14 square units. Verify this area.

Ray, the Boy Character from BrightChamps Saying "Let’s Begin"
Okay, lets begin

We find the calculated area as 14 square units.

Explanation

The quadratic function is y = 4x² - x + 2, and the interval is [-1, 2].

Area = ∫[-1 to 2] (4x² - x + 2) dx = [(4/3)x³ - (1/2)x² + 2x] (from -1 to 2) = [(4/3)(2)³ - (1/2)(2)² + 2(2)] - [(4/3)(-1)³ - (1/2)(-1)² + 2(-1)] = [32/3 - 2 + 4] - [-4/3 - 1/2 - 2] = 14 square units

Max from BrightChamps Praising Clear Math Explanations
Well explained 👍
Max, the Girl Character from BrightChamps

Problem 4

If the parabola y = 5x² + x - 1 is given, find the area from x = -2 to x = 1.

Ray, the Boy Character from BrightChamps Saying "Let’s Begin"
Okay, lets begin

We will find the area as 26.5 square units.

Explanation

The quadratic function is y = 5x² + x - 1, and the interval is [-2, 1].

Area = ∫[-2 to 1] (5x² + x - 1) dx = [(5/3)x³ + (1/2)x² - x] (from -2 to 1) = [(5/3)(1)³ + (1/2)(1)² - (1)] - [(5/3)(-2)³ + (1/2)(-2)² - (-2)] = [5/3 + 1/2 - 1] - [-40/3 + 2 - 2] = 26.5 square units

Max from BrightChamps Praising Clear Math Explanations
Well explained 👍
Max, the Girl Character from BrightChamps

Problem 5

Help Emma find the area under the parabola y = 2x² - 3x + 4 from x = 0 to x = 5.

Ray, the Boy Character from BrightChamps Saying "Let’s Begin"
Okay, lets begin

We will find the area as 95 square units.

Explanation

The quadratic function is y = 2x² - 3x + 4, and the interval is [0, 5].

Area = ∫[0 to 5] (2x² - 3x + 4) dx = [(2/3)x³ - (3/2)x² + 4x] (from 0 to 5) = [(2/3)(5)³ - (3/2)(5)² + 4(5)] - [0] = [250/3 - 75/2 + 20] = 95 square units

Max from BrightChamps Praising Clear Math Explanations
Well explained 👍
Ray Thinking Deeply About Math Problems

FAQs on Area Under a Parabola

1.Is it possible for the area under a parabola to be negative?

No, the area under a parabola is always a positive value as it represents the space enclosed by the curve and the x-axis.

Math FAQ Answers Dropdown Arrow

2.How to find the area under a parabola if the parabola is below the x-axis?

If the parabola is below the x-axis, integrate as usual and take the absolute value of the area to represent positive space.

Math FAQ Answers Dropdown Arrow

3.What if the parabola opens downwards?

For a downward-opening parabola, integrate as usual. The process remains the same, but ensure to consider the limits correctly.

Math FAQ Answers Dropdown Arrow

4.How do you calculate the area if the parabola is symmetric?

If the parabola is symmetric, you can calculate the area for one side and double it. This reduces calculation time and complexity.

Math FAQ Answers Dropdown Arrow

5.What is meant by the area under a parabola?

The area under a parabola is the total space enclosed between the curve and the x-axis over a specified interval.

Math FAQ Answers Dropdown Arrow
Math Teacher Background Image
Math Teacher Image

Seyed Ali Fathima S

About the Author

Seyed Ali Fathima S a math expert with nearly 5 years of experience as a math teacher. From an engineer to a math teacher, shows her passion for math and teaching. She is a calculator queen, who loves tables and she turns tables to puzzles and songs.

Max, the Girl Character from BrightChamps

Fun Fact

: She has songs for each table which helps her to remember the tables

INDONESIA - Axa Tower 45th floor, JL prof. Dr Satrio Kav. 18, Kel. Karet Kuningan, Kec. Setiabudi, Kota Adm. Jakarta Selatan, Prov. DKI Jakarta
INDIA - H.No. 8-2-699/1, SyNo. 346, Rd No. 12, Banjara Hills, Hyderabad, Telangana - 500034
SINGAPORE - 60 Paya Lebar Road #05-16, Paya Lebar Square, Singapore (409051)
USA - 251, Little Falls Drive, Wilmington, Delaware 19808
VIETNAM (Office 1) - Hung Vuong Building, 670 Ba Thang Hai, ward 14, district 10, Ho Chi Minh City
VIETNAM (Office 2) - 143 Nguyễn Thị Thập, Khu đô thị Him Lam, Quận 7, Thành phố Hồ Chí Minh 700000, Vietnam
UAE - BrightChamps, 8W building 5th Floor, DAFZ, Dubai, United Arab Emirates
UK - Ground floor, Redwood House, Brotherswood Court, Almondsbury Business Park, Bristol, BS32 4QW, United Kingdom