Integral Calculus 201-SN3-RE (201-NYB) • CEGEP, Montreal

Revision sheet: the area between two curves, 201-SN3-RE (201-NYB)

This sheet does not give the formula for the area between two curves again: it fits on one line. It deals with what decides the mark in 201-SN3-RE (formerly 201-NYB), that is, setting up the integral, where all the marks of a question are won or lost while the final computation is often elementary.

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

The area between two curves is always the integral of “top minus bottom”. All the work lies elsewhere: finding the bounds that the problem does not give, deciding WHICH curve is on top with a test point, and choosing the direction of the strips that requires the fewest integrals.

This chapter is part of 201-SN3-RE (201-NYB), Integral Calculus

The essentials

Area or integral: two different numbers

• $\int_{a}^{b}f$ is a SIGNED area: what lies below the axis counts negatively.
• $\mathcal{A}=\int_{a}^{b}\left|f\right|$ , computed by splitting at the ZEROS of $f$ and changing the sign of the negative pieces.
• “Integral”, “displacement”, “net change” call for the signed integral.
• “Area”, “distance travelled”, “total amount” call for the integral of the absolute value.

\left[-2, 2\right]

, the cubic

x^{3}-4x

encloses two equal regions on either side of the axis: the integral equals

0

by oddness, the AREA equals

8

A zero integral with a curve clearly away from the axis is never a zero area: it is the signal of a missing split.

The formula, for both directions of the strips

• VERTICAL strips: $\mathcal{A}=\int_{a}^{b}\left(\text{top}-\text{bottom}\right)dx$ , with $a$ and $b$ obtained by solving $f(x)=g(x)$ .
• HORIZONTAL strips: $\mathcal{A}=\int_{c}^{d}\left(\text{right}-\text{left}\right)dy$ , with $c$ and $d$ the $y$ -coordinates of the intersections.
• Both give the SAME number in the SAME unit; only the number of integrals changes.
• A test point strictly inside each subinterval decides which curve is on top, or on the right.

Same region, two ways of cutting it: a vertical strip measures “top − bottom” and is integrated in

dx

, a horizontal strip measures “right − left” and is integrated in

dy

No bound is given in an area problem: they are always found by solving the intersection equation. It is the first line of your paper.

The symmetries that halve the work

• EVEN integrand on a symmetric interval: $\int_{-a}^{a}f=2\int_{0}^{a}f$ .
• ODD integrand on a symmetric interval: $\int_{-a}^{a}f=0$ , but the AREA is not.
• A region symmetric about a line is computed on one half, then doubled.
• A region reflected in $y=x$ keeps the same area, which lets you change the direction of the strips for free.

Even: the two halves are equal, compute one and double it. Odd: the two lobes cancel in the integral, but their areas add up.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Handing in the integral when the question asked for the area

the whole question

What not to write

“ $\int_{-2}^{2}\left(x^{3}-4x\right)dx=0$ , so the area equals $0$ .”

What to write

“The integral equals $0$ by oddness; the area equals $\int_{-2}^{0}f-\int_{0}^{2}f=4+4=8$ .”

Why: A zero area under a curve that clearly moves away from the axis is an immediate contradiction. The word “area” forces the split at the zeros, exactly as the words “distance travelled” force the split where the direction changes.

2. Looking for the bounds anywhere but in the intersection

the whole question; the bounds alone are often worth two marks

What not to write

“The problem gives no bounds, I take $\left[-\infty,+\infty\right]$ .”

What to write

“The region is CLOSED by the curves themselves: I solve $x^{2}=x+2$ , which gives $x=-1$ and $x=2$ .”

The region is closed by the two curves themselves: its bounds are the

x

-coordinates of the two points of intersection, solutions of

x^{2}=x+2

Why: A region bounded by two curves has its bounds written into the intersection equation. It is the first line of every area question, and the grading scheme almost always gives it marks separate from the computation.

3. Deciding which curve is on top without a test point

the whole question, since the result comes out negative

What not to write

“ $x^{2}$ has a larger exponent, so the parabola is on top.”

What to write

“Test point $x=0$ in $\left(-1, 2\right)$ : the line equals $2$ , the parabola equals $0$ , so the LINE is on top.”

x=0

, the line equals

2

and the parabola

0

: the LINE is on top over the whole of

\left(-1, 2\right)

, despite the exponent of

x^{2}

Why: The degree of a formula says nothing about its position on a bounded interval. A single test point, strictly inside, is enough and is worth half a mark on your paper: it is the cheapest argument of the chapter.

