Area between a curve and a line, or two curves

AS · P1 · 15 min

Many Paper 1 area questions are not about the region under a single curve, but about the region trapped between a curve and a straight line, or between two curves. The idea is one subtraction: the area between two graphs is the area under the top one minus the area under the bottom one. The work is in finding where the graphs meet, deciding which is on top, and handling regions whose boundary changes part-way across. The syllabus asks you to find the area of a region bounded by a curve and a line, or by two curves.

Top minus bottom

Take the line $y = x + 2$ and the curve $y = x^2$ . They meet where $x^2 = x + 2$ , that is at $x = -1$ and $x = 2$ . Between these values the line lies above the curve.

y = x + 2 y = x^2 (-1, 1) (2, 4)

The area under the line from $-1$ to $2$ includes the region you want plus the region under the curve. Subtracting the area under the curve leaves exactly the region between them:

\text{area} = \int_{-1}^{2} (x + 2)\, dx - \int_{-1}^{2} x^2\, dx = \int_{-1}^{2} \big[(x + 2) - x^2\big]\, dx

Combining into one integral is quicker and less error-prone. Another way to see it: the area between two curves is the area under the single curve $y = (\text{top}) - (\text{bottom})$ . Here that curve is $y = x + 2 - x^2$ , and the shaded area under it is the same size as the region between the line and the parabola.

y = x + 2 - x^2 fill -1 2 y = x + 2 - x^2

Key result

If $y = f(x)$ is above $y = g(x)$ for $a \le x \le b$ , the area between them is

\int_a^b \big[f(x) - g(x)\big]\, dx \qquad \text{(top minus bottom)}

This holds wherever the region sits: above the $x$ -axis, below it, or crossing it.

The last line is worth dwelling on. With a single curve, a region below the $x$ -axis gives a negative integral and has to be handled separately (see definite integrals and areas). With two graphs, $f(x) - g(x)$ is the vertical gap between them, which is positive whenever $f$ is on top, regardless of where the axis is. So top minus bottom never needs splitting at the $x$ -axis. Example 3 shows this.

Finding the limits and the top curve

The limits are almost always the $x$ -coordinates of the intersection points. Find them by solving the equations simultaneously, as in simultaneous equations.

To decide which graph is on top, substitute any $x$ between the limits into both equations. The larger value is the top graph. Your sketch should agree.

Area between two graphs

Sketch both graphs and shade the region.
Solve the equations simultaneously to find the $x$ -coordinates of the intersection points.
Check which graph is on top by substituting a value between the limits.
Write a single integral of (top) $-$ (bottom), simplify the integrand, then integrate.
Substitute the limits, showing both $F(b)$ and $F(a)$ .
If the answer is negative, you have the order the wrong way round; the area is positive.

A useful check for a parabola and a line

If a curve $y = \pm x^2 + \ldots$ (coefficient of $x^2$ equal to $\pm 1$ ) meets a line at $x = \alpha$ and $x = \beta$ , the area between them is always $\dfrac{(\beta - \alpha)^3}{6}$ . For $y = x^2$ and $y = x + 2$ , $\dfrac{3^3}{6} = \dfrac{9}{2}$ . Use it to check, not as your working: examiners need to see the integration.

Regions whose boundary changes

Some regions are bounded by three graphs, for example a curve, a line and the $x$ -axis. The top boundary or the bottom boundary changes part-way across, so a single "top minus bottom" does not cover the whole region.

There are two ways through:

Split the region at the $x$ -value where the boundary changes, and add the areas of the pieces.
Use a shape you already know. Straight-line boundaries make triangles and trapezia whose areas need no integration. Find the area of the simple shape, then add or subtract the area under the curve.

Example 4 does the second, using a trapezium under a tangent. Example 5 does the first, splitting at an intersection point and using a triangle for the straight part.

Watch out

Squaring or reversing. The integrand is (top) $-$ (bottom), not (bottom) $-$ (top) and certainly not $f(x)^2 - g(x)^2$ (that is for volumes).

Sign errors in the subtraction. $(x - 4) - (x^2 - 4x) = -x^2 + 5x - 4$ . Bracket the bottom function before subtracting, so every term changes sign.

Splitting at the $x$ -axis when you do not need to. Between two graphs, top minus bottom already measures the gap. Splitting at the axis is only for a single curve against the axis.

Using the wrong limits. The limits are where the graphs meet, not where either graph crosses the $x$ -axis, unless the axis is one of the boundaries.

Assuming the line is on top. Check with a value. For $y = 9 - x^2$ and its normal, the curve is on top.

Worked examples

A parabola and a line

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

Solution

Intersections: $x^2 = x + 2$ , so $x^2 - x - 2 = 0$ , $(x - 2)(x + 1) = 0$ , $x = -1$ or $x = 2$ .

