Area Under Curve

A farmer owns a rectangular plot, 100 metres long and 40 metres wide. Its area is 100 \times 40 = 4000 square metres. That computation takes one multiplication.

Now suppose the plot is not rectangular. One edge follows a river, and the river curves. The plot is still 100 metres long, but the width changes at every point — it might be 40 metres wide at one end, 55 metres in the middle, and 30 metres at the other end. What is the area now?

There is no single width to multiply by. The shape has no formula from elementary geometry. You need a new idea.

The idea is simple: slice the region into thin vertical strips. Each strip is so thin that the curve barely changes across it, so each strip is approximately a rectangle. Add up the areas of all those thin rectangles. The thinner you make them, the closer the sum gets to the true area. The exact area is the limit of this sum — and that limit is exactly what a definite integral computes.

This is the connection between integration and area: the definite integral is a machine for adding up infinitely many infinitely thin strips.

Building the picture with strips

Take a concrete curve: y = x^2 from x = 0 to x = 3. The region you want is everything below the parabola and above the x-axis, between x = 0 and x = 3.

Slice this region into n vertical strips, each of width \Delta x = 3/n. The k-th strip sits at position x_k = k \cdot \Delta x and has height approximately f(x_k) = x_k^2. Its area is roughly x_k^2 \cdot \Delta x.

The total area is approximately

Using the standard formula \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}:

Try a few values. With n = 4 strips: S_4 = \frac{27 \cdot 5 \cdot 9}{6 \cdot 16} = \frac{1215}{96} = 12.656. With n = 10: S_{10} = \frac{27 \cdot 11 \cdot 21}{600} = \frac{6237}{600} = 10.395. With n = 100: S_{100} = \frac{27 \cdot 101 \cdot 201}{60000} = \frac{548127}{60000} = 9.1355.

The numbers are settling toward something. As n \to \infty:

The exact area under y = x^2 from 0 to 3 is 9 square units.

The formal definition

What you just computed — the limit of a sum of strip areas — is exactly the definite integral. Here is the precise statement.

Reading the definition. The integral sign \int is a stretched-out S — it stands for "sum." The dx at the end represents the width of each infinitesimal strip. The expression f(x)\,dx is the area of one strip: height times width. The integral adds up all these strip-areas from x = a to x = b.

To actually compute the integral, you find an antiderivative F(x) — a function whose derivative is f(x) — and evaluate:

This is the Fundamental Theorem of Calculus at work. For f(x) = x^2, the antiderivative is F(x) = x^3/3, so

Exactly what the limit-of-sums approach gave.

Vertical strips vs. horizontal strips

So far, every strip has been vertical: a thin rectangle standing upright, with its base on the x-axis and its top touching the curve. This is the natural approach when the curve is written as y = f(x).

But sometimes the curve is easier to describe as x = g(y) — a function of y. In that case, horizontal strips are the natural choice.

A horizontal strip lies on its side. Its "base" is on the y-axis, and it extends rightward to the curve x = g(y). Its height is a thin slice dy, and its width is g(y). The area of one horizontal strip is g(y)\,dy, and the total area is

where y_1 and y_2 are the y-values at the bottom and top of the region.

When to use which. Use vertical strips when the curve is naturally y = f(x). Use horizontal strips when the curve is naturally x = g(y), or when vertical strips would force you to split the region into multiple pieces. Both methods give the same answer — you are just slicing the same region in different directions.

Take the parabola x = y^2 from y = 0 to y = 2. Using horizontal strips, the area between the curve and the y-axis is

If you tried vertical strips here, you would need to rewrite the curve as y = \sqrt{x} and integrate from x = 0 to x = 4: \int_0^4 \sqrt{x}\,dx = \frac{2}{3}x^{3/2}\big|_0^4 = \frac{2}{3}(8) = \frac{16}{3}. But that gives the area under y = \sqrt{x} above the x-axis — which is the region above the parabola inside the rectangle [0,4] \times [0,2]. The area below the parabola (between the curve and the y-axis) is the rectangle's area minus this: 4 \times 2 - 16/3 = 24/3 - 16/3 = 8/3. Both approaches agree.

Signed area: what happens below the axis

There is an important subtlety. The definite integral \int_a^b f(x)\,dx does not always give you "area" in the geometric sense. When the curve dips below the x-axis, the integral subtracts the contribution from the below-axis region instead of adding it.

