# Area Between Curves: Top Minus Bottom, Step by Step

> Find the area between two curves step by step: top minus bottom, finding intersection points, curves that cross, and when to integrate with respect to y.

Source: https://practicecalculus.com/blog/area-between-curves
Updated: 2026-10-14

## Key takeaways

- The area between two curves is $\int_a^b \big[\text{top}(x) - \text{bottom}(x)\big]\,dx$.
- The limits $a$ and $b$ are usually the $x$-values where the curves intersect, found by setting them equal.
- If the curves cross inside the interval, split the integral at each crossing so every piece is top minus bottom.
- When the region is bounded on the left and right, integrate in $y$: $\int_c^d \big[\text{right}(y) - \text{left}(y)\big]\,dy$.

You know how to find the area under one curve. But what about the area trapped between a parabola and a line, or between $\sin x$ and $\cos x$? The idea is simple (subtract the bottom from the top), and the real work is in finding where the curves meet and keeping track of which one is on top.

## The formula for the area between two curves

If $f(x) \ge g(x)$ on $[a, b]$, the area of the region between them is

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

In words: **top minus bottom**, integrated from left to right.

Why does it work? Slice the region into thin vertical strips. Each strip has width $dx$ and height equal to the gap between the curves, $f(x) - g(x)$. Adding up all the strips is exactly what the integral does. It does not matter whether the curves are above or below the $x$-axis: the gap is the gap.

## How to find the area between curves in 5 steps

1. **Sketch** the curves, even roughly.
2. **Find the intersections** by setting the two functions equal and solving. These are usually your limits.
3. **Decide which curve is on top** using a test point between the intersections.
4. **Set up** $\int_a^b (\text{top} - \text{bottom})\,dx$.
5. **Integrate** and evaluate with the [fundamental theorem of calculus](/blog/fundamental-theorem-of-calculus).

## Worked example 1: a line and a parabola

Find the area between $y = x$ and $y = x^2$.

**Step 1. Intersections.** Set $x = x^2$, so $x^2 - x = 0$, $x(x - 1) = 0$, giving $x = 0$ and $x = 1$.

**Step 2. Which is on top?** Test $x = \frac{1}{2}$: the line gives $0.5$, the parabola gives $0.25$. The line is on top.

**Step 3. Set up and integrate.**

$$A = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$$

## Worked example 2: a downward parabola and a line

Find the area of the region bounded by $y = 4 - x^2$ and $y = x + 2$.

```graph
title: 'Region bounded by the parabola y = 4 − x² and the line y = x + 2, which meet at x = −2 and x = 1'
caption: 'The curves intersect at $(-2, 0)$ and $(1, 3)$. Between them the parabola is on top.'
x: [-3.5, 2.5]
y: [-2, 5]
functions:
  - expr: 4 - x^2
    label: y = 4 − x²
  - expr: x + 2
    label: y = x + 2
shade: {expr: 4 - x^2, below: x + 2, from: -2, to: 1}
points:
  - [-2, 0, '(−2, 0)']
  - [1, 3, '(1, 3)']
```

**Step 1. Intersections.** Set $4 - x^2 = x + 2$:

$$x^2 + x - 2 = 0 \quad\Rightarrow\quad (x + 2)(x - 1) = 0 \quad\Rightarrow\quad x = -2,\ x = 1$$

**Step 2. Which is on top?** Test $x = 0$: the parabola gives $4$, the line gives $2$. The parabola is on top.

**Step 3. Set up.** Top minus bottom simplifies first:

$$(4 - x^2) - (x + 2) = 2 - x - x^2$$

**Step 4. Integrate.**

$$A = \int_{-2}^{1} (2 - x - x^2)\,dx = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{1}$$

At $x = 1$: $2 - \frac{1}{2} - \frac{1}{3} = \frac{7}{6}$.

At $x = -2$: $-4 - 2 + \frac{8}{3} = -\frac{10}{3}$.

$$A = \frac{7}{6} - \left(-\frac{10}{3}\right) = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2}$$

The graph below shows the same idea in a different way. The area between two curves equals the area under their **difference** $f(x) - g(x)$, which here is the parabola $y = 2 - x - x^2$.

```graph
title: 'Area under the difference function y = 2 − x − x² from −2 to 1, equal to the area between the curves'
caption: 'Integrating top minus bottom is the same as finding the area under $y = 2 - x - x^2$. Both equal $\frac{9}{2}$.'
x: [-3, 2]
y: [-1.5, 3]
functions:
  - expr: 2 - x - x^2
    label: y = 2 − x − x²
shade: {expr: 2 - x - x^2, from: -2, to: 1}
points:
  - [-2, 0, '−2']
  - [1, 0, '1']
```

## Worked example 3: curves that cross

Find the area between $y = \sin x$ and $y = \cos x$ from $x = 0$ to $x = \frac{\pi}{2}$.

**Step 1. Intersections.** $\sin x = \cos x$ when $\tan x = 1$, so $x = \frac{\pi}{4}$ in this interval. The curves **cross** inside the interval.

**Step 2. Which is on top, piece by piece?**

- On $[0, \frac{\pi}{4}]$: at $x = 0$, $\cos 0 = 1 > \sin 0 = 0$. Cosine is on top.
- On $[\frac{\pi}{4}, \frac{\pi}{2}]$: at $x = \frac{\pi}{2}$, $\sin = 1 > \cos = 0$. Sine is on top.

