How to Solve: Area Between Curves – IIT JEE (Main + Advanced) Guide

How to Solve: Area Between Curves – IIT JEE (Main + Advanced) Guide

Introduction

Mastering Area Between Curves unlocks 10-15 marks in IIT JEE—enough to push you from a 90th to a 99th percentile rank. Whether it’s finding the area between a parabola and a line, or two intersecting curves, this topic appears every year in both Main and Advanced. Real-world applications? Calculating fuel efficiency in rockets, medical imaging, and even architecture—where engineers compute overlapping shapes.

What You Need To Know First

Before diving in, ensure you’re 100% clear on these:
1. Definite Integrals – How to compute the area under a single curve.
2. Graph Sketching – Quickly plotting functions to identify intersection points.
3. Absolute Value in Integrals – Why we take |f(x) - g(x)| and not just f(x) - g(x).

If any of these feel shaky, stop now and review them—this topic builds directly on them.

Key Vocabulary

Formulas To Know

1. Area Between Two Curves (Vertical Slicing)

Formula: A = ∫[a to b] |f(x) - g(x)| dx - f(x) = Upper curve (higher y-value in the region) - g(x) = Lower curve (lower y-value in the region) - a, b = x-coordinates of intersection points (or given bounds) - MEMORISE THIS – It’s the core formula for 90% of problems.

2. Area Between Two Curves (Horizontal Slicing)

Formula: A = ∫[c to d] |f(y) - g(y)| dy - f(y) = Right curve (higher x-value in the region) - g(y) = Left curve (lower x-value in the region) - c, d = y-coordinates of intersection points (or given bounds) - MEMORISE THIS – Used when curves are easier to express as x = f(y).

3. Area Between a Curve and the Y-Axis

Formula: A = ∫[c to d] f(y) dy - f(y) = Rightmost curve (distance from y-axis) - c, d = y-bounds - Given on exam sheet (but understand when to use it).

Step-by-Step Method

Step 1: Sketch the Graphs (Even Roughly!)

• Plot both functions on the same axes.
• Mark intersection points (solve f(x) = g(x)).
• Shade the region whose area you need.

Step 2: Identify Upper and Lower Curves

• For vertical slicing, find which curve is on top in the region.
• For horizontal slicing, find which curve is on the right.

Step 3: Find Bounds of Integration

• Default: Use x-coordinates of intersection points (a to b).
• If bounds are given, use those instead.

Step 4: Set Up the Integral

• Write A = ∫[a to b] |upper - lower| dx (or dy for horizontal slicing).
• Drop the absolute value if upper > lower in the entire region.

Step 5: Compute the Integral

• Integrate term by term.
• Plug in upper and lower bounds.
• Simplify.

Step 6: Check Units and Reasonableness

• Area cannot be negative—if your answer is negative, you swapped upper/lower.
• Compare with a quick estimate (e.g., area of a rectangle around the region).

Worked Examples

Example 1 – Basic (Vertical Slicing)

Problem: Find the area between y = x² and y = 2x from x = 0 to x = 2.

Step 1: Sketch

• y = x² (parabola opening upwards)
• y = 2x (straight line through origin)
• Intersection at x = 0 and x = 2.

Step 2: Identify Upper/Lower

• Between x = 0 and x = 2, 2x > x².
• Upper curve: y = 2x
• Lower curve: y = x²

Step 3: Bounds

• Given: x = 0 to x = 2.

Step 4: Set Up Integral

A = ∫[0 to 2] (2x - x²) dx

Step 5: Compute

= [x² - (x³)/3] from 0 to 2 = (4 - 8/3) - (0 - 0) = 4/3

Step 6: Check

• Area ≈ 4/3 ≈ 1.33 (reasonable for a small region).

What we did and why: - We sketched to confirm 2x is above x². - Used vertical slicing because the curves are y = f(x). - No absolute value needed since 2x > x² in the entire region.

Example 2 – Medium (Horizontal Slicing)

Problem: Find the area between x = y² and x = 4 - y².

Step 1: Sketch

• x = y² (right-opening parabola)
• x = 4 - y² (left-opening parabola)
• Intersection: y² = 4 - y² → 2y² = 4 → y = ±√2

Step 2: Identify Right/Left

• Between y = -√2 and y = √2, 4 - y² > y².
• Right curve: x = 4 - y²
• Left curve: x = y²

Step 3: Bounds

• y = -√2 to y = √2.

Step 4: Set Up Integral

A = ∫[-√2 to √2] [(4 - y²) - y²] dy = ∫[-√2 to √2] (4 - 2y²) dy

Step 5: Compute

= [4y - (2y³)/3] from -√2 to √2 = [4√2 - (2(2√2))/3] - [-4√2 - (2(-2√2))/3] = [4√2 - (4√2)/3] - [-4√2 + (4√2)/3] = (8√2)/3 + (8√2)/3 = (16√2)/3

Step 6: Check

• Area ≈ 7.54 (reasonable for a symmetric region).

What we did and why: - Used horizontal slicing because curves are x = f(y). - Symmetry let us compute from 0 to √2 and double it (but we didn’t need to here). - Absolute value not needed since 4 - y² > y² in the region.

Example 3 – Exam-Style (Disguised Problem)

Problem: Find the area of the region bounded by y = sin x, y = cos x, and the y-axis from x = 0 to x = π/4.

Step 1: Sketch

• y = sin x (starts at 0, peaks at π/2)
• y = cos x (starts at 1, decreases)
• Intersection: sin x = cos x → x = π/4
• Region: Between x = 0 and x = π/4, bounded by y-axis (x = 0).

Step 2: Identify Upper/Lower

• Between 0 and π/4, cos x > sin x.
• Upper curve: y = cos x
• Lower curve: y = sin x

Step 3: Bounds

• Given: x = 0 to x = π/4.

Step 4: Set Up Integral

A = ∫[0 to π/4] (cos x - sin x) dx

Step 5: Compute

= [sin x + cos x] from 0 to π/4 = (sin(π/4) + cos(π/4)) - (sin 0 + cos 0) = (√2/2 + √2/2) - (0 + 1) = √2 - 1

Step 6: Check

• Area ≈ 0.414 (small, as expected near the origin).

What we did and why: - Recognized the region between two trig functions. - Confirmed bounds by solving sin x = cos x. - No absolute value since cos x > sin x in the region.

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—this is your 60-second crash course for Area Between Curves.

1. Sketch the graphs. Always. Even a rough sketch tells you which curve is on top.
2. Find intersection points. Solve f(x) = g(x) to get your bounds.
3. Decide: vertical or horizontal slicing?
4. If curves are y = f(x), use ∫ |upper - lower| dx.
5. If curves are x = f(y), use ∫ |right - left| dy.
6. Set up the integral. Drop the absolute value only if one curve is always above the other.
7. Compute. Integrate term by term, plug in bounds, simplify.
8. Check for negatives. If your answer is negative, you swapped upper and lower.

Pro tip: If curves cross inside the region, split the integral at the intersection point.

You’ve got this. Now go crush that exam!