In short

The area of the region between two curves y = f(x) and y = g(x) from x = a to x = b is \int_a^b |f(x) - g(x)|\,dx. If f(x) \geq g(x) throughout, you integrate f(x) - g(x) directly. If the curves cross each other, you split the integral at the crossing points and handle each piece. Horizontal strips — integrating with respect to y — are sometimes simpler.

You know how to find the area under a single curve. But most interesting regions are not bounded by a curve and an axis — they are bounded by two curves.

Take a concrete case. The line y = x and the parabola y = x^2 enclose a small leaf-shaped region between x = 0 and x = 1. What is the area of that leaf?

The parabola sits below the line for 0 < x < 1 (check: at x = 0.5, the line gives 0.5 and the parabola gives 0.25). So the "height" of the region at any x is the gap between the line and the parabola: x - x^2.

Slice the region into thin vertical strips. Each strip has width dx and height x - x^2. The area is

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}

The leaf has area 1/6 square units. That is remarkably small — only one-sixth of the 1 \times 1 square it sits inside — but the parabola hugs the line closely, leaving a thin sliver.

This is the core idea: the area between two curves is the integral of the gap between them.

The general principle: vertical strips

Suppose f(x) \geq g(x) for all x in [a, b]. Then f is the "upper" curve and g is the "lower" curve throughout.

Area between two curves (vertical strips)

If f(x) \geq g(x) on [a, b], the area between the curves y = f(x) and y = g(x) from x = a to x = b is

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

This is simply the integral of (top minus bottom).

Reading the definition. At each x, the expression f(x) - g(x) is the vertical distance between the two curves — the height of the region at that point. Multiplying by dx gives the area of one thin strip, and integrating adds them all up.

Notice that this formula automatically handles the case where both curves are below the x-axis. If f(x) = -1 and g(x) = -3, the gap is (-1) - (-3) = 2, which is correct. You never need to worry about signs relative to the axis — only the difference between the curves matters.

When curves cross: multiple intersections

The formula above assumes f(x) \geq g(x) everywhere on [a, b]. What if the curves cross each other?

Take f(x) = x^2 and g(x) = 2x. These intersect when x^2 = 2x, i.e., x(x - 2) = 0, so at x = 0 and x = 2. Between x = 0 and x = 2, the line y = 2x is above the parabola (check x = 1: line gives 2, parabola gives 1).

But if you wanted the area between these curves from x = -1 to x = 3, the situation is different. For x < 0, the parabola is above the line (check x = -1: parabola gives 1, line gives -2). For 0 < x < 2, the line is above. For x > 2, the parabola is above again.

The recipe:

  1. Find all intersection points in your interval. Here: x = 0 and x = 2.
  2. On each sub-interval, determine which curve is on top.
  3. Integrate the absolute difference on each sub-interval separately.
A = \int_{-1}^{0} (x^2 - 2x)\,dx + \int_0^2 (2x - x^2)\,dx + \int_2^3 (x^2 - 2x)\,dx

Each integral has (top minus bottom) as its integrand, so each is positive. Add the three pieces for the total area.

The parabola $y = x^2$ and the line $y = 2x$ intersect at $(0, 0)$ and $(2, 4)$. Between the intersection points, the line is above the parabola. Outside this interval, the parabola is on top. The area between them requires handling each sub-interval separately.

Horizontal strips: integrating with respect to y

Sometimes the region is easier to describe using y as the variable.

Consider the region enclosed by x = y^2 and x = y + 2. In terms of y, these are simple expressions. In terms of x, you would need to invert them: y = \sqrt{x} (and its negative branch) for the parabola, and y = x - 2 for the line. Vertical strips would force you to split the integral because the "top" and "bottom" curves change at different x-values.

With horizontal strips, the picture is cleaner. At each height y, the strip runs from the left boundary (x = y^2) to the right boundary (x = y + 2). The width of the strip is (y + 2) - y^2, and you integrate over y.

Area between two curves (horizontal strips)

If the region is bounded on the left by x = g_1(y) and on the right by x = g_2(y) with g_2(y) \geq g_1(y) on [y_1, y_2], the area is

A = \int_{y_1}^{y_2} \bigl[g_2(y) - g_1(y)\bigr]\,dy

This is the integral of (right minus left).

The logic is identical: each horizontal strip has width (right boundary minus left boundary) and height dy. Integrating adds up all the strips.

Two complete worked examples

Example 1: Area between a parabola and a line (vertical strips)

Find the area enclosed between y = 6 - x^2 and y = x.

Step 1. Find the intersection points. Set 6 - x^2 = x, so x^2 + x - 6 = 0, giving (x + 3)(x - 2) = 0. The curves intersect at x = -3 and x = 2.

Why: the intersection points are the limits of integration — the enclosed region starts and ends where the curves meet.

Step 2. Determine which curve is on top. At x = 0: the parabola gives 6, the line gives 0. So 6 - x^2 \geq x on [-3, 2] — the parabola is on top throughout.

Why: you need to know which curve to subtract from which. Testing one interior point is enough when there are no other intersections in the interval.

Step 3. Set up the integral.

A = \int_{-3}^{2} \bigl[(6 - x^2) - x\bigr]\,dx = \int_{-3}^{2} (6 - x - x^2)\,dx

Why: top minus bottom, integrated over the full interval between intersection points.

