Proof of Quick Formulas for Calculating Area Using Integrals
Previously, we learned quick methods for calculating area using integrals. These methods are helpful for solving certain types of problems quickly, but they are limited in scope and don't apply to all problems. As critical thinkers, we shouldn't just accept these formulas without understanding where they come from, even though they give the same results as the fundamental method when applied correctly.
To understand these proofs, you should be familiar with:
- Indefinite and definite integrals of algebraic functions
- Calculating area using integrals
- Operations with roots of quadratic equations
Proof 1: Discriminant Formula
For a quadratic equation \(ax^2 + bx + c = 0\), the discriminant is \(D = b^2 - 4ac\).
Conditions: This formula applies only to regions bounded by:
- Two parabolas, or
- A parabola and a straight line
Case 1: Two parabolas
Let the parabolas be: \(y_1 = a_1x^2 + b_1x + c_1\) and \(y_2 = a_2x^2 + b_2x + c_2\)
Set them equal to find intersection points:
\(a_1x^2 + b_1x + c_1 = a_2x^2 + b_2x + c_2\)
\((a_1 - a_2)x^2 + (b_1 - b_2)x + (c_1 - c_2) = 0\)
Let \(a = a_1 - a_2\), \(b = b_1 - b_2\), \(c = c_1 - c_2\)
Case 2: Parabola and line
Let: \(y_1 = a_1x^2 + b_1x + c_1\) and \(y_2 = b_2x + c_2\)
Set them equal:
\(a_1x^2 + b_1x + c_1 = b_2x + c_2\)
\(a_1x^2 + (b_1 - b_2)x + (c_1 - c_2) = 0\)
Let \(a = a_1\), \(b = b_1 - b_2\), \(c = c_1 - c_2\)
In both cases, we get: \(ax^2 + bx + c = 0\) with roots \(x_1\) and \(x_2\) (intersection points).
Properties of roots:
\(x_1 + x_2 = -\frac{b}{a}\), \(x_1x_2 = \frac{c}{a}\), \(x_2 - x_1 = \frac{\sqrt{D}}{a}\)
\(x_2^2 - x_1^2 = (x_2 - x_1)(x_2 + x_1) = \frac{\sqrt{D}}{a} \cdot \frac{(-b)}{a}\)
\(x_2^3 - x_1^3 = (x_2 - x_1)^3 + 3x_1x_2(x_2 - x_1) = \left(\frac{\sqrt{D}}{a}\right)^3 + 3 \cdot \frac{c}{a} \cdot \frac{\sqrt{D}}{a}\)
Area calculation using integration:
\(\text{Area} = \int_{x_1}^{x_2} (y_1 - y_2) dx = \int_{x_1}^{x_2} (ax^2 + bx + c) dx\)
\(\text{Area} = \left[ \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx \right]_{x_1}^{x_2}\)
\(= \frac{a}{3}(x_2^3 - x_1^3) + \frac{b}{2}(x_2^2 - x_1^2) + c(x_2 - x_1)\)
Substitute root properties:
\(= \frac{a}{3}\left( \frac{D\sqrt{D}}{a^3} + \frac{3c\sqrt{D}}{a^2} \right) + \frac{b}{2}\left( -\frac{b\sqrt{D}}{a^2} \right) + c\left( \frac{\sqrt{D}}{a} \right)\)
\(= \frac{\sqrt{D}}{a^2}\left( \frac{D}{3} + c - \frac{b^2}{2a} + \frac{ac}{a} \right)\)
\(= \frac{\sqrt{D}}{a^2}\left( \frac{2D + 6ac - 3b^2 + 6ac}{6a} \right)\)
\(= \frac{\sqrt{D}}{a^2}\left( \frac{2D - 3(b^2 - 4ac)}{6a} \right)\)
Since \(D = b^2 - 4ac\):
\(= \frac{\sqrt{D}}{a^2}\left( \frac{2D - 3D}{6a} \right) = \frac{\sqrt{D}}{a^2}\left( -\frac{D}{6a} \right)\)
Since area is always positive:
\(\text{Area} = \frac{D\sqrt{D}}{6a^2}\)
Proof 2: Intersection Points Formula
If two curves intersect at \(x_1\) and \(x_2\), and the combined equation is \(ax^2 + bx + c = 0\):
Same conditions as the discriminant formula.
