In short

Three flavours of area problem need extra care. First, regions described by inequalities like y \leq x^2 and y \geq x - 2: you plot all the boundaries, find the bounded region, and integrate the gap. Second, areas for inverse functions: reflecting across y = x preserves area, which gives a shortcut using horizontal strips. Third, odd-shaped regions with multiple boundary pieces: split at crossover points and integrate each piece with the right upper and lower curves.

Sketch the region described by these three conditions all at once: y \geq x^2, y \leq 4, x \geq 0.

The first inequality says "on or above the parabola y = x^2." The second says "on or below the horizontal line y = 4." The third says "to the right of the y-axis." Put them together and you have a curved triangle with a vertical left edge, a flat top at height 4, and a parabolic bottom-right boundary. It lives in the first quadrant, it is bounded, and it has a definite finite area.

What is that area?

This is a region described by inequalities, not by a list of curves. Before you can integrate anything you have to decode the description into a picture, find the corners, decide whether vertical or horizontal strips are easier, and only then set up the integral. That decoding step is where most errors happen, and it is the heart of the problems in this article.

The region from inequalities

Take the example above. The boundaries are y = x^2, y = 4, and x = 0. The parabola meets the horizontal line where x^2 = 4, so at x = 2 (taking the positive root because of x \geq 0). The parabola meets the y-axis at (0, 0). The line y = 4 meets the y-axis at (0, 4). The three corners of the region are (0, 0), (0, 4), and (2, 4).

Slice the region with vertical strips. For a fixed x between 0 and 2, the strip runs from the parabola y = x^2 at the bottom up to the line y = 4 at the top. Strip height is 4 - x^2. Strip width is dx. Adding them up:

A = \int_0^2 (4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}

The curved triangle has area \frac{16}{3} square units.

The region described by $y \geq x^2$, $y \leq 4$, $x \geq 0$. It has three corners: the origin, the point $(0, 4)$ on the y-axis, and the point $(2, 4)$ where the parabola meets the top line. The area is $\frac{16}{3}$.

The workflow is always the same. Plot every inequality. Shade the overlap. Find all corners. Then integrate. The integration is the easy part; the picture is where the thinking happens.

When horizontal strips are easier

For the region above, vertical strips worked cleanly. Sometimes they do not.

Take the region bounded by y^2 = x and y = x - 2. The first is a sideways parabola opening to the right. The second is a straight line of slope 1 passing through (2, 0) and (0, -2). Where do they meet?

Substitute x = y^2 into y = x - 2: y = y^2 - 2, so y^2 - y - 2 = 0, so (y-2)(y+1) = 0. The curves meet at y = 2 (where x = 4) and y = -1 (where x = 1).

Try vertical strips. For x between 0 and 1, the strip runs from the bottom of the parabola y = -\sqrt{x} up to the top of the parabola y = \sqrt{x} — the line y = x - 2 is still below the parabola at that range. For x between 1 and 4, the strip runs from the line y = x - 2 up to the top of the parabola y = \sqrt{x}. You need two separate integrals.

Now try horizontal strips. For a fixed y between -1 and 2, the strip runs from x = y^2 on the left to x = y + 2 on the right. One range, one integral:

A = \int_{-1}^{2}\bigl[(y+2) - y^2\bigr]\,dy
= \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{9}{2}

The area is \frac{9}{2} square units. Horizontal strips saved you from splitting the region.

The rule of thumb: if the region's left and right boundaries change as x varies but its top and bottom boundaries are clean as y varies, use horizontal strips. A sideways parabola with a line cutting through it is the canonical case.

Area using inverse functions

Here is a symmetry you may not have noticed. Take a function f on [a, b] with f(a) = c and f(b) = d, and assume f is increasing. The region under f from x = a to x = b has area \int_a^b f(x)\,dx.

