Area between curves

Area between two curves

In 15 of 20 recent exams
3% of the marks
Paper 1 or 2
3 to 10 marks

How it is set in the exam

Paper 1 (no calculator) (exact) and Paper 2 (calculator) (GDC) about equally, 3–10 marks, often the last parts of a Section B question once the intersections have been found. Very common (15 of 20 exam versions). The shapes: a curve and a line (usually the chord through two points on the curve, less often a tangent), two exponential or two trig graphs, and the graphs of $f$ and $f^{-1}$ (3 versions, all Paper 2 (calculator)).

Spot it in the exam

Exam phrases to spot (4)

“the region enclosed by the graphs of $f$ and $g$ ”
“find the area of the region bounded by the curve and the line”
“the shaded region between the two curves”
“show that the area of $R$ is $\int_0^6 (\ldots)\,dx$ ”

In disguise (5)

an edge of the region is a line from an earlier part (a tangent, a normal, or the chord through two given points): find the line first (Tangent or normal at a given point)
a straight edge whose equation no part asks for: a segment between two labelled points of the graph, such as the origin and a turning point found in an earlier part (on , from the GDC). Find the line's equation yourself, since it carries marks of its own, or use a triangle and the area between the curve and the $x$ -axis
the region enclosed by the graphs of $f$ and $f^{-1}$ : for an $f$ increasing on its whole domain they meet on $y = x$ , and the region is symmetric in that line (the meeting points: The graphs of f and f⁻¹)
" $g'(x) = f'(x)$ for all $x$ ": then $g(x) = f(x) + k$ , a vertical shift. The gap is $\lvert k \rvert$ everywhere, so over $[a, b]$ the area is $\lvert k \rvert(b - a)$ , with no integrating
a region bounded by a curve, a line and the $x$ -axis: its bottom (or top) edge changes part-way, so it splits there

Variants (2)

the curves cross inside the interval: split at each crossing, or $\int \lvert f - g \rvert$ on
an unknown limit or constant: "the area is …; find $k$ "

Looks like it, but isn't (3)

one curve and the $x$ -axis only: Area between a curve and the x-axis
a triangle cut off by a tangent and the axes: no integral (The tangent or normal meets something)
an edge is an arc of a circle: sectors, since $\sqrt{r^2 - x^2}$ has no SL antiderivative (Chords, segments and regions made of sectors and triangles)

Worked example

A parabola and a line

Paper 1 (no calculator)

Question

Let $f(x) = x^2 - 2x + 3$ and $g(x) = x + 7$ .
(a) Find the $x$ -coordinates of the points where the graphs of $f$ and $g$ meet. [3]
(b) Show that the area of the region enclosed by the graphs of $f$ and $g$ is $\displaystyle\int_{-1}^{4} \left(4 + 3x - x^2\right)dx$ . [2]
(c) Find the area of the region. [4]

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Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

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See it move: The region between two curves

At $x = 1.50$ : top minus bottom $= g(x) - f(x) = 8.50 - 2.25 = 6.25$ .
The graphs meet at $x = -1$ and $x = 4$ : the limits.
Between them the line is on top: the gap is $g(x) - f(x) = 4 + 3x - x^{2}$ .
Slid onto the axis, the region is the area under the gap: $\int_{-1}^{4} \left(4 + 3x - x^{2}\right)dx = \frac{125}{6} \approx 20.8$ .

Steps

Where do they meet?
Solve $f(x) = g(x)$ : these are the limits, unless the question gives lines $x = a$ and $x = b$ (: zeros and intersections, with each value stored: keep full-accuracy values).
Which is on top where?
Test one value between each pair of limits, or read a sketch. Where the curves cross inside the interval, the order changes.
Write the expression:
$\int_a^b (\text{top} - \text{bottom})\,dx$ for each piece. If an edge changes part-way (the $x$ -axis, then a curve), split there too; a triangle or a trapezium can replace the integral of a straight line. On , $\int_a^b \lvert f(x) - g(x) \rvert\,dx$ covers crossings in one go (definite integral). A region that is symmetric about a line can be found as twice one half (Example 5.4 uses the line $y = x$ ).
Evaluate
(exactly on ) and add the pieces. The area is positive.

Common mistakes

Bottom minus top, and the order of subtraction.
In "find the area", either order scores if the final area is made positive, so a negative answer left as it is loses the mark. In "show that the area is $\int \ldots$ ", the order is itself the mark. Test a value to see which graph is on top.
One integral across a crossing,
which cancels part of the area (Example 5.5).
Missing where an edge changes
(Example 5.3): the $x$ -axis is an edge for part of the way only.

More mistakes (3)

Losing a meeting point,
for example $x = 0$ by dividing by $x$ , or finding only one on the GDC: the limits are then wrong.
Rounded limits on
use the stored values.
Assuming that $f$ and $f^{-1}$ always meet on $y = x$
that is guaranteed for an $f$ increasing on its whole domain (increasing on each branch separately is not enough), and for a linear-over-linear $f$ that is not self-inverse, but not in general. A decreasing $f$ can meet its inverse off the line: $f(x) = -x^3$ meets it at $(1, -1)$ and $(-1, 1)$ , as well as at the origin, and $f(x) = \frac{1}{x}$ is its own inverse, so the graphs share every point.

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