Mathematics guide, Unit 8, step 26 of 33Previous step Next step

Step 26: Area Between Curves and Average Value

By the EconTutorials team. 11-minute read with 6 practice questions.

SyllabusAP 8.1–8.6 · 9709 P1 1.8 · IB SL 5.11

In 30 seconds

The average value of $ f $ on $ [a, b] $ is $ \dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx $: the height of the rectangle with the same area as the region under the curve. A continuous $ f $ actually reaches that height somewhere.
Integrating a rate gives a change: $ \displaystyle\int_a^b v(t)\,dt $ is displacement, $ \displaystyle\int_a^b |v(t)|\,dt $ is total distance, and amount now $ = $ amount at the start $ + \displaystyle\int $ (rate in $-$ rate out).
The area between two curves is $ \displaystyle\int_a^b (\text{top}-\text{bottom})\,dx $. If the curves cross, split at every crossing. For some regions, horizontal strips are easier: $ \displaystyle\int_c^d (\text{right}-\text{left})\,dy $.
Exam point: an area is never negative or zero. Find every intersection first, test a point in each piece, and say what an integral means in context, with units.

Video lesson coming soonA short animated lesson for this step will appear here.

The average value of a function

What was the average temperature yesterday? You could average 24 hourly readings, but temperature changes continuously, so no finite list captures it. The integral does. Take $ n $ equally spaced readings $ f(x_1), \ldots, f(x_n) $ on $ [a, b] $, with spacing $ \Delta x = \tfrac{b-a}{n} $. Their ordinary average is

\[ \frac{f(x_1) + \cdots + f(x_n)}{n} = \frac{1}{b-a}\sum_{i=1}^{n} f(x_i)\,\Delta x. \]

As $ n \to \infty $ this tends to

\[ f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx, \]

because the sum is a Riemann sum (Step 20). So the average value of a function is its integral divided by the length of the interval.

Key termMean Value Theorem for integrals: if $ f $ is continuous on $ [a, b] $, there is at least one $ c $ in $ [a, b] $ with $ f(c) = \dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx $. The average lies between the least and greatest values of $ f $, and by the Intermediate Value Theorem a continuous function takes every value in between.

Worked example

Find the average value of $ f(x) = 4x-x^2 $ on $ [0, 3] $, and every value of $ c $ in $ [0, 3] $ at which $ f(c) $ equals that average.

Step 1Integrate: $ \displaystyle\int_0^3 (4x-x^2)\,dx = \left[2x^2-\frac{x^3}{3}\right]_0^3 = 18-9 = 9 $.

Step 2Divide by the length of the interval: $ f_{\text{avg}} = \tfrac{9}{3} = 3 $.

Step 3Solve $ 4c-c^2 = 3 $: $ c^2-4c + 3 = (c-1)(c-3) = 0 $, so $ c = 1 $ or $ c = 3 $.

Both values lie in $ [0, 3] $. The theorem promises at least one; here there are two. In Figure 1 the rectangle of height 3 has area 9, the same as the region under the curve.

Average value of f(x) = 4x − x² on [0, 3]

y = 4x − x²rectangle of height f_avg = 3, area 9

Figure 1: the rectangle on [0, 3] with height 3 has the same area, 9, as the region under the curve. The curve reaches the average height at c = 1 and c = 3.

Common mistakeForgetting to divide by $ b-a $, or averaging the end values. $ \displaystyle\int_0^3 (4x-x^2)\,dx = 9 $ is an area, not an average. And $ \tfrac{f(0) + f(3)}{2} = 1.5 $ is the average of two readings, not of the function.

An economics example: an average price. A share price, in dollars, $ t $ months after listing is $ P(t) = 20 + 3\sqrt{t} $. Its average over the first four months is $ \tfrac{1}{4}\displaystyle\int_0^4 (20 + 3t^{1/2})\,dt = \tfrac{1}{4}\left[20t + 2t^{3/2}\right]_0^4 = \tfrac{1}{4}(80 + 16) = 24 $ dollars. Average cost, average inventory and average revenue over a period are all found the same way.

Motion: displacement and total distance

In Step 15 you differentiated position to get velocity. Now run it backwards. Velocity is the rate of change of position, so the integral of velocity is the change in position:

\[ s(b) = s(a) + \int_a^b v(t)\,dt \]

The integral on its own, $ \displaystyle\int_a^b v(t)\,dt $, is the displacement. The total distance travelled is $ \displaystyle\int_a^b |v(t)|\,dt $.

