Area Between Curves

You already know that $\int_a^b f(x)\, dx$ gives the area under a curve (when $f \ge 0$). Now take two curves and ask for the area of the region between them. The idea is to slice the region into thin vertical strips, find the height of one strip, and let the integral add them all up. This “slice and add” thinking is also the starting point for volumes, later in this unit.

Key ideas

Top minus bottom

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

$$A = \int_a^b \big( f(x) - g(x) \big)\, dx = \int_a^b (\text{top} - \text{bottom})\, dx$$

Why: a thin vertical strip at position $x$ has height $f(x) - g(x)$ and width $dx$, so its area is about $\big(f(x) - g(x)\big)\,dx$. The integral adds up all the strips from $a$ to $b$.

This works even if one or both curves dip below the $x$-axis: “top minus bottom” is always the height of the strip.

A representative vertical slice: height $(x + 2) - x^2$, width $dx$.

The steps

1. Sketch both curves (or graph them on a calculator) so you can see the region.
2. Find the intersection points by setting the functions equal. These are usually the limits of integration.
3. Decide which curve is on top on each interval. Test a point if you’re not sure.
4. Integrate top minus bottom.

When the curves cross more than twice

If the curves cross inside the interval, the top and bottom curves swap. Then $\int (f - g)\, dx$ would subtract one piece from another instead of adding. Split the integral at every crossing point and use (top − bottom) on each piece:

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

One integral that does it all is $A = \displaystyle\int_a^b \lvert f(x) - g(x) \rvert\, dx$. That’s the easiest form on a calculator-active question; by hand, you still split at the crossings.

Top and bottom swap at $x = 0$, so the area must be split there.

On the AP exam, an area setup is worth points on its own. Write the integral with the correct limits and integrand before you compute, especially when a calculator does the rest.

Worked examples

Example 1: A line and a parabola

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

Solution. Intersections: $x^2 = x + 2$, so $x^2 - x - 2 = (x - 2)(x + 1) = 0$, and $x = -1$ or $x = 2$.

Test $x = 0$: the line gives $2$ and the parabola gives $0$, so the line is on top.

$$\begin{aligned} A &= \int_{-1}^{2} \big( (x + 2) - x^2 \big)\, dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2} \\ &= \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) = \frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{9}{2} \end{aligned}$$

Example 2: A square root and a line

Find the area between $y = \sqrt{x}$ and $y = \dfrac{x}{2}$.

Solution. $\sqrt{x} = \dfrac{x}{2}$. Square both sides: $x = \dfrac{x^2}{4}$, so $x(x - 4) = 0$ and $x = 0$ or $x = 4$. Both check in the original equation.

At $x = 1$: $\sqrt{1} = 1 \gt \tfrac{1}{2}$, so $\sqrt{x}$ is on top.

$$A = \int_0^4 \left( \sqrt{x} - \frac{x}{2} \right) dx = \left[ \frac{2}{3}x^{3/2} - \frac{x^2}{4} \right]_0^4 = \frac{16}{3} - 4 = \frac{4}{3}$$

Example 3: Three crossings

Find the total area enclosed by $f(x) = x^3 - 3x$ and $g(x) = x$.

Solution. $x^3 - 3x = x$ gives $x^3 - 4x = x(x - 2)(x + 2) = 0$, so $x = -2, 0, 2$.

On $[-2, 0]$, test $x = -1$: $f(-1) = 2$ and $g(-1) = -1$, so $f$ is on top. On $[0, 2]$, test $x = 1$: $f(1) = -2$ and $g(1) = 1$, so $g$ is on top.

$$A = \int_{-2}^{0} \big( (x^3 - 3x) - x \big)\, dx + \int_0^2 \big( x - (x^3 - 3x) \big)\, dx = \int_{-2}^{0} (x^3 - 4x)\, dx + \int_0^2 (4x - x^3)\, dx$$ $$\int_{-2}^{0} (x^3 - 4x)\, dx = \Big[ \tfrac{x^4}{4} - 2x^2 \Big]_{-2}^{0} = 0 - (4 - 8) = 4, \qquad \int_0^2 (4x - x^3)\, dx = \Big[ 2x^2 - \tfrac{x^4}{4} \Big]_0^2 = 8 - 4 = 4$$

The total area is $4 + 4 = 8$. (If you had integrated $f - g$ over $[-2, 2]$ in one go, you’d get $4 - 4 = 0$, which is clearly not the area.)

Example 4: Calculator active

Find the area of the region bounded by $y = 2\cos x$ and $y = x^2$. (Radians.)

Solution. The intersections can’t be found by algebra, so use a calculator: $2\cos x = x^2$ at $x \approx \pm 1.022$. Store the positive value as $r$ to keep full accuracy. Between them, $2\cos x$ is on top (at $x = 0$: $2 \gt 0$).

$$A = \int_{-r}^{r} \big( 2\cos x - x^2 \big)\, dx \approx 2.701$$

Common mistakes

Subtracting in the wrong order. Bottom minus top gives a negative number. If you get a negative “area,” you’ve got the curves the wrong way round.

Using the x-axis as a boundary when it isn’t one. The region is bounded by the curves given. Only use $y = 0$ if the question says the region is bounded by the $x$-axis.

Not splitting at a crossing. When the curves cross inside the interval, one integral of $f - g$ lets pieces cancel. Split, or use $\lvert f - g \rvert$ on a calculator.

Guessing the limits. The limits come from the intersection points (or the given vertical lines). Solve for them; don’t read rough values off a sketch.

Squaring both sides and keeping a fake solution. When you square to solve, check each solution in the original equation.