At $x = 0$ the line gives $2$ and the curve gives $0$ , so the line is on top.

\begin{aligned} \text{Area} &= \int_{-1}^{2} \big(x + 2 - x^2\big)\, dx = \left[\frac{x^2}{2} + 2x - \frac{x^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{27}{6} = \frac{9}{2} \end{aligned}

Two parabolas

Find the area of the region enclosed by the curves $y = x^2$ and $y = 8 - x^2$ .

Solution

Intersections: $x^2 = 8 - x^2$ , so $x^2 = 4$ , $x = \pm 2$ .

At $x = 0$ : $8 - x^2 = 8$ and $x^2 = 0$ , so $y = 8 - x^2$ is on top.

\text{Area} = \int_{-2}^{2} \big(8 - x^2 - x^2\big)\, dx = \int_{-2}^{2} \big(8 - 2x^2\big)\, dx = \left[8x - \frac{2x^3}{3}\right]_{-2}^{2}

= \left(16 - \frac{16}{3}\right) - \left(-16 + \frac{16}{3}\right) = 32 - \frac{32}{3} = \frac{64}{3}

y = x^2 y = 8 - x^2 (-2, 4) (2, 4)

A region below the x-axis

Find the area of the region enclosed by the curve $y = x^2 - 4x$ and the line $y = 2x - 8$ .

Solution

Intersections: $x^2 - 4x = 2x - 8$ , so $x^2 - 6x + 8 = 0$ , $(x - 2)(x - 4) = 0$ , $x = 2$ or $x = 4$ .

At $x = 3$ : the line gives $-2$ and the curve gives $-3$ . The line is on top (less negative).

\begin{aligned} \text{Area} &= \int_2^4 \big[(2x - 8) - (x^2 - 4x)\big]\, dx = \int_2^4 \big(-x^2 + 6x - 8\big)\, dx \\ &= \left[-\frac{x^3}{3} + 3x^2 - 8x\right]_2^4 \\ &= \left(-\frac{64}{3} + 48 - 32\right) - \left(-\frac{8}{3} + 12 - 16\right) \\ &= -\frac{16}{3} - \left(-\frac{20}{3}\right) = \frac{4}{3} \end{aligned}

The whole region lies below the $x$ -axis, yet top minus bottom gives a positive area directly. No splitting is needed.

y = x^2 - 4x y = 2x - 8 (2, -4) (4, 0)

A curve, its tangent and the y-axis

The curve $y = 2\sqrt{x}$ has a tangent at the point $P(4, 4)$ .

(a) Find the equation of the tangent at $P$ .

(b) Find the area of the region bounded by the curve, the tangent and the $y$ -axis.

Solution

(a) $y = 2x^{\frac{1}{2}}$ , so $\dfrac{dy}{dx} = x^{-\frac{1}{2}} = \dfrac{1}{\sqrt{x}}$ , which is $\tfrac{1}{2}$ at $x = 4$ . The tangent is $y - 4 = \tfrac{1}{2}(x - 4)$ , that is $y = \tfrac{1}{2}x + 2$ .

(b) The region runs from $x = 0$ (the $y$ -axis) to $x = 4$ , with the tangent above the curve. (A tangent to this curve lies above it: at $x = 1$ the tangent gives $2.5$ , the curve $2$ .)

Method 1, one integral.

\begin{aligned} \int_0^4 \left(\tfrac{1}{2}x + 2 - 2x^{\frac{1}{2}}\right) dx &= \left[\frac{x^2}{4} + 2x - \frac{4}{3}x^{\frac{3}{2}}\right]_0^4 \\ &= 4 + 8 - \frac{4}{3}(8) - 0 = 12 - \frac{32}{3} = \frac{4}{3} \end{aligned}

Method 2, trapezium minus curve. Under the tangent from $x = 0$ to $x = 4$ is a trapezium with parallel sides $2$ and $4$ and width $4$ : area $\tfrac{1}{2}(2 + 4)(4) = 12$ . Under the curve: $\displaystyle\int_0^4 2x^{\frac{1}{2}}\, dx = \left[\tfrac{4}{3}x^{\frac{3}{2}}\right]_0^4 = \tfrac{32}{3}$ . The area is $12 - \tfrac{32}{3} = \tfrac{4}{3}$ .

y = 2 sqrt(x) y = x/2 + 2 (4, 4) (0, 2)

Curve, line and x-axis

The curve $y = (x - 3)^2$ and the line $y = 9 - 2x$ meet at two points.

(a) Find the area of the region enclosed between the curve and the line.

(b) Find the area of the region bounded by the curve, the line and the $x$ -axis, which lies to the right of $x = 3$ .

