Area Under Curves

01 / UNDERSTAND & APPLY

1. Signed accumulation and geometric area answer different questions

A definite integral accumulates signed vertical contributions. Portions of a graph below the x-axis contribute negatively. Geometric area, by contrast, measures the size of a region and is never negative. The integral of x over [−1,1] is zero because two signed contributions cancel, while the area between its graph and the x-axis is the sum of two triangles and equals 1.

For a continuous function f on [a,b] with a ≤ b, the area between the graph and the x-axis is ∫ₐᵇ|f(x)|dx. If f is nonnegative throughout, the absolute value is unnecessary. If f changes sign, split at its zeros and integrate the nonnegative height on each piece. Taking the absolute value only after integrating is generally wrong because the signed cancellation has already happened.

The same idea applies between two curves. A vertical strip runs from the lower graph to the upper graph, so its nonnegative height is upper minus lower. If the curves cross, their order may swap and the formula for the strip height changes. The area is not determined by which function was written first in the question.

The phrase “bounded by” requires you to identify a closed region. Two curves need not enclose a finite region by themselves. Extra lines such as x = a, x = b, or an axis may provide essential boundaries. Sketch every named boundary and confirm which region is intended before choosing limits.

02 / UNDERSTAND & APPLY

2. Why upper minus lower is the correct height

Partition the x-interval into thin strips. At a sample point ξ, a vertical strip of width Δx has approximate area [f(ξ) − g(ξ)]Δx when f is above g. Adding all strips and shrinking the largest width leads to the definite integral of the difference. This derives the formula from lengths rather than presenting it as an arbitrary subtraction rule.

The lower graph can be below the x-axis. For instance, an upper value 2 and lower value −3 give a strip height of 5, not −1 or 1. Subtracting the lower coordinate automatically includes the region on both sides of the axis. There is no need to split at the x-axis if you are integrating the distance between two graphs and their relative order does not change.

For horizontal strips, solve the boundaries as x = R(y) and x = L(y). The strip length is right minus left, with thickness dy. Its limits are y-coordinates. Mixing x-limits with a dy integral is a conceptual error even when the arithmetic happens to be easy. Label the variable beside each limiting value in your sketch.

A useful schematic has one representative strip drawn all the way across the region. Name its two endpoints on the relevant boundaries. If a single strip crosses a hole or two disconnected pieces, you need more than one length contribution or a different decomposition. The drawing is a test of the formula, not decoration.

03 / UNDERSTAND & APPLY

3. Find every relevant crossing and test the order

Find intersection coordinates by solving f(x) = g(x), retaining only values allowed by both original domains. Then sort the relevant x-values and test the sign of f − g in each interval. A graph helps locate likely crossings, but algebra or a justified monotonicity argument must ensure that none has been missed.

A repeated intersection root may represent tangency, so the order need not change there. Conversely, a boundary can change at a corner even if there is no smooth crossing. When a region is bounded by three or more curves, the active upper or lower boundary may switch at pairwise intersections that are not the outermost endpoints.

Do not integrate between every algebraic intersection automatically. The question may specify a particular quadrant, a fixed interval, or only one component. Intersections outside those restrictions are irrelevant. Squaring radical equations can create extraneous points, so substitute candidates into the original boundary equations.

04 / UNDERSTAND & APPLY

4. Split when a crossing reverses the height

An integral over several lobes can hide cancellation. If f − g is odd on a symmetric interval, its signed integral is zero, but the geometric areas of its lobes may be positive and equal. Symmetry can help you add them: integrate one nonnegative lobe and multiply by the number of congruent pieces.

The absolute-value formulation is concise, but evaluation still requires a sign analysis or another justified method for absolute values. Writing ∫|f − g| does not by itself calculate the area. In a complete solution, show where the sign changes and which expression is used in each piece.

At a crossing, a boundary point has zero strip height and contributes no area by itself. Therefore adjacent integrals can share the crossing as an endpoint without double-counting any positive area. This differs from counting distinct roots, where the shared endpoint is a point that must be counted only once.