Why does distance need the modulus? Over a short time $ dt $ the particle moves $ v\,dt $. When $ v < 0 $ that is a movement backwards, and the integral counts it as negative, cancelling earlier forward movement. Displacement wants exactly that: it asks where you ended up. Distance does not: walking 3 m forward and 3 m back is 6 m of walking. On a velocity–time graph, displacement is the area above the axis minus the area below; distance is the two added.

Worked example

A particle moves along a line with velocity $ v(t) = t^2-5t + 4 $ m/s for $ 0 \le t \le 5 $, and $ s(0) = 1 $ m. Find (a) the displacement, (b) the total distance travelled, (c) the position at $ t = 5 $.

Step 1$ v = (t-1)(t-4) $, so $ v > 0 $ on $ [0, 1) $ and $ (4, 5] $, and $ v < 0 $ on $ (1, 4) $. An antiderivative is $ F(t) = \frac{t^3}{3}-\frac{5t^2}{2} + 4t $, with $ F(0) = 0 $, $ F(1) = \tfrac{11}{6} $, $ F(4) = -\tfrac{8}{3} $, $ F(5) = -\tfrac{5}{6} $.

Step 2(a) Displacement $ = F(5)-F(0) = -\tfrac{5}{6} $ m.

Step 3(b) Split where $ v $ changes sign: the three pieces are $ \tfrac{11}{6} $, $ -\tfrac{8}{3}-\tfrac{11}{6} = -\tfrac{9}{2} $ and $ -\tfrac{5}{6} + \tfrac{8}{3} = \tfrac{11}{6} $. Distance $ = \tfrac{11}{6} + \tfrac{9}{2} + \tfrac{11}{6} = \tfrac{49}{6} \approx 8.17 $ m.

Step 4(c) $ s(5) = 1 + \left(-\tfrac{5}{6}\right) = \tfrac{1}{6} $ m.

The particle travels more than 8 m but finishes less than 1 m from where it started.

Velocity v(t) = (t − 1)(t − 4) m/s

v(t) = t² − 5t + 4

Figure 2: signed areas +11/6, −9/2 and +11/6. Displacement adds them with their signs, giving −5/6 m; distance adds their sizes, giving 49/6 m.

Exam tipAlways find where $ v = 0 $ before a distance question, and check that $ v $ really changes sign there: $ v = (t-2)^2 $ touches zero without the particle turning back. On the AP calculator section you may type $ \int_a^b |v(t)|\,dt $ straight into the calculator. Kinematics with integrals is IB SL 5.9; in Cambridge 9709 it belongs to the Mechanics papers rather than Pure Mathematics.

Accumulation: rate in, rate out

Motion is one case of a general rule, the net change theorem: if you know the rate at which a quantity changes, the integral of the rate is the change in the quantity.

\[ A(t) = A(0) + \int_0^t \big(\text{rate in}-\text{rate out}\big)\,du \]

Start with what you had, add what flowed in, subtract what flowed out. The units of the integral are the rate’s units times time: litres per hour times hours gives litres.

Worked example

A tank holds 200 litres at $ t = 0 $. Water flows in at a steady 48 litres per hour and drains out at $ O(t) = 3t^2 $ litres per hour, for $ 0 \le t \le 6 $. (a) How much water drains out in the 6 hours? (b) How much is in the tank at $ t = 6 $? (c) When is the amount greatest, and what is it?

Step 1(a) $ \displaystyle\int_0^6 3t^2\,dt = \big[t^3\big]_0^6 = 216 $ litres.

Step 2(b) $ A(t) = 200 + \displaystyle\int_0^t (48-3u^2)\,du = 200 + 48t-t^3 $, so $ A(6) = 200 + 288-216 = 272 $ litres.

Step 3(c) $ A'(t) = 48-3t^2 = 0 $ at $ t = 4 $, where $ A’ $ changes from positive to negative. Compare the candidates: $ A(0) = 200 $, $ A(4) = 200 + 192-64 = 328 $, $ A(6) = 272 $.

The tank holds most water, 328 litres, at $ t = 4 $ hours: the moment the outflow rate catches up with the inflow rate.

