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Lesson 1: Area Between Curves

Estimated time: 40-50 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Set up an integral for the area between two curves
Determine which curve is the "top" curve on a given interval
Handle regions where the curves cross by splitting into sub-intervals
Integrate with respect to y when it simplifies the problem

The Basic Area Formula

In Module 6 you learned that ∫_a^b f(x) dx computes the signed area between f and the x-axis. Now we generalize: the area between two curves f(x) and g(x) on [a, b] where f(x) ≥ g(x) is given by:

Area Between Two Curves (integrating with respect to x)

A = ∫_a^b [f(x) − g(x)] dx

where f(x) ≥ g(x) on [a, b]. Here f is the "top" curve and g is the "bottom" curve.

Key idea: At each x, the height of a thin vertical strip is f(x) − g(x). We integrate these heights across [a, b].

Finding the Intersection Points

The limits of integration a and b are usually the x-values where the curves intersect. To find them, set f(x) = g(x) and solve.

Example 1: Area Between a Parabola and a Line

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

Step 1: Find intersections. x² = x + 2 gives x² − x − 2 = 0, so (x − 2)(x + 1) = 0. Intersections at x = −1, x = 2.

Step 2: Determine top curve. Test x = 0: line gives 2, parabola gives 0. So x + 2 ≥ x² on [−1, 2].

Step 3: Integrate.

A = ∫₋₁² [(x + 2) − x²] dx = ∫₋₁² (x + 2 − x²) dx

= [x²/2 + 2x − x³/3]₋₁² = (2 + 4 − 8/3) − (1/2 − 2 + 1/3)

= 10/3 − (−7/6) = 10/3 + 7/6 = 27/6 = 9/2.

When Curves Cross: Splitting the Integral

If the curves switch positions (one is on top on part of the interval, the other is on top for the rest), you must split the integral at each crossing point.

Example 2: Curves That Cross

Find the total area between y = x³ and y = x on [−1, 1].

Step 1: x³ = x gives x(x² − 1) = 0, so x = −1, 0, 1.

Step 2: On (−1, 0): test x = −0.5. x³ = −0.125, x = −0.5. So x³ > x (x³ is above).

On (0, 1): test x = 0.5. x³ = 0.125, x = 0.5. So x > x³ (x is above).

Step 3:

A = ∫₋₁⁰ (x³ − x) dx + ∫₀¹ (x − x³) dx

= [x⁴/4 − x²/2]₋₁⁰ + [x²/2 − x⁴/4]₀¹

= (0 − (1/4 − 1/2)) + ((1/2 − 1/4) − 0) = 1/4 + 1/4 = 1/2.

Integrating with Respect to y

Sometimes horizontal strips are simpler. If we express curves as x = f(y) and x = g(y), we integrate with respect to y:

Area Between Two Curves (integrating with respect to y)

A = ∫_c^d [f(y) − g(y)] dy

where f(y) ≥ g(y) on [c, d]. Here f is the "right" curve and g is the "left" curve.

Example 3: Integrating with Respect to y

Find the area enclosed by x = y² and x = y + 2.

Step 1: Find intersections: y² = y + 2 gives y² − y − 2 = 0, so (y − 2)(y + 1) = 0. y = −1, y = 2.

Step 2: Which is rightmost? At y = 0: y + 2 = 2, y² = 0. So x = y + 2 is to the right.

Step 3:

A = ∫₋₁² [(y + 2) − y²] dy = [y²/2 + 2y − y³/3]₋₁²

= (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 10/3 + 7/6 = 9/2.

Strategy: When to Use dy vs. dx

Use dx when the region is naturally bounded by top/bottom functions of x, and neither curve requires splitting into multiple functions of x.
Use dy when the region is more naturally described by left/right functions of y, or when using dx would require splitting into multiple integrals but dy gives a single integral.

Example 4: Choosing the Better Variable

Find the area enclosed by y = √x, y = 0, and x = 4.

Using dx: A = ∫₀⁴ √x dx = [(2/3)x^3/2]₀⁴ = (2/3)(8) = 16/3.

This is straightforward with dx since the top curve is y = √x and the bottom is y = 0.

Area Between Three or More Curves

When a region is bounded by three or more curves, sketch the region first, then identify which curves form the top and bottom (or left and right) boundaries. You may need to split into sub-regions.

Example 5: Region with Multiple Boundaries

Find the area enclosed by y = x², y = 2 − x, and y = 0 in the first quadrant.

Step 1: Find intersections. x² = 2 − x gives x² + x − 2 = 0, so x = 1 (positive root). Also, 2 − x = 0 at x = 2, and x² = 0 at x = 0.

Step 2: The region splits at x = 1. On [0, 1]: top is y = 2 − x, bottom is y = x². On [1, 2]: top is y = 2 − x, bottom is y = 0.

Step 3:

A = ∫₀¹ [(2 − x) − x²] dx + ∫₁² (2 − x) dx

= [2x − x²/2 − x³/3]₀¹ + [2x − x²/2]₁²

= (2 − 1/2 − 1/3) + [(4 − 2) − (2 − 1/2)] = 7/6 + 1/2 = 5/3.

Key Takeaways

Area = ∫ (top − bottom) dx or ∫ (right − left) dy.
Always find intersection points to determine limits of integration.
If curves cross within the interval, split the integral at each crossing point.
Choose dx or dy based on which gives a simpler single integral.
The integrand must always be non-negative: (top − bottom) ≥ 0.

Check Your Understanding

1. Find the area between y = x² and y = 4.

Answer: Intersections: x² = 4, x = ±2. A = ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = 16 − 16/3 = 32/3.

2. Find the area between y = sin x and y = cos x on [0, π/2].

Answer: They cross at x = π/4. On [0, π/4]: cos x ≥ sin x. On [π/4, π/2]: sin x ≥ cos x. A = ∫₀^π/4 (cos x − sin x) dx + ∫_π/4^π/2 (sin x − cos x) dx = [sin x + cos x]₀^π/4 + [−cos x − sin x]_π/4^π/2 = (√2 − 1) + ((−1) − (−√2)) = 2√2 − 2.

3. Set up (but do not evaluate) the integral for the area between x = y² − 1 and x = 3 − y².

Answer: y² − 1 = 3 − y² gives 2y² = 4, y = ±√2. Right curve: 3 − y². Left curve: y² − 1. A = ∫_−√2^√2 [(3 − y²) − (y² − 1)] dy = ∫_−√2^√2 (4 − 2y²) dy.

4. Find the area enclosed by y = x³ and y = x on [0, 1].

Answer: On [0, 1], x ≥ x³. A = ∫₀¹ (x − x³) dx = [x²/2 − x⁴/4]₀¹ = 1/2 − 1/4 = 1/4.

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