4. Forgetting to split when the roles switch

the whole question

What not to write

$\mathcal{A}=\int_{-1}^{1}\left(x-x^{3}\right)dx=0$

What to write

$\mathcal{A}=\int_{-1}^{0}\left(x^{3}-x\right)dx+\int_{0}^{1}\left(x-x^{3}\right)dx=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

\left(-1, 0\right)

the cubic is above the line, on

\left(0, 1\right)

it is the other way around: a single integral over

\left[-1, 1\right]

subtracts one region instead of adding it.

Why: $x^{3}=x$ has THREE solutions on $\left[-1, 1\right]$ , not two: you lose $x=0$ by dividing by $x$ instead of factoring. This intermediate zero is precisely where top and bottom switch, so where the integral must be cut.

5. Using vertical strips on a region that calls for horizontal strips

no marks lost, but ten minutes of the exam and twice as many chances to make a mistake

What not to write

“ $x=y^{2}$ and $x=y+2$ : I cut along $x$ , which gives me two integrals with a square root.”

What to write

“I integrate with respect to $y$ : $\int_{-1}^{2}\left[(y+2)-y^{2}\right]dy=\frac{9}{2}$ , in ONE single integral.”

With vertical strips, the BOTTOM boundary is

-\sqrt{x}

before

x=1

and

x-2

after: two integrals. The left boundary, on the other hand, never changes.

Why: When the LOWER boundary changes formula along the way, vertical strips force a split. Here the boundary goes from $-\sqrt{x}$ to $x-2$ at $x=1$ . The criterion for choosing is simple: cut in the direction that requires the fewest splits.

6. Redoing the computation when the result is negative

five minutes of the exam

What not to write

“I get $-\frac{9}{2}$ , I start all over again.”

What to write

“A negative sign only signals that I wrote bottom minus top: I change the sign, the area equals $\frac{9}{2}$ .”

Why: An area cannot be negative, and the only way to get a negative number with bounds in increasing order is to have swapped the two functions. The repair takes three seconds, provided you understand what the sign is telling you.

7. Believing an absolute value can be integrated as it stands

the whole question as soon as there is more than one interior intersection

What not to write

$\mathcal{A}=\int_{a}^{b}\left|f-g\right|=\left|\int_{a}^{b}(f-g)\right|$

What to write

Split at the intersections, determine the sign on each piece, then add up the integrals taken in the right order.

Why: The formula with an absolute value is exact but is not a method of computation: no table gives the antiderivative of an absolute value. On $\left[-1, 1\right]$ with $x^{3}$ and $x$ , the absolute value of the integral equals $0$ while the area equals $\frac{1}{2}$ .

8. Changing units by changing the direction of the strips

1 mark for interpretation

What not to write

“Integrating with respect to $y$ , I get an area in a different unit.”

What to write

“Both directions give the same number in the same unit: $\frac{9}{2}$ in both cases.”

Why: A horizontal strip has area $\left(\text{right}-\text{left}\right)dy$ , a length times a length: it is indeed an area. The direction of the strips is a choice of method, not a change of quantity.

Which method to choose

Which direction to cut in

Look at the BOUNDARIES of the region and ask, for each direction, how many times a boundary changes formula.

Four regions, four decisions: a single integral in

x

, a split at the interior intersection, strips in

y

, a split at the vertex.

If both curves are functions of $x$ and only cross at the ends → VERTICAL strips, a single integral

Example: $y=x^{2}$ and $y=x+2$ between $-1$ and $2$
If the curves cross inside the interval → vertical strips with a SPLIT at each intersection

Example: $y=x^{3}$ and $y=x$ on $\left[-1, 1\right]$ , cut at $0$
If a curve is given in the form $x=g(y)$ → HORIZONTAL strips, almost always the economical direction

Example: $x=y^{2}$ and $x=y+2$
If the lower or upper boundary changes formula with vertical strips → try horizontal strips before computing anything

Example: three integrals in $x$ become a single one in $y$
If the region is bounded by three curves or more → split at each VERTEX, in the direction that produces the fewest splits

Example: the triangle of $y=x$ , $y=2x$ and $y=6-x$ requires two integrals

A polygonal region can be checked for free with the determinant formula on its vertices: it is the best check available.

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.

An area between two curves, written up for the grader

When to use it: Any question asking for the area of a region bounded by curves.

