Navigation: Wiki Home > Skills > Area Between Curves (Vertical Rectangles)
| 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
Test yourself on these prerequisites:
1. Can you evaluate a definite integral?
Compute $\int_0^2 (3x^2 - 1) \, dx$
$$\left[x^3 - x\right]_0^2 = (8 - 2) - 0 = 6$$
If this was difficult, review Definite Integrals first.
2. Can you find antiderivatives?
Find $\int (x^3 + 2x) \, dx$
$$\frac{x^4}{4} + x^2 + C$$
If this was difficult, review Antiderivatives first.
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.
Legend: 🟡 Yellow = immediate prerequisites (must master) | 🟢 Green = this skill
| Property | Value |
|---|---|
| Section | Stewart §5.1 |
| Difficulty | Beginner |
| Time | ~20 minutes |
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).
y
| f(x) = top curve
| ╭─────────╮
| ╱ AREA ╲
| ╱ A ╲ A typical rectangle:
| ╱──────────────╲ height = f(x) - g(x)
| ╱ g(x) = bottom width = dx
| ╱ curve ╲
|╱ ╲
└─────────────────────── x
a bThink of slicing the region into thin vertical rectangles:
| 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: $(x^2+1)-(x-1) = x^2-x+2$ |
After computing your answer, verify:
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 $x = 0.5$:
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 $\frac{2}{3}(1)(0.5) \approx 0.33 = \frac{1}{3}$ ✓
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 $y = x^2$.
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.
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?
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?
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.
✅ All boxes checked? You've mastered vertical slicing! Move on to horizontal slicing or curves that cross.
Before you write anything:
Setting up:
Computing:
Verification (crucial for partial credit):
Time tip: Intersection algebra can eat time. If stuck, circle it and move on.
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.
Looking back:
Looking ahead:
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.
| 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