Now reflect this region across the line y = x. Reflection across y = x swaps the roles of x and y; the graph of f becomes the graph of f^{-1}, the inverse function. Reflection is a rigid motion — it preserves area. So the area of the region under f^{-1} from y = c to y = d (which is the reflected region) equals the area under f from x = a to x = b, plus or minus the adjustment for where the axes end up.

The useful consequence is this: you can compute \int_c^d f^{-1}(y)\,dy without ever finding an explicit formula for f^{-1}. Look at the rectangle with corners (0, 0), (b, 0), (b, d), (0, d). Its area is bd. This rectangle is split by the graph of f into two pieces: the area under f (which is \int_a^b f(x)\,dx — assuming a = 0, c = 0 — plus the slab from x = 0 to x = a of height up to the curve), and the area above f (which equals \int_c^d f^{-1}(y)\,dy).

The cleanest form: if f is increasing with f(0) = 0, then for any b > 0 with f(b) = d,

\int_0^b f(x)\,dx + \int_0^d f^{-1}(y)\,dy = b \cdot d

This is called Young's identity. It says the two integrals partition the rectangle [0, b] \times [0, d].

The practical payoff: if f is easy to integrate but f^{-1} is ugly, compute \int_0^b f(x)\,dx and subtract from bd to get \int_0^d f^{-1}(y)\,dy for free.

Try it on f(x) = x^3 between 0 and 2. You have f(2) = 8, so the rectangle is [0, 2] \times [0, 8] with area 16. The area under the cube is

\int_0^2 x^3\,dx = \left[\frac{x^4}{4}\right]_0^2 = 4

So the area under f^{-1}(y) = y^{1/3} from y = 0 to y = 8 is 16 - 4 = 12. You can check this directly: \int_0^8 y^{1/3}\,dy = \left[\frac{3y^{4/3}}{4}\right]_0^8 = \frac{3 \cdot 16}{4} = 12. The identity holds and saved you a cube-root integration.

The graphs of $f(x) = x^3$ and $f^{-1}(y) = y^{1/3}$ are mirror images across $y = x$ (dashed). The area under $x^3$ from $0$ to $2$ is $4$. The area under $y^{1/3}$ from $0$ to $8$ is $12$. Together they fill the $2 \times 8$ rectangle: $4 + 12 = 16$.

Miscellaneous bounded regions

Some regions do not fit any of the standard templates. They have several boundary pieces, maybe a corner, maybe a change in which curve is on top. The strategy is always the same three steps: sketch, split, sum.

Take the region enclosed by y = x, y = 2x, and y = 6 - x. Three straight lines, three intersection points, one triangle.

Find the corners:

From the origin to x = 2, the top boundary is y = 2x and the bottom is y = x. From x = 2 to x = 3, the top changes to y = 6 - x and the bottom stays as y = x. So the area is a sum of two integrals:

A = \int_0^2 (2x - x)\,dx + \int_2^3 \bigl[(6 - x) - x\bigr]\,dx
= \int_0^2 x\,dx + \int_2^3 (6 - 2x)\,dx
= \left[\frac{x^2}{2}\right]_0^2 + \left[6x - x^2\right]_2^3
= 2 + \bigl[(18 - 9) - (12 - 4)\bigr] = 2 + 1 = 3

The triangle has area 3 square units. You can double-check with the formula for the area of a triangle with vertices (0,0), (2, 4), (3, 3):

A = \frac{1}{2}\bigl|0(4 - 3) + 2(3 - 0) + 3(0 - 4)\bigr| = \frac{1}{2}|0 + 6 - 12| = 3

Same answer, and you now know the integral method handles the split correctly.

Worked examples

Example 1: Area of |x| + |y| ≤ 2

Decode the inequality. |x| + |y| \leq 2 means that the sum of the absolute values of x and y is at most 2. The boundary curve |x| + |y| = 2 is what?

Step 1. Break the absolute values into cases by quadrant.

In the first quadrant (x \geq 0, y \geq 0): |x| + |y| = x + y, so x + y = 2. That is a line from (2, 0) to (0, 2).

