Area Between Curves (Vertical Rectangles)
Quick Reference: Essential Formulas
| Formula | When to Use | Memory Aid |
|---|---|---|
| $A = \int_a^b [f(x) - g(x)]\,dx$ | $f(x) \geq g(x)$ on $[a,b]$ | (top − bottom) from left to right |
| $A = \int_a^b \lvert f(x) - g(x) \rvert\,dx$ | Curves may cross | Split at crossings, sum absolute values |
The procedure: Sketch → Find bounds → Identify top/bottom → Integrate → Verify
Before You Start
Test yourself on these prerequisites:
1. Can you evaluate a definite integral?
Compute $\int_0^2 (3x^2 - 1) \, dx$
Check Your Answer
$$\left[x^3 - x\right]_0^2 = (8 - 2) - 0 = 6$$
If this was difficult, review Definite Integrals first.
2. Can you find antiderivatives?
Find $\int (x^3 + 2x) \, dx$
Check Your Answer
$$\frac{x^4}{4} + x^2 + C$$
If this was difficult, review Antiderivatives first.
Why Does This Work?
You already know how to find the area under a single curve using integration. But what if you want to find the area between two curves? Think about it: if you find the area under the top curve and subtract the area under the bottom curve, you get the area of the region enclosed between them.
This is exactly what we do, but instead of computing two separate integrals and subtracting, we combine them into a single integral of the difference.
Prerequisite Map
Legend: 🟡 Yellow = immediate prerequisites (must master) | 🟢 Green = this skill
Quick Reference
| Property | Value |
|---|---|
| Section | Stewart §5.1 |
| Difficulty | Beginner |
| Time | ~20 minutes |
Key Concepts
The Main Formula
If $f(x) \geq g(x)$ for all $x$ in $[a, b]$, then the area between the curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is:
$$\boxed{A = \int_a^b \bigl[f(x) - g(x)\bigr] \, dx}$$
In words: integrate (top minus bottom).
Visualizing the Process
y
| f(x) = top curve
| ╭─────────╮
| ╱ AREA ╲
| ╱ A ╲ A typical rectangle:
| ╱──────────────╲ height = f(x) - g(x)
| ╱ g(x) = bottom width = dx
| ╱ curve ╲
|╱ ╲
└─────────────────────── x
a bStep-by-Step Procedure
- Sketch the region: identify which curve is on top and which is on bottom
- Find intersection points (if not given): solve $f(x) = g(x)$ to get the limits $a$ and $b$
- Set up the integral: $A = \int_a^b (\text{top} - \text{bottom}) \, dx$
- Evaluate: find the antiderivative and apply the Fundamental Theorem
Why “Top Minus Bottom”?
Think of slicing the region into thin vertical rectangles:
- Each rectangle has width $\Delta x$
- Each rectangle has height $(y_{\text{top}} - y_{\text{bottom}}) = f(x) - g(x)$
- Area of each rectangle: $(f(x) - g(x)) \Delta x$
- Total area: sum all rectangles → $\int_a^b [f(x) - g(x)] \, dx$
Common Pitfalls Table
| Mistake | Consequence | Prevention |
|---|---|---|
| Subtracting bottom from top incorrectly | Negative area | Test a point: which $y$-value is larger? |
| Forgetting to find intersections | Wrong bounds | If no bounds given, ALWAYS solve $f(x) = g(x)$ first |
| Sign errors when distributing negatives | Wrong antiderivative | Write $(f(x)) - (g(x))$ with parentheses |
| Using wrong bounds order | Negative or wrong area | Integrate left → right (smaller $x$ to larger $x$) |
| Not simplifying before integrating | Harder integral | Combine like terms: $(x^2+1)-(x-1) = x^2-x+2$ |
💡 Quick Sanity Checks
After computing your answer, verify:
- Positive? Area must be $> 0$. Negative = wrong subtraction order.
- Reasonable size? Region from $x=0$ to $x=2$ with height ~3 should give area ~6, not 600.
- Units match? If problem has units, area should be (length)².
Worked Example
Problem: Find the area enclosed by $y = x^2$ and $y = 2x - x^2$.
Solution:
Step 1: Find where the curves intersect.
Set $x^2 = 2x - x^2$: $$2x^2 - 2x = 0$$ $$2x(x - 1) = 0$$ $$x = 0 \text{ or } x = 1$$
Step 2: Determine which curve is on top.
At $x = 0.5$:
- Top: $y = 2(0.5) - (0.5)^2 = 1 - 0.25 = 0.75$
- Bottom: $y = (0.5)^2 = 0.25$
So $y = 2x - x^2$ is on top.
Step 3: Set up and evaluate the integral.
$$A = \int_0^1 \bigl[(2x - x^2) - x^2\bigr] \, dx = \int_0^1 (2x - 2x^2) \, dx$$
$$= \left[x^2 - \frac{2x^3}{3}\right]_0^1 = \left(1 - \frac{2}{3}\right) - 0 = \frac{1}{3}$$
Step 4: Verify the answer.
Reasonableness check: The region is a “lens” shape between two parabolas on $[0,1]$. Maximum height is at $x = 0.5$: $(0.75 - 0.25) = 0.5$. A lens with width 1 and max height 0.5 has area roughly $\frac{2}{3}(1)(0.5) \approx 0.33 = \frac{1}{3}$ ✓
Practice Problems
Find the area of the region bounded above by $y = 6 - x$, below by $y = 2$, and on the sides by $x = 0$ and $x = 3$.
