8.1 Area Between Curves

8.1 Area Between Curves

Core CLEP Concept: In Chapter 6, the definite integral $\int_a^b f(x) , dx$ was established as the net signed area between a curve $y = f(x)$ and the $x$-axis. When finding the physical area enclosed between two distinct curves $y = f(x)$ and $y = g(x)$, the region is partitioned into infinitesimal approximating strips. On the CLEP Calculus exam, you must fluidly determine whether to integrate vertically with respect to $x$ ($dx$) or horizontally with respect to $y$ ($dy$), locate intersection bounds algebraically, and split integrals when bounding curves cross each other.

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1. The Riemann Sum Foundation & Approximating Strips

To compute the area $A$ of a planar region bounded above by $y = f(x)$ and below by $y = g(x)$ on an interval $[a, b]$ where $f(x) \ge g(x)$, we divide $[a, b]$ into $n$ subintervals of equal width $\Delta x = \frac{b - a}{n}$.

   y ^                                          
     |               f(x) [Top Curve]           
     |           .---*---.                      
     |          /|       |\                     
     |         / |       | \                    
     |        /  |       |  \                   
     |       /   |       |   \                  
     |      /    |       |    \                 
     |     /     |   h   |     \                
     |    /      |       |      \               
     |   /       |       |       \              
     |  .--------*-------.--------.  g(x) [Bottom Curve]
     |           |<-dx-> |                      
   --+-----------+-------+-------------> x      
     0           a       b                      

Each approximating vertical rectangle has:

• Width: $\Delta x$
• Height: $h(x_i^) = f(x_i^) - g(x_i^)$, representing the vertical distance between the upper and lower curves at a sample point $x_i^$.
• Area of single strip: $\Delta A_i = [f(x_i^) - g(x_i^)] , \Delta x$

Summing the areas of all $n$ strips yields the Riemann sum $\sum_{i=1}^n [f(x_i^) - g(x_i^)] , \Delta x$. Taking the limit as $n \to \infty$ ($|\Delta x| \to 0$) produces the exact definite integral:

$A = \lim_{n \to \infty} \sum_{i=1}^n [f(x_i^*) - g(x_i^*)] \, \Delta x = \int_a^b \left[ f(x) - g(x) \right] dx$

Because distance is always positive and $f(x) \ge g(x)$, the height $[f(x) - g(x)] \ge 0$, guaranteeing that the computed area $A$ is strictly non-negative.

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2. Vertical Slices: Integrating with Respect to $x$ ($dx$)

Formal Theorem: Area Between Curves ($dx$)

If $f$ and $g$ are continuous functions on $[a, b]$ and $f(x) \ge g(x)$ for all $x \in [a, b]$, then the area $A$ of the region bounded by the graphs of $y = f(x)$, $y = g(x)$, and the vertical lines $x = a$ and $x = b$ is:

$A = \int_a^b \left[ f(x) - g(x) \right] dx = \int_a^b \left( y_{\text{top}} - y_{\text{bottom}} \right) dx$

4-Step Protocol for Vertical Slicing

1. Find Intersection Points (Limits of Integration): Set $f(x) = g(x)$ and solve algebraically for $x$. The solutions $x = a$ and $x = b$ serve as lower and upper limits if boundary lines are not explicitly provided.
2. Identify Top vs. Bottom: On the interval $(a, b)$, evaluate $f(x_{\text{test}})$ and $g(x_{\text{test}})$ at any convenient test point $x_{\text{test}}$. The function with the larger value is $y_{\text{top}}$.
3. Set Up the Definite Integral: Form the difference $[y_{\text{top}} - y_{\text{bottom}}]$ inside the integral from $x = a$ to $x = b$.
4. Integrate and Evaluate: Find the antiderivative, apply the Fundamental Theorem of Calculus Part 2, and simplify.

Step-by-Step Worked Example 1: Enclosed Parabolic Region

Problem: Find the area of the region enclosed between the parabolas $y = x^2$ and $y = 2x - x^2$.