05 / UNDERSTAND & APPLY

5. Choose the direction that avoids unnecessary splitting

Some regions are awkward as vertical strips because a sideways curve has two y-branches. Horizontal strips may give a single uncomplicated span. The goal is not always the shortest antiderivative; it is the simplest correct description of the region. Compare the number of pieces and the ease of expressing boundaries before deciding.

For a region between x = y² and x = y + 2, horizontal strips naturally have the right endpoint on the line and left endpoint on the parabola. A vertical approach is possible, but solving y = ±√x introduces branch bookkeeping and often an extra split. Both methods describe the same area if done correctly.

When changing perspective, recompute the limits in the new variable rather than reusing numbers from an earlier integral. A point (4,2) contributes x = 4 to a vertical setup and y = 2 to a horizontal one. This simple distinction catches many apparently mysterious factor errors.

06 / UNDERSTAND & APPLY

6. An active boundary can change inside the region

When three curves form the boundary, solve their pairwise intersections and determine which segments actually enclose the chosen region. An upper boundary may be one curve on the left and another on the right. Integrating the higher of all listed curves would describe a different region if that curve lies outside the enclosed boundary.

A clean approach is to trace the boundary once around the region. Mark each vertex or intersection where the governing curve changes. Project those points onto the intended integration axis. Within each resulting interval, draw a strip and identify its endpoints. This turns a complicated picture into a finite list of ordinary upper-minus-lower integrals.

Elementary geometry provides a useful alternative check when one part is triangular or rectangular. You may integrate the curved cap and add a triangle, or subtract a simple outside piece from a larger known region. Such decompositions are valid only when the pieces cover the desired region with no overlap of positive area.

07 / UNDERSTAND & APPLY

7. Symmetry, scale and the limits of a finite-area formula

If a region is symmetric about an axis, calculate one half and double it. If it has symmetry about both axes, a quadrant calculation may be multiplied by four. Justify the symmetry of the whole region, including all boundary lines and restrictions. An even curve alone does not guarantee that a region cut by an asymmetric line has equal halves.

Parameter problems often simplify by scaling. The region under y = a² − x² above the x-axis has endpoints ±a only when a > 0; if a is an arbitrary real parameter, the endpoints are ±|a|. The vertical scale is a² and the horizontal scale is |a|, so the area scales like |a|³. A substitution x = |a|u confirms this and helps predict the form before integration.

An unbounded region may have finite area, but it requires an improper integral and a convergence check. The area under 1/x² from x = 1 to infinity is 1, while the area under 1/x over the same interval diverges. Never substitute infinity into an antiderivative as though it were a number; take a limit. A vertical asymptote inside the interval also requires separate one-sided improper integrals.

For ordinary exam regions bounded by continuous curves on finite intervals, the standard strip formulas suffice. The improper case is a warning about scope: a symbol that resembles an area integral does not automatically produce a finite area, and cancellation across a singularity cannot rescue two divergent positive areas.

08 / UNDERSTAND & APPLY

8. Original practice: set up before integrating

Each answer includes the strip choice and order check. A correct unevaluated setup is meaningful progress; an antiderivative of the wrong height is not.

09 / UNDERSTAND & APPLY

9. A final geometry audit

Before integrating, ask: which region, which strip direction, which two endpoints, which limits, and where does that description change? After integrating, ask: is the result nonnegative, does it have square units, and is its size plausible compared with an enclosing rectangle or triangle? A negative area is evidence of a setup or orientation error, not a final answer to be patched blindly.

Do not confuse total area with the magnitude of a signed integral. Do not assume the curve with a visually larger coefficient is always above another. Do not choose limits before finding relevant intersections and checking the original domains. Do not treat a dy integral as the same expression with a different letter: the geometric boundaries must be expressed in the correct variable.

The best independent check is often a second strip direction. The cap example gives a two-piece vertical setup and a one-piece horizontal setup, and agreement confirms both the region and the arithmetic. Symmetry, scale dependence and elementary area bounds provide additional checks without requiring a plotting tool.

Finally, preserve exact values until the end. Rounding an intersection such as √2 before integration introduces avoidable error in both the limits and antiderivative evaluation. Decimal approximations can be useful for a reasonableness check, but the exact geometric setup is the backbone of the solution.