Find the area enclosed between two curves, or a curve and a line, by integrating the difference of the upper and lower functions between their intersection points

What this dot point is asking

SEAB wants you to find the area of a region trapped between two curves, or between a curve and a line. The method is to find where they cross (the limits), then integrate the upper function minus the lower function across that interval. This builds directly on the definite integral for area under a single curve.

The answer

The upper minus lower principle

The vertical gap between two graphs at each $x$ is (upper function) minus (lower function). Adding up these gaps across the region gives the enclosed area:

This automatically handles regions partly below the $x$-axis, because it measures the gap between the curves, not the distance to the axis.

Finding the limits

The limits $a$ and $b$ are the $x$-coordinates where the two graphs intersect. Set the two expressions equal, solve the resulting equation, and use the solutions as the limits.

Identifying upper and lower

Between the intersection points, decide which graph is on top, for example by testing a value of $x$ in between or by sketching. Subtracting in the wrong order gives a negative answer; if you do get a negative, take its modulus, as it just means the order was reversed.

A clean single integral

Because you subtract before integrating, the whole region is handled by one definite integral; you do not need to compute two separate areas and subtract them, although that also works.

When the curves cross more than twice

If two graphs intersect at three or more points, the upper and lower curves swap between consecutive intersections, so a single integral over the whole span would wrongly cancel parts of the area. The fix is to split the region at every intersection and integrate each piece separately with the correct upper-minus-lower order, then add the (positive) pieces. For a cubic crossing a line three times, you would compute two separate integrals, one for each enclosed loop, and sum them. Always find all the intersection points first and check, by testing a value in each subinterval, whether the same curve stays on top throughout.

Integrating with respect to y instead

Some regions are far easier to integrate horizontally, treating $x$ as a function of $y$. When the boundaries are naturally written as $x = \mathrm{f}(y)$ (for example a sideways parabola), the area becomes $\int_c^d (x_{\text{right}} - x_{\text{left}})\,dy$, the mirror image of the usual formula. The limits are then $y$-values of the intersections. Recognising when a region is bounded more simply left-and-right than top-and-bottom, and switching the variable of integration accordingly, can turn an awkward two-part vertical integral into a single clean horizontal one.

Examples in context

Example 1. Cross-section of a channel. The area of a water channel's cross-section, bounded above by the water surface and below by the curved bed, is the area between two curves, used to compute flow capacity.

Example 2. Profit between cost and revenue. Plotting revenue and cost against output, the area between the two curves over a range represents accumulated profit, an economic reading of the enclosed area.

Try this

Q1. State the integral for the area between $y = x^2$ and $y = 4$ from their intersections. [2 marks]

• Cue. Intersections $x = \pm 2$: $\displaystyle\int_{-2}^{2}(4 - x^2)\,dx$.

Q2. Find where $y = x^2$ meets $y = 2x$. [2 marks]

• Cue. $x^2 = 2x$ gives $x = 0$ or $x = 2$.

Q3. Find the area between $y = x^2$ and $y = 2x$. [4 marks]

• Cue. Line is upper: $\displaystyle\int_0^2 (2x - x^2)\,dx = \left[x^2 - \dfrac{x^3}{3}\right]_0^2 = 4 - \dfrac{8}{3} = \dfrac{4}{3}$ square units.