Area of Parabola
2026-02-28 13:58 Diff
https://brightchamps.com/en-us/math/measurement/area-of-parabola

181 Learners

Last updated on September 17, 2025

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.

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.

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]

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

Explore Our Programs

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.

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.

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.

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.

Problem 1

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

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

Well explained 👍

Problem 2

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

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

Well explained 👍

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.

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

Well explained 👍

Problem 4

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

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

Well explained 👍

Problem 5

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

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

Well explained 👍

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.

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.

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.

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.

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.

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.

Fun Fact

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