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Unit 8.4 – Finding the Area Between Curves Expressed as Functions of x

AP® Calculus AB & BC | Integration to Find Areas

Why This Matters: Area between curves is one of the most important geometric applications of integration! While single integrals give area under one curve, this topic finds the area between TWO curves. The key concept: integrate the difference of the functions. This appears on virtually every AP® exam and has applications in economics, physics, engineering, and more. Master this and you've conquered a fundamental calculus skill!

📐 The Fundamental Formula

Area Between Two Curves

THE FORMULA:
\[ A = \int_a^b [\text{top} - \text{bottom}] \, dx \]

Or more formally:

\[ A = \int_a^b [f(x) - g(x)] \, dx \]
Where:
  • \(f(x)\) = upper (top) function
  • \(g(x)\) = lower (bottom) function
  • \([a, b]\) = interval where curves bound the region
  • \(f(x) \geq g(x)\) on \([a, b]\)

📝 Key Insight: Always subtract the lower function from the upper function. The order matters! Think: "Top minus Bottom" or "Upper minus Lower"

💡 Why the Formula Works

Geometric Understanding:

Area between curves = (Area under top curve) - (Area under bottom curve)

\[ A = \int_a^b f(x)\,dx - \int_a^b g(x)\,dx = \int_a^b [f(x) - g(x)]\,dx \]

By subtracting the integrals, we remove the overlap and keep only the region between the curves!

🔍 Finding Limits of Integration

Finding Intersection Points:

When curves intersect, they bound a region. To find intersection points:

Set the functions equal:

\[ f(x) = g(x) \]

Solve for \(x\) to find the boundaries \(a\) and \(b\)

📋 Step-by-Step Process

Complete Method

The 6-Step Approach:

  1. Sketch the curves: (if possible) to visualize the region
  2. Find intersection points: Solve \(f(x) = g(x)\) for \(a\) and \(b\)
  3. Determine which is on top: Test a point or graph to see which function is higher
  4. Set up integral: \(\int_a^b [\text{top} - \text{bottom}]\,dx\)
  5. Evaluate the integral: Find antiderivative and apply FTC
  6. State answer with units: Area = ___ square units

📖 Comprehensive Worked Examples

Example 1: Basic Area Between Curves

Problem: Find the area between \(f(x) = x^2\) and \(g(x) = x\) from \(x = 0\) to \(x = 1\).

Solution:

Step 1: Determine which is on top

Test \(x = 0.5\):

\(f(0.5) = 0.25\) and \(g(0.5) = 0.5\)

So \(g(x) = x\) is on top!

Step 2: Set up integral

\[ A = \int_0^1 [x - x^2] \, dx \]

Step 3: Evaluate

\[ A = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 \]
\[ = \left(\frac{1}{2} - \frac{1}{3}\right) - 0 = \frac{3-2}{6} = \frac{1}{6} \]

ANSWER: Area = \(\frac{1}{6}\) square units

Example 2: Finding Intersection Points

Problem: Find the area between \(y = x^2 - 4x\) and \(y = x - 4\).

Step 1: Find intersection points

\[ x^2 - 4x = x - 4 \]
\[ x^2 - 5x + 4 = 0 \]
\[ (x-1)(x-4) = 0 \]

Intersection points: \(x = 1\) and \(x = 4\)

Step 2: Determine which is on top

Test \(x = 2\):

\(x^2 - 4x = 4 - 8 = -4\)

\(x - 4 = 2 - 4 = -2\)

So \(y = x - 4\) is on top

Step 3: Set up and evaluate

\[ A = \int_1^4 [(x-4) - (x^2-4x)] \, dx \]
\[ = \int_1^4 (5x - x^2 - 4) \, dx \]
\[ = \left[\frac{5x^2}{2} - \frac{x^3}{3} - 4x\right]_1^4 \]
\[ = \left(40 - \frac{64}{3} - 16\right) - \left(\frac{5}{2} - \frac{1}{3} - 4\right) \]
\[ = \frac{9}{2} = 4.5 \]

ANSWER: Area = 4.5 square units

Example 3: Curves That Switch Positions

Problem: Find the area between \(y = \sin x\) and \(y = \cos x\) from \(x = 0\) to \(x = \pi\).

Step 1: Find where they intersect

\[ \sin x = \cos x \quad \Rightarrow \quad x = \frac{\pi}{4} \]

(in the interval \([0, \pi]\))

Step 2: Determine position on each interval

  • On \([0, \frac{\pi}{4}]\): \(\cos x > \sin x\)
  • On \([\frac{\pi}{4}, \pi]\): \(\sin x > \cos x\)

Must split into TWO integrals!

Step 3: Set up and evaluate

\[ A = \int_0^{\pi/4} (\cos x - \sin x)\,dx + \int_{\pi/4}^{\pi} (\sin x - \cos x)\,dx \]

After evaluation:

\[ A = 2\sqrt{2} \]

Example 4: Between a Curve and the x-axis

Problem: Find the area between \(y = x^2 - 4\) and the x-axis.