Solution

(a) $(x - 3)^2 = 9 - 2x$ gives $x^2 - 6x + 9 = 9 - 2x$ , so $x^2 - 4x = 0$ , $x = 0$ or $x = 4$ . The intersection points are $(0, 9)$ and $(4, 1)$ . At $x = 2$ the line gives $5$ and the curve $1$ , so the line is on top.

\int_0^4 \big[9 - 2x - (x - 3)^2\big]\, dx = \int_0^4 \big(4x - x^2\big)\, dx = \left[2x^2 - \frac{x^3}{3}\right]_0^4 = 32 - \frac{64}{3} = \frac{32}{3}

(b) The curve touches the $x$ -axis at $(3, 0)$ , meets the line at $(4, 1)$ , and the line meets the $x$ -axis at $\left(\tfrac{9}{2}, 0\right)$ . The region has the curve as its upper boundary from $x = 3$ to $x = 4$ , then the line from $x = 4$ to $x = \tfrac{9}{2}$ . Split at $x = 4$ .

Under the curve:

\int_3^4 (x - 3)^2\, dx = \left[\frac{(x - 3)^3}{3}\right]_3^4 = \frac{1}{3}

Under the line: a triangle with base $\tfrac{9}{2} - 4 = \tfrac{1}{2}$ and height $1$ , area $\tfrac{1}{2} \times \tfrac{1}{2} \times 1 = \tfrac{1}{4}$ .

Total area: $\tfrac{1}{3} + \tfrac{1}{4} = \tfrac{7}{12}$ .

y = (x - 3)^2 y = 9 - 2x fill 3 4 y = (x - 3)^2 fill 4 4.5 y = 9 - 2x (4, 1) (0, 9)

The shaded region in part (b) is the curved piece from $x = 3$ to $4$ plus the small triangle from $x = 4$ to $4.5$ .

Exam tip

Show the intersections. Solving for the limits usually earns its own marks. Give them as exact values.
Write the integrand before integrating. " $\int_0^4 (4x - x^2)\,dx$ " shows the subtraction and simplification. An examiner can then award the method mark even if you slip later.
Show the substitution of both limits. Writing $\left(32 - \tfrac{64}{3}\right) - 0$ is safer than jumping to $\tfrac{32}{3}$ .
Triangles and trapezia from straight boundaries can be found by geometry. That is quicker and is fully accepted, as long as you state the shape and its dimensions.
"Find the area of the shaded region" questions often need two or three parts added or subtracted. Say in words what you are doing: "area = trapezium $-$ area under curve".
Exact answers. Leave $\tfrac{7}{12}$ , not $0.583$ , unless told otherwise.

Summary

Area between two graphs $=\displaystyle\int_a^b (\text{top} - \text{bottom})\, dx$ .
The limits are the $x$ -coordinates of the intersection points; find them by solving simultaneously.
Decide which graph is on top by substituting a value between the limits.
Top minus bottom is positive even if the region is below or across the $x$ -axis: no splitting at the axis.
Regions bounded by three graphs (e.g. curve, line, $x$ -axis) are split where the boundary changes.
Use triangles and trapezia for straight-line pieces, then add or subtract the area under the curve.
For a parabola with $x^2$ coefficient $\pm 1$ and a line meeting at $\alpha$ , $\beta$ , the area is $\tfrac{(\beta - \alpha)^3}{6}$ : a quick check.

Practice questions

Question

Find the area of the region enclosed by the curve $y = x^2$ and the line $y = 3x$ .
Find the area of the region enclosed by the curve $y = 6 - x^2$ and the line $y = x$ .
Find the area of the region enclosed by the curves $y = x^2$ and $y = \sqrt{x}$ .
Find the area of the region enclosed by the curves $y = x^2 - 2x$ and $y = 2x - x^2$ .
Find the area of the region in the first quadrant enclosed by the curve $y = x^3$ and the line $y = 4x$ .
Find the equation of the tangent to the curve $y = x^2 + 1$ at the point $(1, 2)$ , and find the area of the region bounded by the curve, the tangent and the $y$ -axis.
The normal to the curve $y = 9 - x^2$ at the point $(1, 8)$ meets the curve again at $Q$ . Find the $x$ -coordinate of $Q$ and the area of the region enclosed between the curve and the normal.
The curve $y = (x - 1)^2$ and the line $y = x + 1$ meet at $(0, 1)$ and at one other point. Find the other point, and find the area of the region bounded by the curve, the line and the $x$ -axis.