In the second quadrant (x \leq 0, y \geq 0): -x + y = 2, so y = x + 2. A line from (0, 2) to (-2, 0).

In the third quadrant (x \leq 0, y \leq 0): -x - y = 2, so y = -x - 2. A line from (-2, 0) to (0, -2).

In the fourth quadrant (x \geq 0, y \leq 0): x - y = 2, so y = x - 2. A line from (0, -2) to (2, 0).

Why: the absolute values turn one equation into four line segments, each valid in its own quadrant. Together they form a closed shape.

Step 2. Identify the shape.

Four line segments joining (2, 0), (0, 2), (-2, 0), (0, -2). That is a square rotated 45° from the axes — a diamond with diagonals of length 4 along the x- and y-axes.

Why: opposite sides of the shape are parallel (both lines in the first and third quadrants have slope -1; both in the second and fourth have slope +1), and adjacent sides are perpendicular. Equal side lengths make it a square.

Step 3. Compute the side length and the area.

Each side goes from one axis-intersection to the next, say from (2, 0) to (0, 2). Its length is \sqrt{2^2 + 2^2} = 2\sqrt{2}. The area of a square with side 2\sqrt{2} is (2\sqrt{2})^2 = 8.

Why: once you have the shape, Pythagoras gives the side length and squaring gives the area. No integration needed for a polygon.

Step 4. Verify with integration.

Use the symmetry: the diamond is symmetric in both axes, so compute the first-quadrant piece and multiply by 4. In the first quadrant the region is the triangle bounded by x = 0, y = 0, and x + y = 2. Strip height at a given x is 2 - x:

A_{\text{Q1}} = \int_0^2 (2 - x)\,dx = \left[2x - \frac{x^2}{2}\right]_0^2 = 4 - 2 = 2

Total area: 4 \times 2 = 8. It matches.

Why: a symmetric region lets you compute one piece and multiply. Integration confirms the elementary answer.

Result: The region |x| + |y| \leq 2 has area 8.

The region $|x| + |y| \leq 2$ is a square tilted $45°$ from the axes, with vertices at $(\pm 2, 0)$ and $(0, \pm 2)$. Its side length is $2\sqrt 2$ and its area is $8$.

The big takeaway: absolute-value inequalities produce polygons. Always crack the absolute values by quadrant, and you will find the shape is something familiar.

Example 2: Area between y = ln x and its inverse from y = 0 to y = 1

Compute \int_0^1 \ln(x + 1)\,dx using the inverse-function trick — not direct integration.

Step 1. Rename the boundary cleanly. The function is f(x) = \ln(x + 1) on [0, 1]. At x = 0, f = 0. At x = 1, f = \ln 2 \approx 0.693. So you want the area under f on [0, 1], which is a region that fits inside the rectangle [0, 1] \times [0, \ln 2] of area \ln 2.

Why: Young's identity relates this area to the area under f^{-1}. Finding the inverse is much easier than integrating \ln(x+1) by parts.

Step 2. Find the inverse.

If y = \ln(x + 1), then e^y = x + 1, so x = e^y - 1. The inverse is f^{-1}(y) = e^y - 1.

Why: e^y - 1 is a standard integral. You exchange a logarithm for an exponential by using the inverse.

Step 3. Apply Young's identity.

The rectangle [0, 1] \times [0, \ln 2] has area \ln 2. It is split by the curve y = \ln(x+1) (equivalently x = e^y - 1) into two pieces:

\int_0^1 \ln(x+1)\,dx + \int_0^{\ln 2}(e^y - 1)\,dy = \ln 2

Why: the curve passes through both corners of the rectangle, (0, 0) and (1, \ln 2), so it cleanly splits the rectangle into the two complementary regions.

Step 4. Compute the easy integral.

\int_0^{\ln 2}(e^y - 1)\,dy = \bigl[e^y - y\bigr]_0^{\ln 2} = (2 - \ln 2) - (1 - 0) = 1 - \ln 2

Step 5. Solve for the hard integral.

