MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: areas between curves and average value (MATH 141)

Q: How do I know which curve is on top when finding the area between two curves?

Find every intersection first, by setting the curves equal and factoring. Between two consecutive intersections the curves cannot swap, so pick one test value in each piece and compare the two heights there. The larger one is the top on that whole piece. Do not trust a quick sketch on a thin piece: the test value takes ten seconds.

Q: When should I integrate with respect to y instead of x?

Trace a few vertical slices and a few horizontal ones on your sketch. Choose the direction in which every slice starts on the same curve and ends on the same curve, so one integral covers the region. If a boundary is written as x in terms of y, horizontal slices are usually that direction. With y, the bounds are y-values and the integrand is right minus left.

Q: Why is the integral of f minus g not the area when the curves cross?

Because the integral counts the piece where f is on top as positive and the piece where g is on top as negative, and the two cancel. The result is a net, the difference of the pieces. The area is their sum, so you must split at every crossing and integrate top minus bottom on each piece. Taking the absolute value at the end does not fix it.

Q: What is the average value of a function, and how is it different from the average of two values?

The average value of f on an interval is its integral divided by the length of the interval: the height of the rectangle with the same signed area. It takes every point of the graph into account. The average of the two endpoint values only sees two points, and agrees with the true average value only when the function is linear.

Q: What does the mean value theorem for integrals say, and when can I use it?

If a function is continuous on a closed interval, it actually takes its average value at some point of that interval, at least once. You can use it only after checking continuity: a function that jumps can have an average that it never reaches. When solving for that point, keep only the solutions that lie inside the interval, and list all of them.

This sheet is not a summary of sections 6.1 and 6.5 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on areas between curves and average value in MATH 141 at McGill University, and which precise gesture avoids each loss.

As on the exam, there is no calculator: every intersection below factors, every answer is exact, and every number quoted has been checked twice, once by the antiderivative and once by an independent computation.

The thread of the chapter

An area between curves is a sum of slice LENGTHS, and a length is top minus bottom (or right minus left) on EACH piece: find every intersection first, check the order on each piece, and choose the variable in which one formula describes the slice. The average value is the one integral of the chapter that keeps its sign.

This chapter is part of MATH 141, Calculus 2 (McGill)

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Area between two curves
Vertical and horizontal slices
Points of intersection
Integration with respect to y
Net signed area
Average value of a function
Mean value theorem for integrals
Comparison property of the integral
Extraneous roots

The essentials

An area is a sum of slice lengths

• Vertical slices: $A = \int_a^b \left[\text{top}(x) - \text{bottom}(x)\right] dx$ . Horizontal slices: $A = \int_c^d \left[\text{right}(y) - \text{left}(y)\right] dy$ .
• The bounds are the INTERSECTIONS, in the variable of integration: set the curves equal, bring everything to one side, FACTOR.
• Between two consecutive intersections the order of two continuous curves cannot change: one test value per piece decides which is on top.
• Where the curves cross, the top changes: split there, one integral per piece. Every piece must come out positive.
• The $x$ -axis plays no role: a negative bottom height is subtracted, which lengthens the slice. Only the crossings of the two curves matter.

y = x^3 - 3x

and

y = x

cross at

-2

0

and

2

. The two lobes have signed areas

+4

and

-4

: the net integral is

0

, the area is

8

A negative result is never an area: it is an alarm that top and bottom were swapped. And $\left|\int_a^b (f - g)\right|$ is not $\int_a^b |f - g|$ as soon as the curves cross.

The average value keeps the sign

• $f_{\text{ave}} = \frac{1}{b - a}\int_a^b f(x)\,dx$ : the height of the rectangle on $[a, b]$ with the same SIGNED area as the graph.
• It is not $\frac{f(a) + f(b)}{2}$ , except for a linear $f$ .
• If $m \le f \le M$ on $[a, b]$ , then $m \le f_{\text{ave}} \le M$ : the five-second check.
• Mean value theorem for integrals: if $f$ is CONTINUOUS on $[a, b]$ , some $c$ in $[a, b]$ has $f(c) = f_{\text{ave}}$ . At least one, possibly several, and only inside $[a, b]$ .
• An average DISTANCE between two curves uses $|f - g|$ ; an average value of $f - g$ does not.

“For $-1 \le x \le 0$ the slice goes from $-\sqrt{x + 1}$ to $\sqrt{x + 1}$ , for $0 \le x \le 3$ from $x - 1$ to $\sqrt{x + 1}$ : $A = \frac{4}{3} + \frac{19}{6} = \frac{9}{2}$ .”