Step 1: Determine intersection boundaries. Set the two equations equal: $x^2 = 2x - x^2 \implies 2x^2 - 2x = 0 \implies 2x(x - 1) = 0$ $\text{Intersection points: } x = 0 \quad \text{and} \quad x = 1$

Step 2: Establish Top and Bottom curves on $(0, 1)$. Select test point $x = 0.5$:

• $y_1 = 2(0.5) - (0.5)^2 = 1 - 0.25 = 0.75$
• $y_2 = (0.5)^2 = 0.25$

Since $0.75 > 0.25$, the upper curve is $y_{\text{top}} = 2x - x^2$ and the lower curve is $y_{\text{bottom}} = x^2$.

Step 3: Construct the integral. $A = \int_0^1 \left[ (2x - x^2) - x^2 \right] dx = \int_0^1 (2x - 2x^2) \, dx$

Step 4: Antidifferentiate and evaluate. $A = \left[ x^2 - \frac{2}{3}x^3 \right]_0^1 = \left( (1)^2 - \frac{2}{3}(1)^3 \right) - (0 - 0) = 1 - \frac{2}{3} = \frac{1}{3}$

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3. Horizontal Slices: Integrating with Respect to $y$ ($dy$)

When to Integrate with Respect to $y$

On the CLEP exam, certain regions are either impossible to integrate as a single $dx$ integral or require awkward piecewise multi-region splits. Integrating with respect to $y$ is advantageous when:

1. Bounding curves are given as functions of $y$, such as sideways parabolas $x = y^2 - 4$.
2. The top or bottom boundary changes equation midway across the $x$-interval, whereas the left and right boundaries remain single, unbroken functions throughout the $y$-interval.

   y ^                                          
   d |----------.------------.  x = f(y) [Right Curve]
     |          |\          /                   
     |          | \        /                    
     |          |  \      /                     
     |          |   \    /                      
     |      dy  |====w===|                      
     |          |   /    \                      
     |          |  /      \                     
   c |----------.-'--------`  x = g(y) [Left Curve]
     |                                          
   --+---------------------------------> x      
     0                                          

Formal Theorem: Area Between Curves ($dy$)

If $f(y)$ and $g(y)$ are continuous on $[c, d]$ and $f(y) \ge g(y)$ for all $y \in [c, d]$, then the area $A$ of the region bounded by $x = f(y)$, $x = g(y)$, $y = c$, and $y = d$ is:

$A = \int_c^d \left[ f(y) - g(y) \right] dy = \int_c^d \left( x_{\text{right}} - x_{\text{left}} \right) dy$

Step-by-Step Worked Example 2: $dx$ Multi-Region Split vs. $dy$ Single Integral

Problem: Find the area of the region in the first quadrant bounded by $y = \sqrt{x}$, the line $y = x - 2$, and the $x$-axis ($y = 0$).

Comparison: Method A ($dx$) vs. Method B ($dy$)

Intersection Analysis:

• $y = \sqrt{x}$ intersects the $x$-axis at $(0, 0)$.
• $y = x - 2$ intersects the $x$-axis at $(2, 0)$.
• Curves intersect where $\sqrt{x} = x - 2 \implies x = (x - 2)^2 = x^2 - 4x + 4 \implies x^2 - 5x + 4 = 0 \implies (x - 4)(x - 1) = 0$. Since $\sqrt{1} \neq 1 - 2 = -1$, the extraneous root is $x = 1$; the true intersection is $(4, 2)$.

Method A: Vertical Slicing ($dx$) — Requires 2 Separate Integrals

Between $x = 0$ and $x = 2$, the bottom boundary is the $x$-axis ($y = 0$). Between $x = 2$ and $x = 4$, the bottom boundary switches to the line $y = x - 2$.

$A_1 = \int_0^2 (\sqrt{x} - 0) \, dx = \left[ \frac{2}{3}x^{3/2} \right]_0^2 = \frac{2}{3}(2\sqrt{2}) = \frac{4\sqrt{2}}{3}$

$A_2 = \int_2^4 \left[ \sqrt{x} - (x - 2) \right] dx = \left[ \frac{2}{3}x^{3/2} - \frac{x^2}{2} + 2x \right]_2^4$ $= \left( \frac{2}{3}(8) - 8 + 8 \right) - \left( \frac{4\sqrt{2}}{3} - 2 + 4 \right) = \frac{16}{3} - \frac{4\sqrt{2}}{3} - 2 = \frac{10}{3} - \frac{4\sqrt{2}}{3}$

$A_{\text{total}} = A_1 + A_2 = \frac{4\sqrt{2}}{3} + \left( \frac{10}{3} - \frac{4\sqrt{2}}{3} \right) = \frac{10}{3}$

Method B: Horizontal Slicing ($dy$) — Solved with 1 Single Integral

Convert both equations into functions of $y$:

• $y = \sqrt{x} \implies x_{\text{left}} = y^2$
• $y = x - 2 \implies x_{\text{right}} = y + 2$
• Vertical interval in $y$: from $y = 0$ ($x$-axis) to $y = 2$ (intersection point).