Take f(x) = x^2 - 4 on the interval [0, 3]. This parabola sits below the x-axis between x = 0 and x = 2 (since x^2 - 4 < 0 there) and above the axis from x = 2 to x = 3.

If you blindly compute the integral:

A negative answer — which cannot be a geometric area. What happened? The below-axis piece contributed -16/3 and the above-axis piece contributed 7/3, and the integral added them with their signs: -16/3 + 7/3 = -3.

To find the actual geometric area, you must handle the below-axis and above-axis parts separately, taking the absolute value of each:

The geometric area is 23/3, not -3. The rule is: find where the curve crosses the axis, split the integral at those points, and take the absolute value of each piece.

Computing areas: the workflow

Here is the process for finding the area of a region bounded by a curve and the axes.

1. Sketch the curve. Identify where it crosses the x-axis (set f(x) = 0 and solve).
2. Determine the interval [a, b] over which you need the area.
3. Check the sign of f(x) on each sub-interval. If f(x) \geq 0 throughout, integrate directly. If f(x) changes sign, split at the zeros.
4. Integrate each piece and take absolute values of the below-axis contributions.
5. Add the pieces to get the total geometric area.

Common confusions

• "The integral always gives area." Not quite. The integral gives signed area: positive above the axis, negative below. To get geometric area, you must split the integral at zeros and take absolute values. Forgetting this is the most common error in area problems.

• "I always use vertical strips." Vertical strips work when the curve is y = f(x), but for curves like x = y^2, horizontal strips are much simpler. Choosing the right strip direction can turn a hard problem into an easy one.

• "Area must be positive, so I just put absolute value bars around the whole integral." This does not work in general. \left|\int_a^b f(x)\,dx\right| is not the same as \int_a^b |f(x)|\,dx. The first takes the absolute value of the net signed area; the second takes the absolute value at each point before integrating. The second is what you want for geometric area.

• "The area under y = x^2 from -3 to 3 is zero because the function is symmetric." No — x^2 is always non-negative, so both sides contribute positive area. The integral \int_{-3}^{3} x^2\,dx = 18, not zero. You might be confusing this with an odd function like x^3, where the signed integral from -a to a is indeed zero — but even then, the geometric area is 2\int_0^a x^3\,dx, not zero.

Going deeper

Left sums, right sums, and the squeeze

In the strip computation above, you used right endpoints: each strip's height was f(x_k) where x_k is the right edge of the strip. You could equally have used the left edge, giving a left Riemann sum. For an increasing function, the left sum underestimates and the right sum overestimates. For a decreasing function, it is reversed.

The key fact is: as n \to \infty, both the left and right sums converge to the same number — the definite integral. This is not obvious, and the proof uses the fact that continuous functions on closed intervals are uniformly continuous. The result is that you can use any point within each strip (left endpoint, right endpoint, midpoint, or any other choice) and the limit is the same.

Area as the antiderivative's increment

Why does the Fundamental Theorem of Calculus work? Define A(x) = \int_a^x f(t)\,dt — the area under the curve from a to a variable right endpoint x. When you move the right endpoint by a tiny amount h, the area increases by approximately f(x) \cdot h (one more thin strip). So A'(x) = f(x) — the rate at which area accumulates is the height of the curve. That means A(x) is an antiderivative of f(x), and the total area from a to b is A(b) - A(a) = F(b) - F(a) for any antiderivative F.

This is the deep reason integration and differentiation are reverse operations: differentiation measures rates, and integration accumulates. The area function is the bridge between them.

Curves that touch the axis

When f(x) = 0 at a single point but does not change sign, no splitting is needed — the integral is still positive. For example, y = (x-1)^2 touches the axis at x = 1 but never goes below it. The integral \int_0^2 (x-1)^2\,dx = 2/3 is the correct geometric area without any adjustments.

The splitting rule applies only when f actually changes sign — when it crosses from positive to negative or vice versa.

Where this leads next

You now know how to find the area of a region bounded by a single curve and the axes. The next set of articles extends this to more complex regions.

• Area Between Curves — when the region is bounded by two curves instead of a curve and an axis.
• Area - Advanced — area using parametric curves, polar coordinates, and variable boundaries.
• Area - Special Cases — inequalities, inverse functions, and other unusual bounding regions.
• Volumes of Revolution — spin a curve around an axis and find the volume of the resulting solid.
• Arc Length and Surface Area — measuring how long a curve is, and the surface area when you revolve it.