Left of

x = 0

a vertical slice runs from branch to branch; right of it, from the line to the upper branch. Two formulas in

x

, one in

y

Why: Solved for $y$ , the parabola $x = y^2 - 1$ is TWO functions, $\pm\sqrt{x + 1}$ . On $[-1, 0]$ the line is outside the region and the slice runs branch to branch. Trace three slices on the sketch before writing anything; here horizontal slices avoid the split altogether.

6. The average of the endpoint values

the whole part

What not to write

“The average value of $x^2$ on $[0, 2]$ is $\frac{f(0) + f(2)}{2} = 2$ .”

What to write

“ $f_{\text{ave}} = \frac{1}{2}\int_0^2 x^2\,dx = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}$ .”

The shaded rectangle of height

\frac{4}{3}

has the same area as the region under

y = x^2

; the dashed line at

2

, the endpoint average, is too high.

Why: The endpoint average only sees two points of the graph. For a curve that bends upward the chord lies above it, so the endpoint average overestimates. The two agree only for a linear function.

7. Forgetting to divide by b minus a

1 to 2 marks

What not to write

“The average value of $\sin x$ on $[0, \pi]$ is $\int_0^\pi \sin x\,dx = 2$ .”

What to write

“ $f_{\text{ave}} = \frac{1}{\pi}\int_0^\pi \sin x\,dx = \frac{2}{\pi}$ .”

Why: An average is a total divided by a width. The check catches it at once: $\sin x \le 1$ , so its average cannot be $2$ . On $[0, 1]$ the factor is $1$ and disappears, which is how the habit of forgetting it is formed.

8. Keeping a c outside the interval

1 mark for the conclusion

What not to write

“ $f(x) = x^2$ on $[0, 3]$ has $f_{\text{ave}} = 3$ , so $c^2 = 3$ and $c = \pm\sqrt 3$ .”

What to write

“ $c^2 = 3$ with $c$ in $[0, 3]$ gives $c = \sqrt 3$ only; $-\sqrt 3$ is rejected.”

Why: The theorem places $c$ in $[a, b]$ : a root of the equation outside it is not an answer. The theorem also says at least one, so when two roots fall inside, both are listed.

9. Invoking the mean value theorem for a function that jumps

1 mark, and the whole justification

What not to write

“ $f = 1$ on $[0, 1)$ and $f = 3$ on $[1, 2]$ has average $2$ , so $f(c) = 2$ for some $c$ .”

What to write

“ $f$ is not continuous at $1$ , the theorem does not apply, and indeed $f$ never equals $2$ .”

Why: The average exists for any integrable function; the theorem that it is REACHED needs continuity. Write the hypothesis, $f$ is continuous on $[a, b]$ , before using it.

Which method to choose

dx or dy: decide on the FORM of the boundaries

Trace three vertical slices and three horizontal slices on the sketch, and ask whether each family always starts and ends on the same curves

If both curves are graphs $y = f(x)$ and the vertical slices never change their ends → integrate in $x$ , one integral

Example: $y = x + 4$ over $y = x^2 - 2x$ on $[-1, 4]$ : $\frac{125}{6}$
If a boundary is given as $x = g(y)$ , a sideways parabola → integrate in $y$ , right minus left, with $y$ -bounds

Example: $x = y + 1$ right of $x = y^2 - 1$ on $[-1, 2]$ : $\frac{9}{2}$
If three boundaries, and the bottom (or top) changes formula at a corner → try the other variable: it often sees one curve on each side

Example: $y = \sqrt x$ , $y = x - 2$ , $y = 0$ : two integrals in $x$ , one in $y$ , $\frac{10}{3}$
If the curves cross inside the region → split at every crossing, whatever the variable

Example: $\sin 2x$ and $\sin x$ on $[0, \pi]$ : pieces $\frac{1}{4}$ and $\frac{9}{4}$
If both variables give one integral → pick the easier antiderivative

Example: under $y = 3$ , above $y = e^x$ : $\int (3 - e^x)\,dx$ beats $\int \ln y\,dy$
If the question asks for an average → divide by the width; take $|f - g|$ only for an average DISTANCE

Example: $\sin x$ on $[0, 2\pi]$ averages $0$ , while the area is $4$

Whichever branch applies, the first line on the copy is the list of intersections, obtained by factoring. Every other mark depends on it.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Area between two curves, set up for full marks

When to use it: Any question that says find the area of the region enclosed by, or bounded by, two or more curves

1 Write the equation of the intersections, bring everything to one side, factor, and list ALL the roots in the relevant range.
2 State the variable of integration and why, in a few words (one formula for each end of the slice).
3 On each piece between consecutive intersections, name the top and the bottom (or right and left) with a test value.
4 Write one integral per piece, top minus bottom, then the antiderivative with its bracket and the substituted values.
5 Add the pieces, check that each one is positive, and compare the total with a simple shape that contains the region.