Rounding intersection points early. On a calculator question, store the intersection values. Using $1.02$ instead of $1.0217\ldots$ can change the third decimal of the area.

Practice

1. (Warm-up) Find the area between $y = x + 3$ and $y = 1$ from $x = 0$ to $x = 2$.

Solution

On $[0, 2]$, $x + 3 \ge 3 \gt 1$, so the line $y = x + 3$ is on top.

$$\int_0^2 \big( (x + 3) - 1 \big)\, dx = \Big[ \tfrac{x^2}{2} + 2x \Big]_0^2 = 2 + 4 = 6$$

2. (Warm-up) Find the area of the region bounded by $y = x$ and $y = x^2$.

Solution

$x = x^2$ at $x = 0$ and $x = 1$. At $x = \tfrac{1}{2}$, $x \gt x^2$, so $y = x$ is on top.

$$\int_0^1 (x - x^2)\, dx = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$$

3. (Warm-up) Find the area of the region bounded by $y = x^2$ and $y = 4$.

Solution

$x^2 = 4$ at $x = \pm 2$, and $y = 4$ is on top.

$$\int_{-2}^{2} (4 - x^2)\, dx = \Big[ 4x - \tfrac{x^3}{3} \Big]_{-2}^{2} = \left( 8 - \tfrac{8}{3} \right) - \left( -8 + \tfrac{8}{3} \right) = \frac{32}{3}$$

4. (Core) Find the area of the region bounded by $y = 6 - x^2$ and $y = x$.

Solution

$6 - x^2 = x$ gives $x^2 + x - 6 = (x + 3)(x - 2) = 0$, so $x = -3$ or $x = 2$. At $x = 0$, the parabola ($6$) is above the line ($0$).

$$\begin{aligned} A &= \int_{-3}^{2} (6 - x^2 - x)\, dx = \left[ 6x - \frac{x^3}{3} - \frac{x^2}{2} \right]_{-3}^{2} \\ &= \left( 12 - \frac{8}{3} - 2 \right) - \left( -18 + 9 - \frac{9}{2} \right) = \frac{22}{3} + \frac{27}{2} = \frac{125}{6} \end{aligned}$$

5. (Core) Find the total area between $y = \sin x$ and $y = \cos x$ from $x = 0$ to $x = \pi$ (radians).

Solution

They cross where $\sin x = \cos x$, at $x = \dfrac{\pi}{4}$ in this interval. On $\left[0, \tfrac{\pi}{4}\right]$, $\cos x$ is on top; on $\left[\tfrac{\pi}{4}, \pi\right]$, $\sin x$ is on top.

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

Total: $(\sqrt{2} - 1) + (1 + \sqrt{2}) = 2\sqrt{2} \approx 2.828$.

6. (Core) Find the total area enclosed by $y = x^3$ and $y = x$.

Solution

$x^3 = x$ gives $x(x - 1)(x + 1) = 0$, so $x = -1, 0, 1$.

On $[-1, 0]$, $x^3$ is on top (at $x = -\tfrac{1}{2}$: $-\tfrac{1}{8} \gt -\tfrac{1}{2}$). On $[0, 1]$, $x$ is on top.

$$\int_{-1}^{0} (x^3 - x)\, dx = 0 - \left( \tfrac{1}{4} - \tfrac{1}{2} \right) = \frac{1}{4}, \qquad \int_0^1 (x - x^3)\, dx = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$

Total area $= \dfrac{1}{2}$.

7. (Core) Find the area of the region bounded by $y = e^x$, $y = 1$, and $x = 2$.

Solution

$e^x = 1$ at $x = 0$, so the region runs from $x = 0$ to $x = 2$, with $e^x$ on top.

$$\int_0^2 (e^x - 1)\, dx = \Big[ e^x - x \Big]_0^2 = (e^2 - 2) - 1 = e^2 - 3 \approx 4.389$$

8. (Challenge) Find the area of the region in the first quadrant bounded by $y = x$, $y = \dfrac{x}{4}$, and $y = \dfrac{1}{x}$.

Solution

Sketch it: both lines start at the origin; the curve $y = \dfrac{1}{x}$ closes the region off.

• $y = x$ meets $y = \dfrac{1}{x}$ at $x = 1$.
• $y = \dfrac{x}{4}$ meets $y = \dfrac{1}{x}$ at $x^2 = 4$, so $x = 2$.

The bottom is always $y = \dfrac{x}{4}$, but the top changes at $x = 1$: it’s $y = x$ on $[0, 1]$ and $y = \dfrac{1}{x}$ on $[1, 2]$.

$$A = \int_0^1 \left( x - \frac{x}{4} \right) dx + \int_1^2 \left( \frac{1}{x} - \frac{x}{4} \right) dx = \frac{3}{8} + \left( \ln 2 - \frac{3}{8} \right) = \ln 2 \approx 0.693$$

(Second integral: $\Big[ \ln x - \tfrac{x^2}{8} \Big]_1^2 = \left( \ln 2 - \tfrac{1}{2} \right) - \left( 0 - \tfrac{1}{8} \right) = \ln 2 - \tfrac{3}{8}$.)

9. (Challenge) For $a \gt 0$, the region between $y = x^2$ and $y = ax$ has area $36$. Find $a$.

Solution

$x^2 = ax$ at $x = 0$ and $x = a$, with the line on top in between.

$$\int_0^a (ax - x^2)\, dx = \frac{a^3}{2} - \frac{a^3}{3} = \frac{a^3}{6}$$

Set $\dfrac{a^3}{6} = 36$: $a^3 = 216$, so $a = 6$.