Step 4. Evaluate.

A = \left[6x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-3}^{2} = \left(12 - 2 - \frac{8}{3}\right) - \left(-18 - \frac{9}{2} + 9\right)
= \left(\frac{22}{3}\right) - \left(-\frac{27}{2}\right) = \frac{22}{3} + \frac{27}{2} = \frac{44 + 81}{6} = \frac{125}{6}

Why: careful arithmetic with fractions. A useful check: the intersection points span a length of 5 units, and the maximum gap between the curves (at the vertex of the parabola relative to the line) is about 6.25, so the area should be roughly \frac{2}{3} \times 5 \times 6.25 \approx 20.8, which is close to 125/6 \approx 20.83.

Result: The enclosed area is \dfrac{125}{6} square units.

The downward-opening parabola $y = 6 - x^2$ and the line $y = x$. They enclose a region from $x = -3$ to $x = 2$. The parabola is above the line throughout this interval. The enclosed area is $125/6$.

The graph confirms the geometry: a clean leaf-shaped region with the parabola forming the upper boundary and the line forming the lower boundary. The area 125/6 \approx 20.83 is reasonable for a region roughly 5 units wide with a maximum gap of about 6.25 units.

Example 2: Area using horizontal strips

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

Step 1. Find intersection points. Set y^2 = y + 2, so y^2 - y - 2 = 0, giving (y - 2)(y + 1) = 0. The curves meet at y = -1 and y = 2.

Why: these y-values define where the enclosed region starts and ends vertically. At y = -1: both give x = 1. At y = 2: both give x = 4.

Step 2. Determine which curve is on the right. At y = 0: the parabola gives x = 0, the line gives x = 2. So the line x = y + 2 is to the right of the parabola x = y^2 throughout.

Why: for horizontal strips, you need (right minus left), so you must know which curve is farther to the right.

Step 3. Set up the integral.

A = \int_{-1}^{2} \bigl[(y + 2) - y^2\bigr]\,dy = \int_{-1}^{2} (y + 2 - y^2)\,dy

Why: each horizontal strip at height y runs from x = y^2 (left boundary) to x = y + 2 (right boundary). Its width is the difference.

Step 4. Evaluate.

A = \left[\frac{y^2}{2} + 2y - \frac{y^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{10}{3} + \frac{7}{6} = \frac{20 + 7}{6} = \frac{27}{6} = \frac{9}{2}

Result: The enclosed area is \dfrac{9}{2} square units.

The parabola $x = y^2$ and the line $x = y + 2$ (drawn as $y = x - 2$). They enclose a region from $y = -1$ to $y = 2$. Horizontal strips make this problem straightforward: the line is to the right of the parabola throughout, and the enclosed area is $9/2$.

If you tried this with vertical strips, you would need to split the integral at x = 1 (where the line enters the region), handle two different "top" curves for 0 \leq x \leq 1 vs. 1 \leq x \leq 4, and deal with the two branches of y = \pm\sqrt{x}. Horizontal strips avoided all of that in one clean integral.

Common confusions

Going deeper

If you came here to learn how to find areas between curves, you have the tools — you can stop here. The rest of this section explores a few subtleties that appear in harder problems.

The absolute-value shortcut

If you don't want to determine which curve is on top in each sub-interval, you can write

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

This always gives the correct geometric area, regardless of which curve is on top. The absolute value ensures every strip contributes positively.

In practice, you still need to find the intersection points to split the integral (you cannot evaluate \int |h(x)|\,dx by antidifferentiation without first removing the absolute value). But this form is a good way to set up the problem before doing the case analysis.

Regions with more than two boundaries

Some regions are bounded by three or more curves. For example, the region bounded by y = x^2, y = 4, and x = 0 (in the first quadrant). The strategy is the same: at each x (or each y), identify the top and bottom (or right and left) boundaries. If the boundaries change at some point, split the integral there.

A useful technique: sketch the region first, identify all boundary curves, and then choose the direction of integration that minimises the number of splits. Often, one direction requires one integral while the other requires two or three.

Symmetric regions

When a region is symmetric about the x-axis or the y-axis, you can compute the area of one half and double it. For example, the area enclosed by x^2 + y^2 = r^2 (a circle of radius r) can be computed as

A = 2\int_{-r}^{r} \sqrt{r^2 - x^2}\,dx = 4\int_0^r \sqrt{r^2 - x^2}\,dx

using the symmetry about both axes. The substitution x = r\sin\theta then gives A = \pi r^2, the familiar formula. Integration proves the area formula for a circle — a result that was known for millennia but never proved rigorously until calculus existed.

Area of a circle segment

A common application: find the area cut off from the circle x^2 + y^2 = r^2 by the line x = a (where 0 < a < r). The region is between the vertical line and the circle, from x = a to x = r. Using vertical strips:

A = 2\int_a^r \sqrt{r^2 - x^2}\,dx

This integral requires the substitution x = r\sin\theta, and the result involves \arcsin — connecting area to inverse trigonometric functions in a way that appears throughout JEE problems.

Where this leads next

You now know how to find areas enclosed by two curves using both vertical and horizontal strips.