Concluding sentence

“On $[-1, 0]$ the cubic is on top, since at $x = -\frac{1}{2}$ it gives $-\frac{1}{8}$ against $-\frac{3}{4}$ ; on $[0, 2]$ the parabola is on top. So $A = \int_{-1}^{0} (x^3 - x^2 - 2x)\,dx + \int_0^2 (x^2 + 2x - x^3)\,dx = \frac{37}{12}$ .”

The trap: Skipping the test value and trusting the sketch: a sketch drawn quickly swaps top and bottom on a thin piece, and the piece then comes out negative.

Marking: Typically 1 to 2 marks for the intersections, 2 for the correct integrals with the split, 1 for the antiderivatives and 1 for the value. The set-up is worth more than the arithmetic.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Every piece is positive

Evaluate each piece separately before adding. A negative piece means top and bottom were swapped on that piece; a zero net with visible lobes means a crossing was ignored.

$\frac{5}{12} > 0$ and $\frac{8}{3} > 0$ : consistent. A piece of $-\frac{5}{12}$ would flag the swap.
The box test

The area is at most the width of the region times its longest slice. For a parabola cut by a chord, Archimedes gives the exact value: two thirds of the enclosing rectangle.

$y = x + 4$ over $y = x^2 - 2x$ : width $5$ , longest slice $\frac{25}{4}$ , and $\frac{2}{3} \cdot 5 \cdot \frac{25}{4} = \frac{125}{6}$ .
The average lies between the extreme values

If $m \le f \le M$ on $[a, b]$ , then $m \le f_{\text{ave}} \le M$ . It catches the forgotten $\frac{1}{b - a}$ on any long interval.

$\frac{1}{1 + x^2}$ on $[0, 1]$ lies between $\frac{1}{2}$ and $1$ , so its average $\frac{\pi}{4}$ must too: $2 \le \pi \le 4$ .
A root found by squaring is tested in the original equation

Squaring $\sqrt x = x - 2$ adds the solutions of $-\sqrt x = x - 2$ . Substitute each root back before using it as a bound.

$x = 4$ : $2 = 2$ , kept. $x = 1$ : $1 \ne -1$ , rejected.

The typical problem, taken apart

Two curves that cross three times, from the intersections to the area

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

No calculator. Every step must be justified as on a MATH 141 final.

A thin lobe on

[-1, 0]

where the cubic is on top, a large one on

[0, 2]

where the parabola is: two integrals, never one.

Step 1

$x^3 = x^2 + 2x \iff x^3 - x^2 - 2x = 0 \iff x(x - 2)(x + 1) = 0$ : intersections at $x = -1$ , $0$ and $2$ , points $(-1, -1)$ , $(0, 0)$ , $(2, 8)$ .

Why

Factoring, not dividing by $x$ , keeps the root $0$ , which is where the order changes. This line alone carries the intersection marks.

Step 2

Let $h = x^3 - (x^2 + 2x)$ . At $x = -\frac{1}{2}$ : $h = -\frac{1}{8} + \frac{3}{4} = \frac{5}{8} > 0$ , cubic on top on $(-1, 0)$ . At $x = 1$ : $h = -2 < 0$ , parabola on top on $(0, 2)$ .

Why

One test value per piece is enough, because $h$ is continuous and vanishes only at the three roots. Writing the test is what justifies the split.

Step 3

$A = \int_{-1}^{0} (x^3 - x^2 - 2x)\,dx + \int_0^2 (x^2 + 2x - x^3)\,dx$ , with $H(x) = \frac{x^4}{4} - \frac{x^3}{3} - x^2$ an antiderivative of $h$ .

Why

Top minus bottom on each piece. One antiderivative serves both pieces, with the sign flipped on the second: fewer computations, fewer slips.

Step 4

$H(0) = 0$ , $H(-1) = \frac{1}{4} + \frac{1}{3} - 1 = -\frac{5}{12}$ , $H(2) = 4 - \frac{8}{3} - 4 = -\frac{8}{3}$ . First piece $H(0) - H(-1) = \frac{5}{12}$ ; second $-(H(2) - H(0)) = \frac{8}{3}$ .

Why

Both pieces come out positive, which confirms the order found in step 2. A negative piece here would send you back to the test values.