An economics example: stocks, flows and present value. Economists separate stocks, measured at a moment (capital, inventories, debt), from flows, measured per unit of time (investment, production, borrowing). A flow is the rate of change of a stock, so a stock is accumulated flow. The same slicing values a stream of income. Income of $ R(t) $ dollars per year earned between $ t $ and $ t + dt $ is worth $ R(t)e^{-rt}\,dt $ today, so the present value of the whole stream is $ PV = \displaystyle\int_0^T R(t)e^{-rt}\,dt $. For $2000 a year for 10 years at $ r = 0.04 $: $ PV = 50{,}000\left(1-e^{-0.4}\right) $, about $16,484, against $20,000 undiscounted.

The area between two curves

Slice the region into thin vertical strips. A strip at $ x $ has height (top curve) minus (bottom curve) and width $ dx $. Add the strips:

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

Here $ f(x) \ge g(x) $ on $ [a, b] $, so $ f $ is the top curve.

This works even when both curves are below the $ x $-axis. The height of a strip is a difference of $ y $-values, and a difference does not care where the axis is: a strip from $ y = -5 $ up to $ y = -1 $ is 4 units tall.

Worked example

Find the area of the region enclosed by $ y = 2x $ and $ y = x^2-3 $.

Step 1Intersections: $ x^2-3 = 2x $, so $ x^2-2x-3 = (x-3)(x + 1) = 0 $, giving $ x = -1 $ and $ x = 3 $.

Step 2Which is on top? At $ x = 0 $ the line gives 0 and the parabola $ -3 $, so the line is on top throughout.

Step 3$ A = \displaystyle\int_{-1}^{3} \big(2x-x^2 + 3\big)\,dx = \left[x^2-\frac{x^3}{3} + 3x\right]_{-1}^{3} = 9-\left(-\tfrac{5}{3}\right) = \tfrac{32}{3} $.

The area is $ \tfrac{32}{3} \approx 10.7 $ square units.

When the curves cross more than once, the one on top changes. Split at every crossing and make each piece top minus bottom.

Worked example

Find the total area enclosed between $ y = x^3 $ and $ y = x^2 + 2x $.

Step 1$ x^3 = x^2 + 2x $ gives $ x^3-x^2-2x = x(x-2)(x + 1) = 0 $: crossings at $ x = -1, 0, 2 $.

Step 2Test a point in each piece. At $ x = -\tfrac{1}{2} $: cubic $ -\tfrac{1}{8} $, quadratic $ -\tfrac{3}{4} $, so the cubic is on top on $ [-1, 0] $. At $ x = 1 $: cubic 1, quadratic 3, so the quadratic is on top on $ [0, 2] $.

Step 3$ \displaystyle\int_{-1}^{0} (x^3-x^2-2x)\,dx = \tfrac{5}{12} $ and $ \displaystyle\int_{0}^{2} (x^2 + 2x-x^3)\,dx = \tfrac{8}{3} $.

Total area $ = \tfrac{5}{12} + \tfrac{8}{3} = \tfrac{37}{12} \approx 3.08 $ square units.

The area between y = x³ and y = x² + 2x

y = x³y = x² + 2x

Figure 3: the cubic is on top on [−1, 0] and the quadratic on [0, 2]. Each piece is integrated as top minus bottom and the two areas are added: 5/12 + 8/3 = 37/12.

Common mistakeIntegrating once across all the crossings. $ \displaystyle\int_{-1}^{2} \big(x^2 + 2x-x^3\big)\,dx = \tfrac{9}{4} $, which is $ \tfrac{8}{3}-\tfrac{5}{12} $: the left-hand region has been subtracted instead of added. When two pieces are equal, a single integral even gives 0. Find every intersection, test a point in each piece, and add the pieces.

An economics example: consumer and producer surplus. Demand is $ P = 120-2Q^2 $ and supply is $ P = 30 + 8Q $. They meet where $ 2Q^2 + 8Q-90 = 0 $, that is $ (Q + 9)(Q-5) = 0 $, so $ Q_0 = 5 $ and $ P_0 = 70 $ dollars. Consumer surplus is the area between the demand curve and the price line: $ \displaystyle\int_0^5 \big(120-2Q^2-70\big)\,dQ = 250-\tfrac{250}{3} = \tfrac{500}{3} $, about $166.67. Producer surplus is the area between the price line and supply: $ \displaystyle\int_0^5 (70-30-8Q)\,dQ = 100 $ dollars. Both are areas between curves, with the horizontal line $ P = P_0 $ as one of them.

