6.1 Areas between Curves

Big idea

The definite integral was built to answer one geometric question: how much area sits between a graph and the horizontal axis. That question stood in for a much larger family. The real content of the integral is not area but a procedure, and this is where the procedure gets used on something the original setup did not cover.

The procedure has three steps, and every section in this chapter is those three steps applied to a different quantity. Cut the thing you want into thin slices. Approximate one slice by a shape whose size you can write down exactly. Add the approximations and let the slices thin out, which turns the sum into an integral. For area between two graphs a slice is a thin vertical strip, approximated by a rectangle of width $dx$ whose height is the vertical gap between the curves. The gap is top minus bottom, so the area is the integral of that difference.

Seen that way, the area under a single curve stops being a separate formula. It is the case where the bottom curve is the line $y = 0$, and the gap is just $f(x)$. So you never again need to remember whether a region sits above or below the axis: subtract bottom from top and the sign takes care of itself.

What is genuinely new here is the bookkeeping. You have to know which curve is on top, and over what interval, and that can change partway across the region. You also get a choice the original area problem never offered: slice vertically and integrate in $x$, or slice horizontally and integrate in $y$. Picking the wrong one turns a one-line computation into a three-piece computation, so seeing which way the strips should run is most of the skill.

Decoder

Let $f$ and $g$ be continuous on $[a, b]$.

If $f(x) \ge g(x)$ for every $x$ in the interval, the region caught between the two graphs from $x = a$ to $x = b$ has area equal to the integral of $f(x) - g(x)$ over that interval.

The phrase carrying the content is “for every $x$ in the interval.” The integral of a difference exists no matter which function is larger, but it measures an area only when the difference never changes sign. If the curves cross inside $[a, b]$, the part where $g$ is on top is counted as negative and the result is a net total, not an area.

The fix is not a new formula but new intervals. Find the crossings, break the region at them, decide again on each piece which curve is on top, and add. Writing the answer as the integral of $|f(x) - g(x)|$ says exactly this, but you cannot evaluate an absolute value directly; you still have to split.

The classic mistake is subtracting in the order the problem happens to name the functions. A region described as lying between $y = x^2$ and $y = 2x$ does not tell you which comes first. Sketch, or test one interior point, before you write the integrand.

Definitions and results

The region between two graphs. Given continuous $f$ and $g$ on $[a, b]$, the region between them is the set of points $(x, y)$ with $a \le x \le b$ and $y$ between $g(x)$ and $f(x)$. It is bounded by the two curves above and below, and by the lines $x = a$ and $x = b$ on the sides unless the curves meet there.

Area from vertical strips. When $f(x) \ge g(x)$ on $[a, b]$,

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

One strip contributes height $f(x) - g(x)$ and width $dx$.

Area under a single curve. Taking $g(x) = 0$ gives the familiar statement: if $f(x) \ge 0$ on $[a, b]$, the area between the graph and the axis is $\int_a^b f(x)\,dx$. If $f(x) \le 0$, the axis is the top curve and the area is $-\int_a^b f(x)\,dx$.

Finding the limits. When only the curves are given, the endpoints are the roots of $f(x) = g(x)$, taken in increasing order.

Curves that cross. If $f - g$ changes sign at interior points $c_1 < c_2 < \cdots$, split at those points and integrate the correct difference on each subinterval:

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

Horizontal strips. If the region is bounded on the right by $x = p(y)$ and on the left by $x = q(y)$ for $c \le y \le d$, slice horizontally instead:

$$ A = \int_c^d \big[ p(y) - q(y) \big] \, dy $$

The same rule with the axes exchanged: right minus left, thickness $dy$.

Choosing the variable. Slice in the direction that lets one formula describe every strip. If the top boundary is one curve for part of the interval and a different curve for the rest, vertical strips force a split and horizontal strips may not. A sideways parabola such as $x = y^2$ is the standard signal to integrate in $y$.

Signed area. The number $\int_a^b f(x)\,dx$ is a signed total: area above the axis counts positive, area below negative. Area itself is never negative, so an answer that comes out negative means you subtracted in the wrong order.