**Step 3. Split and integrate.**

$$\int_0^{\pi/4} (\cos x - \sin x)\,dx = \Big[\sin x + \cos x\Big]_0^{\pi/4} = \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - (0 + 1) = \sqrt{2} - 1$$

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

**Step 4. Add.**

$$A = 2(\sqrt{2} - 1) = 2\sqrt{2} - 2 \approx 0.828$$

If you had integrated $\cos x - \sin x$ over the whole interval without splitting, the two pieces would cancel and you would get $0$, which is clearly not the area.

> **Tip:** a compact way to write this is $A = \int_a^b |f(x) - g(x)|\,dx$. The absolute value is a reminder to split wherever the curves cross.

## Integrating with respect to y

Some regions are awkward to slice vertically. If the region is bounded on the **left and right** by curves of the form $x = h(y)$, slice it horizontally instead:

$$A = \int_c^d \big[\text{right}(y) - \text{left}(y)\big]\,dy$$

where $c$ and $d$ are $y$-values.

### Worked example 4: a sideways parabola and a line

Find the area between $x = y^2$ and $y = x - 2$.

```graph
title: 'Region between the sideways parabola x = y² and the line y = x − 2'
caption: 'The upper half $y = \sqrt{x}$ and lower half $y = -\sqrt{x}$ make up the parabola $x = y^2$. It meets the line at $(1, -1)$ and $(4, 2)$.'
x: [-1, 5]
y: [-2.5, 3]
functions:
  - expr: sqrt(x)
    label: y = √x
  - expr: -sqrt(x)
    label: y = −√x
  - expr: x - 2
    label: y = x − 2
points:
  - [1, -1, '(1, −1)']
  - [4, 2, '(4, 2)']
```

**Step 1. Rewrite both as $x$ in terms of $y$.** The parabola is $x = y^2$ and the line is $x = y + 2$.

**Step 2. Intersections.** $y^2 = y + 2$ gives $y^2 - y - 2 = 0$, so $(y - 2)(y + 1) = 0$ and $y = -1$ or $y = 2$.

**Step 3. Which is on the right?** Test $y = 0$: the line gives $x = 2$, the parabola gives $x = 0$. The line is on the right.

**Step 4. Integrate.**

$$A = \int_{-1}^{2} \big[(y + 2) - y^2\big]\,dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^{2}$$

At $y = 2$: $2 + 4 - \frac{8}{3} = \frac{10}{3}$. At $y = -1$: $\frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6}$.

$$A = \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$$

In $x$, the bottom boundary changes at $x = 1$ (from $-\sqrt{x}$ to the line), so you would need two integrals: $\int_0^1 2\sqrt{x}\,dx + \int_1^4 (\sqrt{x} - x + 2)\,dx = \frac{4}{3} + \frac{19}{6} = \frac{9}{2}$. Same answer, twice the work.

## Quick reference: dx or dy?

| Situation                                           | Slice      | Formula                                     |
| --------------------------------------------------- | ---------- | ------------------------------------------- |
| One curve on top, one on the bottom                 | Vertical   | $\int_a^b (\text{top} - \text{bottom})\,dx$ |
| One curve on the right, one on the left             | Horizontal | $\int_c^d (\text{right} - \text{left})\,dy$ |
| Top or bottom changes partway across                | Try $dy$, or split the $dx$ integral | |
| Curves cross inside the interval                    | Either     | Split at each crossing                      |

## Common mistakes with area between curves

1. **Bottom minus top.** You get the right size with the wrong sign. If an area comes out negative, this is the first thing to check.
2. **Not splitting at crossings.** Pieces cancel and the answer is too small, sometimes zero.
3. **Using $x$-limits in a $dy$ integral.** When integrating in $y$, the limits must be $y$-values of the intersection points.
4. **Losing a sign when simplifying.** $(4 - x^2) - (x + 2)$ is $2 - x - x^2$. Distribute the minus sign over every term of the bottom curve.
5. **Guessing the limits.** The limits come from solving $f(x) = g(x)$, not from where the curves hit the axis, unless the problem says so.

## Practice finding areas between curves

These problems follow the same five steps every time, and the hard part is usually the algebra of finding intersections. Practice builds the habit of sketching first. The interactive lesson below checks each setup and each answer instantly, so you learn to catch a flipped top and bottom before it costs you. Next, the same slicing idea gives you [volumes of revolution](/blog/volumes-of-revolution).

## Frequently asked questions

### How do you find the area between two curves?

Find where the curves intersect to get the limits, decide which curve is on top, then integrate top minus bottom between the limits. The result is always positive if you subtracted in the right order.

### How do you know which curve is on top?

Pick any test point between the intersection points and plug it into both functions. The larger value is the top curve on that whole stretch, as long as the curves do not cross in between. A quick sketch also helps.

### What if the curves cross in the middle of the interval?

Split the integral at every crossing point. On each piece, use whichever curve is on top there. Adding the pieces gives the total area. Integrating without splitting lets the pieces cancel, which gives the wrong answer.

### When should you integrate with respect to y?

Integrate in $y$ when the region is easier to describe from left to right, for example when one boundary is a sideways parabola $x = y^2$, or when integrating in $x$ would need two or more separate integrals. Then use right minus left with $y$-limits.

### Can the area between curves be negative?

No. Area is always positive. If you get a negative number, you subtracted the curves in the wrong order or the curves cross inside your interval and you need to split the integral.