Analysis:

x-axis is \(y = 0\)

Intersection: \(x^2 - 4 = 0 \Rightarrow x = \pm 2\)

Between \(x = -2\) and \(x = 2\), the parabola is BELOW the x-axis

Setup:

\[ A = \int_{-2}^2 [0 - (x^2-4)] \, dx = \int_{-2}^2 (4 - x^2) \, dx \]
\[ = \left[4x - \frac{x^3}{3}\right]_{-2}^2 = \frac{32}{3} \]

🔑 Special Cases and Situations

Common Scenarios:

1. Curves Switch Positions:

Split the integral at intersection point(s)

\[ A = \int_a^c [f-g]\,dx + \int_c^b [g-f]\,dx \]
2. Multiple Regions:

Calculate each region separately and add

3. Area vs. Net Signed Area:

For area, always use \(|\text{top} - \text{bottom}|\) (absolute value)

💡 Essential Tips & Strategies

✅ Success Strategies:

  • Sketch first: A rough graph prevents major errors
  • Top minus bottom: ALWAYS this order
  • Find intersections: Set functions equal
  • Test a point: To determine which is on top
  • Check for switching: Do curves cross in the interval?
  • Area is positive: If you get negative, you switched top/bottom
  • Simplify before integrating: Combine like terms
  • Use symmetry: Can save calculation time

🔥 Quick Checks:

  • If answer is negative: You subtracted wrong way!
  • Vertical line test: Each x-value has one top, one bottom
  • Units: Area is always square units
  • Calculator friendly: Can often use numerical integration

❌ Common Mistakes to Avoid

  • Mistake 1: Subtracting in wrong order (bottom - top)
  • Mistake 2: Not finding intersection points
  • Mistake 3: Not checking which function is on top
  • Mistake 4: Missing a region when curves cross
  • Mistake 5: Forgetting to split integral when curves switch
  • Mistake 6: Integration errors (algebra mistakes)
  • Mistake 7: Wrong limits of integration
  • Mistake 8: Not simplifying before integrating
  • Mistake 9: Confusing area with net signed area
  • Mistake 10: Arithmetic errors in final calculation

📝 Practice Problems

Find the area between the curves:

  1. \(y = x^2\) and \(y = 2x\)
  2. \(y = x^2 - 2x\) and \(y = 0\) (x-axis)
  3. \(y = e^x\) and \(y = e^{-x}\) from \(x = 0\) to \(x = 1\)
  4. \(y = \sqrt{x}\) and \(y = x^2\)

Answers:

  1. \(\frac{4}{3}\) square units (intersect at \(x=0\) and \(x=2\))
  2. \(\frac{4}{3}\) square units
  3. \(e - \frac{1}{e} - 2\) square units
  4. \(\frac{1}{3}\) square units (intersect at \(x=0\) and \(x=1\))

✏️ AP® Exam Success Tips

What AP® Graders Look For:

  • Show intersection points: Solve \(f(x) = g(x)\)
  • Correct integral setup: \(\int_a^b [\text{top} - \text{bottom}]\,dx\)
  • Show which is on top: Test or state reasoning
  • For split regions: Show separate integrals
  • Show antiderivative: Before applying bounds
  • Show substitution: When evaluating at bounds
  • Simplify answer: Exact or decimal as requested
  • Include units: "square units" in context problems

💯 Exam Strategy:

  1. Sketch curves if time permits (helps avoid errors)
  2. Find intersection points first
  3. Determine which function is on top (test a point)
  4. Check if curves switch positions in interval
  5. Write integral setup clearly
  6. Show all integration work
  7. Check: Is answer positive? (area must be!)
  8. State answer with units

⚡ Quick Reference Guide

AREA BETWEEN CURVES ESSENTIALS

The Main Formula:

\[ A = \int_a^b [\text{top} - \text{bottom}] \, dx \]

Finding Bounds:

  • Solve \(f(x) = g(x)\) for intersection points
  • These are your limits \(a\) and \(b\)

Key Steps:

  1. Find intersection points (limits)
  2. Determine which is on top
  3. Set up: \(\int_a^b [\text{top} - \text{bottom}]\,dx\)
  4. Evaluate integral
  5. Answer must be positive!

Master Area Between Curves! The fundamental formula: Area = \(\int_a^b [\text{top} - \text{bottom}]\,dx\) where top function minus bottom function over the interval. To find limits \(a\) and \(b\), solve \(f(x) = g(x)\) for intersection points. Always determine which function is on top—test a point in the interval or analyze the functions. The area is the integral of the DIFFERENCE of the functions, which equals (area under top curve) - (area under bottom curve). If curves switch positions, SPLIT the integral at the crossing point. Area must be positive—if negative, you subtracted wrong way! Common setup: (1) find intersections, (2) determine top/bottom, (3) write \(\int_a^b [f(x)-g(x)]\,dx\), (4) evaluate, (5) state with units. Special cases: area between curve and x-axis (use \(y=0\) as bottom), multiple regions (sum areas), switching curves (split integral). This is a major AP® exam topic—appears every year! Practice until automatic! 🎯✨