We'll prove this using the discriminant formula and properties of roots.
From root properties: \(x_2 - x_1 = \frac{\sqrt{D}}{a}\)
Therefore: \(\sqrt{D} = a(x_2 - x_1)\) and \(D = a^2(x_2 - x_1)^2\)
Using the discriminant formula:
\(\text{Area} = \frac{D\sqrt{D}}{6a^2} = \frac{a^2(x_2 - x_1)^2 \cdot a(x_2 - x_1)}{6a^2}\)
Simplify: \(= \frac{a(x_2 - x_1)^3}{6}\)
Since area is always positive: \(= \frac{a}{6}|x_1 - x_2|^3\)
Proof complete. This shows the connection between the two formulas.
Proof 3: Rectangle Ratio Formula
For quadratic functions (parabolas) where the region's sides pass through the vertex:
Condition: The "fatter" part inside the curve has 2:1 area ratio with the "thinner" part outside.
We'll use the simplest quadratic function \(y = ax^2\). Other quadratics \(y = ax^2 + bx + c\) are just translations of this basic form and don't change area ratios.
Consider the parabola \(y = ax^2\) and the horizontal line \(y = ak^2\) (where \(k > 0\)):
Area of larger part (A):
Bounded by \(y = ak^2\) (top) and \(y = ax^2\) (bottom) from 0 to \(k\):
\(\text{Area A} = \int_0^k (ak^2 - ax^2) dx\)
\(= \left[ ak^2x - \frac{a}{3}x^3 \right]_0^k\)
\(= \left( ak^3 - \frac{a}{3}k^3 \right) - 0\)
\(= \frac{2}{3}ak^3\)
Area of smaller part (B):
Bounded by \(y = ax^2\) and x-axis from 0 to \(k\):
\(\text{Area B} = \int_0^k ax^2 dx\)
\(= \left[ \frac{a}{3}x^3 \right]_0^k\)
\(= \frac{a}{3}k^3\)
Area ratio:
\(\frac{\text{Area A}}{\text{Area B}} = \frac{\frac{2}{3}ak^3}{\frac{1}{3}ak^3} = \frac{2}{1}\)
So Area A : Area B = 2 : 1
Total rectangle area:
The rectangle formed has width \(k\) and height \(ak^2\):
Rectangle area = \(k \times ak^2 = ak^3\)
Relation to rectangle area:
Area A = \(\frac{2}{3}ak^3 = \frac{2}{3} \times \text{rectangle area}\)
Area B = \(\frac{1}{3}ak^3 = \frac{1}{3} \times \text{rectangle area}\)
Key insight: The ratio 2:1 between the "inside" and "outside" areas of a parabola relative to a horizontal line through a point on the parabola is constant and independent of the specific quadratic coefficients (as long as we're measuring from the vertex).
Summary of Proofs
Proof Strategy: Start with general quadratic forms, find intersection points, use root properties and integration.
Key Step: \(x_2 - x_1 = \frac{\sqrt{D}}{a}\) connects geometry to algebra.
Proof Strategy: Derived directly from Formula 1 using root distance.
Key Insight: Area depends on cube of distance between intersections.
Proof Strategy: Use simplest parabola \(y = ax^2\), calculate areas with integration.
Key Result: Constant 2:1 ratio for areas inside/outside parabola relative to horizontal line.
1. These proofs show why the quick formulas work, but remember they have limited applicability.
2. The fundamental integration method works for all cases, while these formulas work only for specific situations.
3. Understanding these proofs helps develop mathematical intuition and connects algebra, geometry, and calculus.
These proofs beautifully connect:
- Algebra: Discriminant and root properties of quadratics
- Geometry: Area measurements and spatial relationships
- Calculus: Integration as a tool for area calculation
- Analytic Geometry: Equations of curves and their intersections
This demonstrates the power of mathematics as an interconnected discipline.