Find the area enclosed by $y = x + 2$ and $y = x^2$.
🔄 Still confused about basic area setup?
- Can’t identify top/bottom? → Pick any $x$ between bounds, compute both $y$-values
- Struggling with intersections? → Review Finding Intersection Points
- Integration errors? → Review Definite Integrals
- Not sure why we subtract? → Think of it as (area under top) − (area under bottom)
Find the area of the region bounded by $y = x^3 - x$ and $y = 3x$ in the first quadrant.
Find the area of the region bounded by the parabola $y = x^2$, the tangent line to this parabola at the point $(2, 4)$, and the $y$-axis.
For what value of $c > 0$ does the area of the region bounded by the parabolas $y = x^2 - c^2$ and $y = c^2 - x^2$ equal $72$?
✅ Checkpoint: If you can solve Level 5 problems involving parameters, you have strong mastery of this skill. You’re ready for Curves That Cross and Horizontal Integration.
CCI-Style Conceptual Questions
Question 1: The area between $y = f(x)$ and $y = g(x)$ from $x = 1$ to $x = 4$ equals $12$. The area under $y = f(x)$ from $x = 1$ to $x = 4$ is $20$. What is the area under $y = g(x)$ from $x = 1$ to $x = 4$, assuming $f(x) \geq g(x)$ on this interval?
Answer
Since $\int_1^4 [f(x) - g(x)] \, dx = 12$ and $\int_1^4 f(x) \, dx = 20$, we have:
$$\int_1^4 g(x) \, dx = 20 - 12 = 8$$
Question 2: If you accidentally compute $\int_a^b [g(x) - f(x)] \, dx$ instead of $\int_a^b [f(x) - g(x)] \, dx$ when $f(x) > g(x)$, what happens to your answer?
Answer
You get the negative of the correct area. The area formula requires (top $-$ bottom) to ensure a positive result. Swapping the order just negates the integral.
Common Misconceptions
the area between two curves is found by integrating each curve separately and then subtracting the results.
This is the concept-image-conflicts-definition error. The formula $\int_a^b [f(x) - g(x)]\,dx$ computes both in a single integral, but the subtraction must happen inside the integral sign, not outside it. For example, integrating $y = x+2$ and $y = x^2$ separately over $[-1, 2]$ and subtracting gives the correct answer only when one area happens to cancel correctly, which is a coincidence of setup. If either curve dips below the $x$-axis, computing each area separately and subtracting can yield a wrong result because the individual integrals carry sign information.
Mastery Checklist
✅ All boxes checked? You’ve mastered vertical slicing! Move on to horizontal slicing or curves that cross.
Exam Strategy Tips
🎯 How an expert approaches area problems on exams
Before you write anything:
- Read the problem twice. Are bounds given? If not, you need intersections.
- Quick sketch (30 seconds max). Label which curve is which.
Setting up:
- Test ONE point between bounds to confirm top/bottom.
- Write the integral with explicit parentheses: $\int_a^b \bigl[(f(x)) - (g(x))\bigr]\,dx$
Computing:
- Simplify the integrand BEFORE finding the antiderivative.
- Double-check signs when evaluating at bounds.
Verification (crucial for partial credit):
- Is the answer positive? If negative, you subtracted wrong.
- Quick estimate: width × average height ≈ area.
- If unsure, show your setup clearly: partial credit!
Time tip: Intersection algebra can eat time. If stuck, circle it and move on.
Mental Model
The Layered Cake Principle:
Imagine the region as a slice of layered cake between two frosting layers. The area of the filling is found by taking the top layer minus the bottom layer. Each thin vertical strip has height (top $-$ bottom) and thickness $dx$. Sum up all the strips from left to right.
Connections
Looking back:
- This extends the definite integral $\int_a^b f(x) \, dx$ (area under one curve) to area between two curves
- Intersection points come from solving equations, an algebra skill applied in a calculus context
Looking ahead:
- Area Between Curves (Horizontal): sometimes slicing horizontally is easier
- Curves That Cross: what if the curves switch positions?
- Volumes by Disks/Washers: rotating these regions creates 3D solids
📚 Historical Note
The “slice into rectangles” approach dates back to Archimedes (287-212 BCE), who computed the area of a parabolic segment using triangular approximations. The modern integral notation $\int$ was introduced by Leibniz in 1675: it’s an elongated “S” for “summa” (sum), reflecting the idea of summing infinitely many infinitesimal rectangles.
The area-between-curves formula is a natural extension: instead of summing rectangles from the $x$-axis to one curve, we sum rectangles between two curves. This simple shift in perspective is what makes calculus so powerful: the same technique generalizes to many new problems.
Summary
| Concept | Key Point |
|---|---|
| Formula | $A = \int_a^b [f(x) - g(x)]\,dx$ |
| Direction | Always (top) − (bottom), integrate left → right |
| Finding bounds | Solve $f(x) = g(x)$ if not given |
| Verification | Answer must be positive; estimate width × height |
| When to use | When top/bottom are clear single functions of $x$ |
| When NOT to use | When boundaries change mid-region (try $dy$ instead) |
| Previous | Up | Next |
|---|---|---|
| Section 5.1 | Section 5.1 | Area (Horizontal) |
Last updated: 2026-01-22