Step 5

$A = \frac{5}{12} + \frac{32}{12} = \frac{37}{12}$ . Net check: $H(2) - H(-1) = -\frac{8}{3} + \frac{5}{12} = -\frac{9}{4} = \frac{5}{12} - \frac{8}{3}$ .

Why

The net integral is the difference of the pieces: computing it confirms both values at once, and shows why $\frac{9}{4}$ is not the area.

The conclusion, written out

“The curves meet at $x = -1$ , $0$ , $2$ ; the cubic is on top on $[-1, 0]$ and the parabola on $[0, 2]$ , so $A = \frac{5}{12} + \frac{8}{3} = \frac{37}{12}$ .”

The classic mistake on this problem: Dividing by $x$ and integrating $x^3 - x^2 - 2x$ from $-1$ to $2$ in one piece, which gives $-\frac{9}{4}$ , then reporting $\frac{9}{4}$ : off by $\frac{5}{6}$ , twice the small lobe.

Learn by heart

• $A = \int (\text{top} - \text{bottom})\,dx$ or $A = \int (\text{right} - \text{left})\,dy$ , bounds in the variable of integration.
• Intersections first: one side, FACTOR, never divide by $x$ or $\sin x$ .
• Crossing inside: split, one integral per piece, every piece positive.
• The $x$ -axis plays no role; $\left|\int (f - g)\right| \ne \int |f - g|$ once the curves cross.
• Choose the variable in which each end of the slice is one formula; tie broken by the easier antiderivative.
• $f_{\text{ave}} = \frac{1}{b - a}\int_a^b f$ ; keeps the sign; lies between $\min f$ and $\max f$ .
• Mean value theorem for integrals: $f$ continuous, some $c$ in $[a, b]$ with $f(c) = f_{\text{ave}}$ ; list every $c$ inside, reject the others.

Frequently asked questions

How do I know which curve is on top when finding the area between two curves?

Find every intersection first, by setting the curves equal and factoring. Between two consecutive intersections the curves cannot swap, so pick one test value in each piece and compare the two heights there. The larger one is the top on that whole piece. Do not trust a quick sketch on a thin piece: the test value takes ten seconds.

When should I integrate with respect to y instead of x?

Trace a few vertical slices and a few horizontal ones on your sketch. Choose the direction in which every slice starts on the same curve and ends on the same curve, so one integral covers the region. If a boundary is written as x in terms of y, horizontal slices are usually that direction. With y, the bounds are y-values and the integrand is right minus left.

Why is the integral of f minus g not the area when the curves cross?

Because the integral counts the piece where f is on top as positive and the piece where g is on top as negative, and the two cancel. The result is a net, the difference of the pieces. The area is their sum, so you must split at every crossing and integrate top minus bottom on each piece. Taking the absolute value at the end does not fix it.

What is the average value of a function, and how is it different from the average of two values?

The average value of f on an interval is its integral divided by the length of the interval: the height of the rectangle with the same signed area. It takes every point of the graph into account. The average of the two endpoint values only sees two points, and agrees with the true average value only when the function is linear.

What does the mean value theorem for integrals say, and when can I use it?

If a function is continuous on a closed interval, it actually takes its average value at some point of that interval, at least once. You can use it only after checking continuity: a function that jumps can have an average that it never reaches. When solving for that point, keep only the solutions that lie inside the interval, and list all of them.

Practise it

Corrected exercises: Areas between curves and average value, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: areas between curves and average value (MATH 141)

The essentials

An area is a sum of slice lengths

The average value keeps the sign

The mistakes that cost marks

1. One integral across a crossing

2. The absolute value put outside the integral

3. Dividing by x, or by sin x, and losing an intersection

4. x-bounds on a dy integral

5. Vertical slices on a sideways parabola, with the wrong bottom

6. The average of the endpoint values

7. Forgetting to divide by b minus a

8. Keeping a c outside the interval

9. Invoking the mean value theorem for a function that jumps

Which method to choose

dx or dy: decide on the FORM of the boundaries

How the answer is expected to be written

Area between two curves, set up for full marks

Check before you hand in

The typical problem, taken apart

Two curves that cross three times, from the intersections to the area

Learn by heart

Frequently asked questions

How do I know which curve is on top when finding the area between two curves?

When should I integrate with respect to y instead of x?

Why is the integral of f minus g not the area when the curves cross?

What is the average value of a function, and how is it different from the average of two values?

What does the mean value theorem for integrals say, and when can I use it?

Corrected exercises: Areas between curves and average value, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?