Area Between Two Curves: Class 12 Integration Methods Explained

Finding the area under a single curve is a warm-up. The real test in Application of Integrals is finding the area between two curves — a region enclosed above by one curve and below by another. This is one of the highest-value topics in the chapter for CBSE board exams, and this guide breaks it down into a method you can apply to almost any pair of curves.

What Does “Area Between Two Curves” Mean?

When two curves intersect, they enclose a bounded region between their points of intersection. Rather than finding the area under each curve separately and subtracting, there’s a more direct approach: integrate the difference between the upper curve and the lower curve across the shared interval.

The Step-by-Step Method

Step 1: Find the Points of Intersection

Set the two curve equations equal to each other and solve. These intersection points become your limits of integration — this step is almost always the starting point of the entire problem.

Step 2: Sketch Both Curves

Draw a rough sketch showing both curves and shade the enclosed region. This tells you immediately which curve is “on top” (has the larger y-value) across the interval — a detail you cannot afford to get wrong.

Step 3: Set Up the Integral

If f(x) is the upper curve and g(x) is the lower curve between x = a and x = b:

Area = ∫ₐᵇ [f(x) − g(x)] dx

If the upper and lower curves swap partway through the interval, split the integral at the point where they cross and handle each piece separately.

Step 4: Integrate and Simplify

Evaluate the integral using the Fundamental Theorem of Calculus, substituting the intersection points found in Step 1.

Worked Example 1: Parabola and a Line

Find the area enclosed between the parabola y² = 4x and the line y = x.

Step 1: Express both in terms of y, since the parabola is naturally x = y²/4. The line gives x = y. Setting them equal:

y²/4 = y → y² − 4y = 0 → y = 0 or y = 4

Step 2: Between y = 0 and y = 4, the line x = y lies to the right of the parabola x = y²/4 (check with a test point, such as y = 2: line gives x = 2, parabola gives x = 1).

Step 3: Set up the integral using horizontal strips:

Area = ∫₀⁴ [y − y²/4] dy = [y²/2 − y³/12]₀⁴ = (8 − 64/12) = 8 − 16/3 = 8/3 square units

Worked Example 2: Two Parabolas

Find the area enclosed between y = 6x − x² and y = x² − 2x.

Step 1: Set the equations equal: 6x − x² = x² − 2x → 8x = 2x² → x² − 4x = 0 → x = 0 or x = 4

Step 2: Test x = 2: first curve gives y = 8, second gives y = 0, so the first curve is on top throughout the interval.

Step 3: Area = ∫₀⁴ [(6x − x²) − (x² − 2x)] dx = ∫₀⁴ (8x − 2x²) dx = [4x² − 2x³/3]₀⁴ = 64 − 128/3 = 64/3 square units

Worked Example 3: Parabola and Latus Rectum

Find the area bounded by the parabola y² = 4x and its latus rectum x = 1.

Since the parabola is symmetric about the x-axis, find the area of the upper half and double it:

Area = 2∫₀¹ √(4x) dx = 2 × 2∫₀¹ √x dx = 4 × [(2/3)x^(3/2)]₀¹ = 4 × 2/3 = 8/3 square units

This “8a²/3” pattern for a parabola y² = 4ax bounded by its latus rectum (x = a) is worth memorizing as a shortcut, since it appears often in board papers.

Common Mistakes to Avoid

The single biggest error is skipping the sketch and guessing which curve is on top — always verify with a test point inside the interval rather than assuming. A close second is forgetting that the upper and lower curves can swap positions partway through the interval, which requires splitting the integral into two parts. Finally, students sometimes integrate f(x) − g(x) using the wrong variable (x instead of y, or vice versa) when the curves are more naturally expressed the other way — check which variable makes the algebra simplest before starting.

Related Reading

• Area of an Ellipse Using Integration: Formula, Derivation & Solved Examples

• How to Find Area Under a Simple Curve Using Integration – Class 12 Maths

Ready to Practice?

Area-between-curves questions carry some of the highest marks in this chapter — and the highest risk of sign errors. Get instant doubt resolution and topic-wise practice on ChampionsPrep for just ₹10 per use. Practice on ChampionsPrep →

Test Your Knowledge

Q1.The area enclosed between y² = 4x and the line y = x is:

8/3 square units16/3 square units4 square units8 square units

Q2.The first step in finding the area between two curves is:

Integrating the difference immediatelySketching the curves and finding intersectionsChoosing vertical strips alwaysDoubling the area under the upper curve at once

Q3.Between y = 0 and y = 4 for y² = 4x and y = x, horizontal strips are used because:

Vertical strips are not allowed by CBSE in any questionThe curves are naturally expressed as x in terms of yThe region is unboundedIntegration in x is impossible

Q4.The area bounded by y = 6x − x² and y = x² − 2x is:

64/3 square units32/3 square units16 square units64 square units

Q5.For a parabola y² = 4ax bounded by its latus rectum x = a, the enclosed area is:

4a²/32a²/38a²/316a²/3