Answers

$x^2 = 3x$ gives $x = 0, 3$ . Line on top. $\displaystyle\int_0^3 (3x - x^2)\, dx = \left[\tfrac{3x^2}{2} - \tfrac{x^3}{3}\right]_0^3 = \tfrac{27}{2} - 9 = \tfrac{9}{2}$ .
$6 - x^2 = x$ gives $x^2 + x - 6 = 0$ , $x = -3$ or $2$ . Curve on top (at $x = 0$ : $6 > 0$ ). $\displaystyle\int_{-3}^{2} (6 - x - x^2)\, dx = \left[6x - \tfrac{x^2}{2} - \tfrac{x^3}{3}\right]_{-3}^{2} = \left(12 - 2 - \tfrac{8}{3}\right) - \left(-18 - \tfrac{9}{2} + 9\right) = \tfrac{22}{3} + \tfrac{27}{2} = \tfrac{125}{6}$ .
$x^2 = \sqrt{x}$ gives $x^4 = x$ , $x = 0$ or $1$ . At $x = \tfrac{1}{4}$ , $\sqrt{x} = \tfrac{1}{2} > \tfrac{1}{16}$ , so $\sqrt{x}$ is on top. $\displaystyle\int_0^1 \left(x^{\frac{1}{2}} - x^2\right) dx = \left[\tfrac{2}{3}x^{\frac{3}{2}} - \tfrac{x^3}{3}\right]_0^1 = \tfrac{2}{3} - \tfrac{1}{3} = \tfrac{1}{3}$ .
$x^2 - 2x = 2x - x^2$ gives $2x^2 - 4x = 0$ , $x = 0$ or $2$ . At $x = 1$ : $2x - x^2 = 1$ , $x^2 - 2x = -1$ , so $y = 2x - x^2$ is on top. $\displaystyle\int_0^2 (4x - 2x^2)\, dx = \left[2x^2 - \tfrac{2x^3}{3}\right]_0^2 = 8 - \tfrac{16}{3} = \tfrac{8}{3}$ .
$x^3 = 4x$ gives $x(x^2 - 4) = 0$ ; in the first quadrant $x = 0$ to $x = 2$ . At $x = 1$ , $4x = 4 > 1$ , line on top. $\displaystyle\int_0^2 (4x - x^3)\, dx = \left[2x^2 - \tfrac{x^4}{4}\right]_0^2 = 8 - 4 = 4$ .
$\dfrac{dy}{dx} = 2x = 2$ at $x = 1$ , so the tangent is $y - 2 = 2(x - 1)$ , $y = 2x$ . The curve is above the tangent. Area $= \displaystyle\int_0^1 (x^2 + 1 - 2x)\, dx = \int_0^1 (x - 1)^2\, dx = \left[\tfrac{(x - 1)^3}{3}\right]_0^1 = 0 - \left(-\tfrac{1}{3}\right) = \tfrac{1}{3}$ .
$\dfrac{dy}{dx} = -2x = -2$ at $x = 1$ , so the normal gradient is $\tfrac{1}{2}$ and the normal is $y - 8 = \tfrac{1}{2}(x - 1)$ , $y = \tfrac{1}{2}x + \tfrac{15}{2}$ . Meeting the curve: $9 - x^2 = \tfrac{1}{2}x + \tfrac{15}{2}$ , so $2x^2 + x - 3 = 0$ , $(2x + 3)(x - 1) = 0$ , and $Q$ has $x = -\tfrac{3}{2}$ . The curve is on top (at $x = 0$ : $9 > 7.5$ ). $\int_{-3/2}^{1} \left(\tfrac{3}{2} - \tfrac{1}{2}x - x^2\right) dx = \left[\tfrac{3x}{2} - \tfrac{x^2}{4} - \tfrac{x^3}{3}\right]_{-3/2}^{1} = \left(\tfrac{3}{2} - \tfrac{1}{4} - \tfrac{1}{3}\right) - \left(-\tfrac{9}{4} - \tfrac{9}{16} + \tfrac{9}{8}\right) = \tfrac{11}{12} + \tfrac{27}{16} = \tfrac{125}{48}$
$(x - 1)^2 = x + 1$ gives $x^2 - 3x = 0$ , so the other point is $(3, 4)$ . The region bounded by the curve, the line and the $x$ -axis has corners at $(-1, 0)$ (line meets axis), $(0, 1)$ (intersection) and $(1, 0)$ (curve touches axis). From $x = -1$ to $0$ the top is the line: a triangle of area $\tfrac{1}{2} \times 1 \times 1 = \tfrac{1}{2}$ . From $x = 0$ to $1$ the top is the curve: $\displaystyle\int_0^1 (x - 1)^2\, dx = \tfrac{1}{3}$ . Total $\tfrac{1}{2} + \tfrac{1}{3} = \tfrac{5}{6}$ .

Top minus bottom

Finding the limits and the top curve

Regions whose boundary changes

Worked examples

Practice questions

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