Integrating with respect to $ y $

Some regions are awkward to slice vertically, because the curve at the top or bottom of a strip changes part-way across. Slice horizontally instead. A strip at height $ y $ has length (right curve) minus (left curve) and thickness $ dy $:

\[ A = \int_c^d \big[p(y)-q(y)\big]\,dy = \int_c^d (\text{right}-\text{left})\,dy. \]

Worked example

Find the area enclosed by the parabola $ x = y^2-4 $ and the line $ x = y + 2 $.

Step 1Intersections: $ y^2-4 = y + 2 $, so $ y^2-y-6 = (y-3)(y + 2) = 0 $: $ y = -2 $ and $ y = 3 $, at the points $ (0, -2) $ and $ (5, 3) $.

Step 2Which is on the right? At $ y = 0 $ the line gives $ x = 2 $ and the parabola $ x = -4 $, so the line is on the right.

Step 3$ A = \displaystyle\int_{-2}^{3} \big(y + 2-y^2 + 4\big)\,dy = \left[6y + \frac{y^2}{2}-\frac{y^3}{3}\right]_{-2}^{3} = \tfrac{27}{2}-\left(-\tfrac{22}{3}\right) = \tfrac{125}{6} $.

With vertical strips you would need two integrals, because to the left of $ x = 0 $ every strip runs between the two halves of the parabola, while to the right it runs from the line up to the parabola.

Horizontal strips: x = y² − 4 and x = y + 2

x = y² − 4x = y + 2

Figure 4: every horizontal strip runs from the parabola on the left to the line on the right, so one integral in y, from y = −2 to y = 3, covers the whole region.

Exam tipOnce you integrate with respect to $ y $, the limits are $ y $-values and every curve must be written as $ x $ in terms of $ y $. Areas between curves are on every syllabus: AP 8.4 to 8.6, Cambridge 9709 P1 1.8 (a curve and a line, two curves, or a curve and the $ y $-axis) and IB SL 5.11. In IB, area bounded by a curve and the $ y $-axis is HL (AHL 5.17).

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Find the average value of $ f(x) = 3x^2 $ on $ [1, 3] $. Show worked answer

$ \displaystyle\int_1^3 3x^2\,dx = \big[x^3\big]_1^3 = 27-1 = 26 $.
Divide by $ 3-1 = 2 $: $ f_{\text{avg}} = 13 $.

EasyQ2. Find the area of the region enclosed by $ y = x^2 $ and $ y = 4 $. Show worked answer

Intersections: $ x^2 = 4 $, so $ x = \pm 2 $. The line is on top.
$ A = \displaystyle\int_{-2}^{2} (4-x^2)\,dx = \left[4x-\frac{x^3}{3}\right]_{-2}^{2} = \tfrac{16}{3}-\left(-\tfrac{16}{3}\right) = \tfrac{32}{3} $.

MediumQ3. A particle has velocity $ v(t) = 3t^2-12t + 9 $ m/s for $ 0 \le t \le 4 $. Find its displacement and the total distance it travels. Show worked answer

$ v = 3(t-1)(t-3) $, which changes sign at $ t = 1 $ and $ t = 3 $.
$ F(t) = t^3-6t^2 + 9t $: $ F(0) = 0 $, $ F(1) = 4 $, $ F(3) = 0 $, $ F(4) = 4 $.
Displacement $ = F(4)-F(0) = 4 $ m.
Pieces: $ 4 $, $ 0-4 = -4 $, $ 4-0 = 4 $. Distance $ = 4 + 4 + 4 = 12 $ m.

MediumQ4. Find the total area between $ y = \sin x $ and $ y = \cos x $ for $ 0 \le x \le \pi $. Show worked answer

They cross where $ \tan x = 1 $: $ x = \tfrac{\pi}{4} $. Cosine is on top before it, sine after.
$ \displaystyle\int_0^{\pi/4} (\cos x-\sin x)\,dx = \big[\sin x + \cos x\big]_0^{\pi/4} = \sqrt{2}-1 $.
$ \displaystyle\int_{\pi/4}^{\pi} (\sin x-\cos x)\,dx = \big[-\cos x-\sin x\big]_{\pi/4}^{\pi} = 1 + \sqrt{2} $.
Total $ = 2\sqrt{2} \approx 2.83 $.