$A = \int_0^2 \left( x_{\text{right}} - x_{\text{left}} \right) dy = \int_0^2 \left[ (y + 2) - y^2 \right] dy$ $A = \left[ \frac{y^2}{2} + 2y - \frac{y^3}{3} \right]_0^2 = \left( \frac{4}{2} + 2(2) - \frac{8}{3} \right) - 0 = 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$

Strategic Takeaway: Integrating with respect to $y$ collapsed a complex, two-part radical split into a single basic polynomial integral, saving valuable exam time and eliminating algebra errors.

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4. Multiple Intersections & Sign Reversals (Crossing Curves)

When curves intersect at intermediate points $c \in (a, b)$, the relative position of the curves reverses (e.g., $f(x) \ge g(x)$ on $[a, c]$, but $g(x) \ge f(x)$ on $[c, b]$). Integrating $\int_a^b [f(x) - g(x)] , dx$ without splitting yields net signed cancellation rather than geometric area.

Total Geometric Area Formula

$A = \int_a^b |f(x) - g(x)| \, dx = \int_a^c [f(x) - g(x)] \, dx + \int_c^b [g(x) - f(x)] \, dx$

Step-by-Step Worked Example 3: Trigonometric Crossing

Problem: Determine the total area enclosed between $y = \sin x$ and $y = \cos x$ on the interval $[0, \pi/2]$.

Step 1: Find intermediate intersection. $\sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4} \in \left[0, \frac{\pi}{2}\right]$

Step 2: Partition the interval and check curve dominance.

• On $\left[0, \frac{\pi}{4}\right]$: $\cos(0) = 1 > \sin(0) = 0 \implies y_{\text{top}} = \cos x, , y_{\text{bottom}} = \sin x$.
• On $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$: $\sin(\pi/2) = 1 > \cos(\pi/2) = 0 \implies y_{\text{top}} = \sin x, , y_{\text{bottom}} = \cos x$.

Step 3: Evaluate each subregion integral. $A_1 = \int_0^{\pi/4} (\cos x - \sin x) \, dx = \left[ \sin x + \cos x \right]_0^{\pi/4} = \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right) - (0 + 1) = \sqrt{2} - 1$

$A_2 = \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx = \left[ -\cos x - \sin x \right]_{\pi/4}^{\pi/2} = (0 - 1) - \left( -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \right) = -1 + \sqrt{2} = \sqrt{2} - 1$

Step 4: Sum the subregions. $A_{\text{total}} = A_1 + A_2 = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2$

(Note: If evaluated blindly as $\int_0^{\pi/2} (\sin x - \cos x) , dx$, the result is $0$, which is geometrically false!)

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5. Strategic Decision Matrix: $dx$ vs. $dy$ Slicing

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6. Common CLEP Pitfalls & Traps

1. Inverting Subtraction Order: Integrating $\int (y_{\text{bottom}} - y_{\text{top}}) , dx$ produces a negative number. Geometric area must always be strictly positive. If your result is negative, you subtracted in the wrong direction.
2. Failing to Split at Intermediate Intersections: Integrating across a crossing point without splitting causes positive and negative regions to cancel out, yielding an incorrect net signed area.
3. Mixing Variables in the Integrand or Limits: When setting up a $dy$ integral, every variable in the integrand must be $y$, and the integration limits must be $y$-coordinates. Leaving $x$ terms inside a $dy$ integral is an immediate algebraic violation.
4. Extraneous Intersection Roots: When squaring both sides to solve equations involving radicals (e.g., $\sqrt{x} = x - 2$), always check candidate solutions in the original un-squared equation to discard false algebraic artifacts.
5. Omitting Differential Multipliers: Forgetting parentheses around multi-term expressions before $dx$ or $dy$ frequently leads to distributing errors when subtracting polynomials: $(f(x) - g(x)) \neq f(x) - g_1(x) + g_2(x)$.