1 Look for the intersections and write them down: “ $x^{2}=x+2$ gives $x=-1$ and $x=2$ .”
2 Choose the direction of the strips and JUSTIFY it in one sentence.
3 Use an interior test point and state which curve is on top: “At $x=0$ , the line equals $2$ and the parabola $0$ : the line is on top.”
4 Write the complete integral with its bounds, then the antiderivative in brackets.
5 Evaluate and conclude with a sentence giving the area with its unit.

Concluding sentence

The area of the region bounded by $y=x^{2}$ and $y=x+2$ equals $\int_{-1}^{2}\left[(x+2)-x^{2}\right]dx=\frac{9}{2}$ square units.

The trap: Writing the integral without having stated the test point. The grader looks for the sentence that justifies the order “top minus bottom”; without it, the integral looks as if it were written at random, even if it is right.

Marking: Usually 1 mark for the bounds, 1 mark for the test point and the order, 2 marks for setting up the integral, 1 mark for the computation and the conclusion.

Check before you hand in

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

Check that the result is positive

With bounds in increasing order and the order “top minus bottom”, the area is always positive. A negative result signals a swap, not a computation error.

$-\frac{9}{2}$ is corrected to $\frac{9}{2}$ without redoing a single line.
Enclose the region in a rectangle

Bound the area above by that of the smallest rectangle containing the region: the result must be smaller, and usually of the order of half.

The region between $y=x^{2}$ and $y=x+2$ fits in a rectangle $3$ by $4$ , that is $12$ ; the answer $4.5$ is plausible, $45$ would not be.
Redo the computation with strips in the other direction

When time allows, redo the computation with horizontal strips: the number must be identical. It is the most complete check of the chapter.

$x=y^{2}$ and $x=y+2$ give $\frac{9}{2}$ in one integral in $y$ and $\frac{4}{3}+\frac{19}{6}=\frac{9}{2}$ in two integrals in $x$ ✓.
Count the intersections before concluding

Solve the intersection equation by FACTORING and count the solutions. Each solution inside the interval forces a split.

$x^{3}=x$ gives $x(x-1)(x+1)=0$ , so three solutions. Dividing by $x$ loses one, and the split at $0$ disappears.
Check a polygonal region with the determinant

For a triangle with known vertices, the area is half the absolute value of the determinant built on the coordinates. Ten seconds, and it is independent of the integral calculus.

Vertices $(0, 0)$ , $(2, 4)$ , $(3, 3)$ : $\frac{1}{2}\left|2(3)+3(-4)\right|=3$ , which confirms the two integrals.

The typical problem, taken apart

A region that calls for horizontal strips

Compute the area of the region bounded by the parabola $x=y^{2}$ and the line $x=y+2$ .

The LEFT boundary of the region is the parabola over its whole height, but with vertical strips the lower boundary changes formula at

x=1

: that change is what forces horizontal strips.

Step 1

Intersections: set the two expressions of $x$ equal, that is $y^{2}=y+2$ , so $y^{2}-y-2=0$ and $(y-2)(y+1)=0$ . The $y$ -coordinates are $y=-1$ and $y=2$ , and the points are $(1, -1)$ and $(4, 2)$ .

Why

You set the $x$ equal because it is $x$ that is expressed as a function of $y$ . The variable of integration is thus chosen by the problem itself.

Step 2

Interior test point, $y=0$ : the parabola gives $x=0$ , the line gives $x=2$ . The LINE is therefore on the right over the whole interval.

Why

With horizontal strips, the formula is “right minus left”, exactly like “top minus bottom” with vertical strips. The test point decides, and you must write it.

Step 3

$\mathcal{A}=\int_{-1}^{2}\left[(y+2)-y^{2}\right]dy=\left[\frac{y^{2}}{2}+2y-\frac{y^{3}}{3}\right]_{-1}^{2}$ .

Why

A single integral, with no square root and no split. That is the gain from the right direction of the strips, and it shows as soon as you write it.

Step 4

At $2$ : $2+4-\frac{8}{3}=\frac{10}{3}$ . At $-1$ : $\frac{1}{2}-2+\frac{1}{3}=-\frac{7}{6}$ . So $\mathcal{A}=\frac{10}{3}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$ .

Why

Evaluate at the two bounds separately before subtracting: that is where sign errors hide, and writing them one under the other makes them visible.