Worked examples

A line above a parabola

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

No interval is given, so the curves supply it. Setting $x + 2 = x^2$ gives $(x - 2)(x + 1) = 0$, so the region runs from $x = -1$ to $x = 2$. At $x = 0$ the line is at $2$ and the parabola at $0$, so the line is on top throughout.

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

At $x = 2$ the bracket is $2 + 4 - \tfrac{8}{3} = \tfrac{10}{3}$. At $x = -1$ it is $\tfrac{1}{2} - 2 + \tfrac{1}{3} = -\tfrac{7}{6}$. The area is $\tfrac{10}{3} + \tfrac{7}{6} = \tfrac{9}{2}$.

A rough check: the region sits inside the rectangle $-1 \le x \le 2$, $0 \le y \le 4$ of area $12$, and fills something under half of it, so $4.5$ is plausible.

Area under a curve that dips below the axis

Find the area between $y = x^2 - 1$ and the horizontal axis for $0 \le x \le 2$.

The graph is below the axis on $[0, 1]$ and above it on $[1, 2]$, so top and bottom swap at $x = 1$ and the region splits:

$$ A = \int_0^1 \big[ 0 - (x^2 - 1) \big] dx + \int_1^2 \big[ (x^2 - 1) - 0 \big] dx $$

The first integral is $\left[ x - \tfrac{x^3}{3} \right]_0^1 = \tfrac{2}{3}$. The second is $\left[ \tfrac{x^3}{3} - x \right]_1^2 = \left( \tfrac{8}{3} - 2 \right) - \left( \tfrac{1}{3} - 1 \right) = \tfrac{4}{3}$. So $A = 2$.

Compare with the unsplit integral $\int_0^2 (x^2 - 1)\,dx = \tfrac{8}{3} - 2 = \tfrac{2}{3}$. That smaller number is the signed total, area above the axis minus area below. Both are correct answers to different questions.

Curves that cross in the middle

Find the area enclosed by $y = x^3$ and $y = x$.

Setting $x^3 = x$ gives $x(x^2 - 1) = 0$, so the graphs meet at $x = -1$, $0$ and $1$. At $x = -\tfrac{1}{2}$ the cubic is $-\tfrac{1}{8}$ and the line is $-\tfrac{1}{2}$, so the cubic is on top on $[-1, 0]$; at $x = \tfrac{1}{2}$ the line is on top.

$$ A = \int_{-1}^{0} (x^3 - x)\, dx + \int_{0}^{1} (x - x^3)\, dx $$

The first is $\left[ \tfrac{x^4}{4} - \tfrac{x^2}{2} \right]_{-1}^0 = \tfrac{1}{4}$ and the second is $\left[ \tfrac{x^2}{2} - \tfrac{x^4}{4} \right]_0^1 = \tfrac{1}{4}$, so the total is $\tfrac{1}{2}$. The two pieces coming out equal is not luck: both functions are odd, so the region is symmetric through the origin.

Slicing horizontally

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

Vertical strips would be painful, since each parabola supplies both the top and the bottom boundary over part of the region. Horizontal strips see one right curve and one left curve. The curves meet where $y^2 = 2 - y^2$, so $y$ runs from $-1$ to $1$, and at $y = 0$ the curve $x = 2 - y^2$ is the right boundary.

$$ A = \int_{-1}^{1} \big[ (2 - y^2) - y^2 \big] \, dy = \int_{-1}^{1} (2 - 2y^2)\, dy $$

The integrand is even, so this is $2\int_0^1 (2 - 2y^2)\,dy = 2\left[ 2y - \tfrac{2y^3}{3} \right]_0^1 = 2\left( 2 - \tfrac{2}{3} \right) = \tfrac{8}{3}$.

Check the scale: the region fits inside the box $0 \le x \le 2$, $-1 \le y \le 1$ of area $4$, and fills two thirds of it, which suits the lens shape two facing parabolas make.

Practice

Start with the two-curve setup. Each problem gives a pair of graphs, or a pair plus an interval, and asks for the enclosed area. Solve for the crossings first, then decide which curve is on top before writing an integral sign.

Then the single-curve case, including regions that dip below the axis. Watch the difference between the area of a region and the signed value of an integral over the same interval.

Quiz

Five items on setting up area integrals, locating crossings, and choosing between vertical and horizontal strips.