MediumQ5. Find the area of the region bounded by $ y = \sqrt{x} $, the $ y $-axis and the line $ y = 3 $. Show worked answer

Use horizontal strips: $ x = y^2 $, from the $ y $-axis ($ x = 0 $) to the curve, for $ 0 \le y \le 3 $.
$ A = \displaystyle\int_0^3 y^2\,dy = \left[\frac{y^3}{3}\right]_0^3 = 9 $.
Check: the rectangle $ 0 \le x \le 9 $, $ 0 \le y \le 3 $ has area 27, and the area under $ y = \sqrt{x} $ is $ \displaystyle\int_0^9 \sqrt{x}\,dx = 18 $; $ 27-18 = 9 $.

ExamQ6. (AP style, no calculator) A museum has 1000 visitors inside at $ t = 0 $. For $ 0 \le t \le 6 $ hours, visitors arrive at a constant 300 per hour and leave at $ L(t) = 30t^2 + 30 $ per hour. (a) How many visitors leave in the first 2 hours? [2] (b) Is the number inside increasing or decreasing at $ t = 4 $? Give a reason. [2] (c) Find the greatest number of visitors inside during $ 0 \le t \le 6 $. Justify your answer. [3] (d) Find the average rate at which visitors leave over $ 0 \le t \le 6 $. [2] Show worked answer

(a) $ \displaystyle\int_0^2 (30t^2 + 30)\,dt = \big[10t^3 + 30t\big]_0^2 $ [M1] $ = 80 + 60 = 140 $ visitors. [A1]
(b) Net rate at $ t = 4 $: $ 300-L(4) = 300-510 = -210 $. [M1] This is negative, so the number inside is decreasing. [A1]
(c) $ N(t) = 1000 + \displaystyle\int_0^t (270-30u^2)\,du = 1000 + 270t-10t^3 $. $ N'(t) = 0 $ when $ 30t^2 = 270 $, so $ t = 3 $. [M1]
Candidates: $ N(0) = 1000 $, $ N(3) = 1540 $, $ N(6) = 460 $. [M1]
Greatest number: 1540 visitors, at $ t = 3 $. [A1]
(d) $ \dfrac{1}{6}\displaystyle\int_0^6 (30t^2 + 30)\,dt = \dfrac{1}{6}(2160 + 180) $ [M1] $ = 390 $ visitors per hour. [A1]
Total: 9 marks.

Check yourself

1. What is the average value of $ f(x) = x $ on $ [0, 4] $?

$ \tfrac{1}{4}\displaystyle\int_0^4 x\,dx = \tfrac{1}{4} \times 8 = 2 $. The 8 is the area; you must divide by the length of the interval.

2. A particle has $ v(t) = t-2 $ for $ 0 \le t \le 4 $. What are its displacement and total distance?

The signed areas are $ -2 $ on $ [0, 2] $ and $ +2 $ on $ [2, 4] $. Displacement adds them: 0. Distance adds their sizes: 4.

3. What is the area enclosed by $ y = x $ and $ y = x^2 $?

They meet at $ x = 0 $ and $ x = 1 $, with the line on top: $ \displaystyle\int_0^1 (x-x^2)\,dx = \tfrac{1}{2}-\tfrac{1}{3} = \tfrac{1}{6} $.

Your score: 0 out of 3

The short version

Average value: $ f_{\text{avg}} = \dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx $. A continuous function takes its average value at least once (the Mean Value Theorem for integrals).
Displacement $ = \displaystyle\int_a^b v\,dt $; distance $ = \displaystyle\int_a^b |v|\,dt $, found by splitting where $ v $ changes sign.
Accumulation: amount $ = $ starting amount $ + \displaystyle\int $ (rate in $-$ rate out). The maximum is where rate in $ = $ rate out, but compare with the end points too.
Area between curves: $ \displaystyle\int (\text{top}-\text{bottom})\,dx $, split at every crossing; or $ \displaystyle\int (\text{right}-\text{left})\,dy $ with $ y $-limits.
Consumer and producer surplus are areas between a demand or supply curve and the price line.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Step 26: Area Between Curves and Average Value

The average value of a function

Motion: displacement and total distance

Accumulation: rate in, rate out

The area between two curves

Integrating with respect to \( y \)

Practise: 6 questions

Check yourself

The short version

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