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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.
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.
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]
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
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.
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.
To ensure accurate results while calculating the area under a parabola, consider these tips:
Errors can occur when calculating the area under a parabola. Let’s examine some common mistakes.
Calculate the area under the parabola y = x² + 2x + 3 from x = 1 to x = 4.
We will find the area as 39 square units.
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
Find the area under the parabola y = 3x² - 2x + 1 from x = 0 to x = 3.
We will find the area as 24.5 square units.
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
The area under the parabola y = 4x² - x + 2 from x = -1 to x = 2 is given as 14 square units. Verify this area.
We find the calculated area as 14 square units.
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
If the parabola y = 5x² + x - 1 is given, find the area from x = -2 to x = 1.
We will find the area as 26.5 square units.
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
Help Emma find the area under the parabola y = 2x² - 3x + 4 from x = 0 to x = 5.
We will find the area as 95 square units.
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
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.
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