Step 5

Cross-check with vertical strips: from $0$ to $1$ , the region runs from $-\sqrt{x}$ to $\sqrt{x}$ ; from $1$ to $4$ , it runs from $x-2$ to $\sqrt{x}$ . This gives $\int_{0}^{1}2\sqrt{x}\,dx+\int_{1}^{4}\left(\sqrt{x}-x+2\right)dx=\frac{4}{3}+\frac{19}{6}=\frac{9}{2}$ ✓.

Why

Two integrals instead of one, with square roots: the comparison shows concretely what the right direction of the strips saves, and it checks the result.

Step 6

Order-of-magnitude check: the region fits in the rectangle $\left[0, 4\right]\times\left[-1, 2\right]$ , of area $12$ ; $\frac{9}{2}$ fills a little more than a third of it, which matches the drawing ✓.

Why

A rough bound catches errors by a factor of $10$ , the ones that rereading the algebra never sees.

The conclusion, written out

The area of the region bounded by $x=y^{2}$ and $x=y+2$ equals $\frac{9}{2}$ square units, obtained in a single integral with horizontal strips.

The classic mistake on this problem: Setting $y^{2}$ equal to $y+2$ and then integrating with respect to $x$ with those bounds. The numbers $-1$ and $2$ are $y$ -COORDINATES: using them as bounds of an integral in $x$ gives a number unrelated to the region.

Learn by heart

• Bounds: they are never given, they are found by solving $f=g$ .
• Vertical: $\int\left(\text{top}-\text{bottom}\right)dx$ . Horizontal: $\int\left(\text{right}-\text{left}\right)dy$ .
• A test point STRICTLY inside decides the order, on each subinterval.
• Each interior intersection forces a SPLIT.
• Negative result: reversed order. Change the sign, do not redo the computation.
• The direction of the strips changes neither the value nor the unit, only the number of integrals.
• Odd integrand on a symmetric interval: zero integral, nonzero area.
• Area of a triangle with known vertices: half the absolute value of the determinant.

Frequently asked questions

How do you find the bounds for the area between two curves?

By solving the intersection equation, that is by setting the two expressions equal. The solutions you get are the bounds of the integral. An area problem almost never gives the bounds explicitly, because the region is closed by the curves themselves.

How do you know which curve is on top?

Choose a point strictly inside the interval and evaluate both functions. The one that gives the larger value is on top, and it stays there as long as the curves do not cross again. The degree or the shape of the formula says nothing about the position.

When should you integrate with respect to y?

When the lower or upper boundary of the region changes formula with vertical strips, which would force a split. This is almost always the case when one of the curves is given in the form x equals a function of y, for example a parabola lying on its side.

Why is my area negative?

Because the two functions were written in the wrong order: you integrated bottom minus top. Just change the sign of the result, without redoing the computation. It is the only way to get a negative number when the bounds are in increasing order.

Are the integral and the area the same thing?

Only if the function keeps a constant sign on the interval. Otherwise the integral counts what lies below the axis negatively, and it can even be zero while the area is not. For an area, split at the zeros and change the sign of the negative pieces.

Practise it

Corrected exercises: Area between curves, 201-SN3-RE (201-NYB)

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.

16 corrected exercises
160 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Definite and improper integrals Next sheet Numerical integration

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-area-between-curves. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Looking for a 201-SN3-RE (201-NYB) tutor in Montreal?

Get in touch for a first session. We go back over the moves that cost marks on the integral calculus exam, then we test them on exercises at the real level of CEGEP tests.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: the area between two curves, 201-SN3-RE (201-NYB)

The essentials

Area or integral: two different numbers

The formula, for both directions of the strips

The symmetries that halve the work

The mistakes that cost marks

1. Handing in the integral when the question asked for the area

2. Looking for the bounds anywhere but in the intersection

3. Deciding which curve is on top without a test point

4. Forgetting to split when the roles switch

5. Using vertical strips on a region that calls for horizontal strips

6. Redoing the computation when the result is negative

7. Believing an absolute value can be integrated as it stands

8. Changing units by changing the direction of the strips

Which method to choose

Which direction to cut in

How the answer is expected to be written

An area between two curves, written up for the grader

Check before you hand in

The typical problem, taken apart

A region that calls for horizontal strips

Learn by heart

Frequently asked questions

How do you find the bounds for the area between two curves?

How do you know which curve is on top?

When should you integrate with respect to y?

Why is my area negative?

Are the integral and the area the same thing?

Corrected exercises: Area between curves, 201-SN3-RE (201-NYB)

See also

Looking for a 201-SN3-RE (201-NYB) tutor in Montreal?