\int_0^1 \ln(x + 1)\,dx = \ln 2 - (1 - \ln 2) = 2\ln 2 - 1

Why: subtract the easy piece from the rectangle's total area. The answer is exact, not a decimal.

Result: \int_0^1 \ln(x + 1)\,dx = 2\ln 2 - 1 \approx 0.386.

$y = \ln(x+1)$ in black and its inverse $y = e^x - 1$ in red are reflections across $y = x$. The areas under them on the strips $[0, 1]$ and $[0, \ln 2]$ together equal the rectangle of area $\ln 2$, giving $\int_0^1 \ln(x+1)\,dx = 2\ln 2 - 1$.

You have just computed an integral without integrating the function itself. The inverse-function move is not always available — both f and f^{-1} must be increasing, and the boundary of integration must match the rectangle — but when it is available, it bypasses hard integration.

Common confusions

Going deeper

If your goal is to handle exam-style area problems with inequalities and inverses, you have it. The rest of this section pulls back for a wider view — the general philosophy of region decomposition, and the geometric meaning of Young's identity.

Why splitting always works

Every bounded planar region, no matter how strange its boundary, can be written as a finite union of simple regions — regions that look like \{(x, y) : a \leq x \leq b, g(x) \leq y \leq f(x)\} or \{(x, y) : c \leq y \leq d, p(y) \leq x \leq q(y)\}. This is a theorem about piecewise smooth boundaries, and it is the formal justification for the "split and sum" strategy.

In practice, the splitting happens at two kinds of points: corners (where two boundary curves meet at an angle) and crossover points (where two curves cross, and the roles of upper and lower swap). Once you have identified all of these special points, the region breaks into pieces on which the strip method is straightforward.

Young's identity is a Legendre transform in disguise

If f: [0, a] \to [0, b] is continuous, strictly increasing, and f(0) = 0, then for any x \in [0, a] and y \in [0, b],

xy \leq \int_0^x f(t)\,dt + \int_0^y f^{-1}(t)\,dt

with equality if and only if y = f(x). This is Young's inequality, and the version you used above is the equality case.

The geometric picture: the left side xy is the area of a rectangle with corners (0, 0) and (x, y). The right side is the sum of the area under f up to x and the area under f^{-1} up to y. When y = f(x), the rectangle is exactly filled by those two pieces. When y \neq f(x), the two pieces cover the rectangle and more — there is overlap or extra area, which is why you get an inequality.

A famous corollary: take f(x) = x^{p-1} for p > 1. Then f^{-1}(y) = y^{1/(p-1)}. If you let q = p/(p-1), so that \frac{1}{p} + \frac{1}{q} = 1, the integrals simplify and you get

xy \leq \frac{x^p}{p} + \frac{y^q}{q}

This is the classical form of Young's inequality — the foundation for the Hölder and Minkowski inequalities that appear in analysis. Your area problem is connected to the theory of inner product spaces through a single picture.

Monotone rearrangements and hidden symmetries

Some difficult area problems become easy when you swap variables or rotate the region. The underlying principle: area is invariant under rigid motions (translation, rotation, reflection) and more generally under measure-preserving maps. A transformation that preserves area can turn an awkward region into a standard one.

Example. To find the area of the region bounded by xy = 1, xy = 4, y = x, and y = 4x in the first quadrant, try the substitution u = xy, v = y/x. The region becomes 1 \leq u \leq 4, 1 \leq v \leq 4 — a square in (u, v) coordinates. The Jacobian of the transformation lets you compute the original area from the simple integral over the square. This kind of change-of-variable is the subject of multivariable calculus and will become central when you meet double integrals.

Where this leads next

You now know how to compute the area of any reasonable planar region — with curves, inequalities, and inverses. The same slice-and-sum idea extends to higher dimensions and to lengths along curves.