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

Revision sheet: the definite integral and improper integrals, 201-SN3-RE (201-NYB)

Q: What is the difference between a definite integral and an area?

The integral counts what lies below the x-axis as negative, while the area counts everything as positive. The two agree only if the function keeps the same sign. For an area, split the interval at the zeros of the function and change the sign of the negative pieces.

Q: What is an improper integral?

It is an integral with an infinite limit of integration, or whose integrand is undefined somewhere on the interval. It is defined as a limit: replace the troublesome endpoint with a variable, compute the ordinary integral, then let that variable tend to the endpoint. If the limit is finite, the integral converges.

Q: How do you spot a singularity hidden inside the interval?

Look for zeros of the denominator, negative radicands and zero arguments of a logarithm, over the whole interval and not only at the endpoints. The classic example is the integral of one over x squared from minus one to one, where the singularity at zero is invisible in the way the integral is written.

Q: When does an integral of one over x to the power p converge?

At infinity it converges if p is strictly greater than one, because the function must decrease fast enough. Near zero it is the other way round: it converges if p is strictly less than one, because the function must blow up slowly enough. The case p equal to one diverges in both situations.

Q: Which way round is the Comparison Test used?

A positive function smaller than a function whose integral converges also has a convergent integral. A positive function larger than a function whose integral diverges also has a divergent integral. Both functions must be positive, otherwise you first have to bound the absolute value.

This sheet does not restate the Fundamental Theorem: it is in your notes. It deals with what decides the mark on the third 201-SN3-RE (formerly 201-NYB) exam, where almost every loss comes from applying the formula $F(b)-F(a)$ where it does not apply, for want of checking that the function stays defined on the whole interval.

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.

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The thread of the chapter

A definite integral is not an area but a SIGNED area, and an improper integral is not even an integral: it is a LIMIT, which must be written with its variable limit of integration before anything is computed.

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

The essentials

What the Fundamental Theorem says, and what it requires

• $\int_{a}^{b}f(x)\,dx=F(b)-F(a)$ , provided $f$ is CONTINUOUS on the whole of $\left[a,b\right]$ .
• Derivative form: $\frac{d}{dx}\int_{a}^{x}f(t)\,dt=f(x)$ . An integral with a variable limit is an antiderivative.
• A definite integral is a NUMBER; an indefinite integral is a family of functions. No $+C$ in the first one.
• Additivity: $\int_{a}^{c}=\int_{a}^{b}+\int_{b}^{c}$ , and $\int_{b}^{a}=-\int_{a}^{b}$ .
• Average value: $\frac{1}{b-a}\int_{a}^{b}f$ , reached at least once if $f$ is continuous.

\left[0,\frac{3\pi}{2}\right]

, the positive arch counts

+2

and the negative part counts

-1

: the integral is

1

, the AREA is

3

. The two questions do not have the same answer.

The continuity hypothesis is not decoration: it is what separates an ordinary integral from an improper integral in disguise.

The three ways to be improper

• INFINITE limit of integration: $\int_{a}^{+\infty}f=\lim_{t\to+\infty}\int_{a}^{t}f$ .
• Singularity AT AN ENDPOINT: $\int_{a}^{b}f=\lim_{t\to b^{-}}\int_{a}^{t}f$ if $f$ blows up at $b$ .
• Singularity INSIDE the interval: split at $c$ , handle the two pieces SEPARATELY, and require BOTH to converge.
• Doubly improper: split at any point and handle each half on its own. Both must converge.
• Converges means: the limit exists and is FINITE. Otherwise the integral is said to diverge.

Three areas without an edge: towards infinity, against an asymptote at an endpoint, against an asymptote in the middle. Each one is written with a limit, the third one by splitting the interval.

Two divergent pieces that “cancel out” do NOT make a convergent integral: the definition requires each one to converge.

The p-test, to know by heart

• $\int_{1}^{+\infty}\frac{dx}{x^{p}}$ CONVERGES if $p>1$ , DIVERGES if $p\le 1$ .
• $\int_{0}^{1}\frac{dx}{x^{p}}$ CONVERGES if $p<1$ , DIVERGES if $p\ge 1$ . The two conditions are reversed.
• Boundary case $p=1$ : both diverge, and $\int\frac{dx}{x}=\ln x$ shows it.
• Comparison: if $0\le f\le g$ and $\int g$ converges, then $\int f$ converges; if $\int f$ diverges, then $\int g$ diverges.

At infinity,

\frac{1}{x}

(red) diverges and

\frac{1}{x^{2}}

(green) converges; at

0

, the test is reversed:

\frac{1}{\sqrt{x}}

(green) converges,

\frac{1}{x}

(red) diverges.

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. Confusing the integral with the area

the whole question

What not to write

“ $\int_{0}^{2\pi}\sin x\,dx=0$ , so the area under the curve is zero.”

What to write

“The integral is $0$ ; the AREA is $\int_{0}^{\pi}\sin x\,dx-\int_{\pi}^{2\pi}\sin x\,dx=2+2=4$ .”

Why: The integral counts what lies below the axis as negative. The word “area” calls for splitting at the ZEROS of the function and changing the sign of the negative pieces. A zero area with a curve visibly away from the axis is an immediate contradiction.

2. Applying the Fundamental Theorem across a singularity

the whole question, plus a negative area under a positive function

What not to write

$\int_{-1}^{1}\frac{dx}{x^{2}}=\left[-\frac{1}{x}\right]_{-1}^{1}=-1-1=-2$

What to write

The integrand is not defined at $0$ : the integral is IMPROPER, it must be split, and each of the two pieces DIVERGES.

\frac{1}{x^{2}}

blows up at

0

, INSIDE

\left[-1, 1\right]

: applying

F(b)-F(a)

over it gives

-2

, a negative number under a curve that is entirely positive.

Why: The Fundamental Theorem requires continuity on the whole interval. The result $-2$ is absurd since $\frac{1}{x^{2}}>0$ everywhere: the integral of a positive function cannot be negative. This free sign check is what reveals the forgotten singularity.

3. Computing an improper integral without writing the limit

1 mark every time, even when the value is right

What not to write

$\int_{1}^{+\infty}\frac{dx}{x^{2}}=\left[-\frac{1}{x}\right]_{1}^{+\infty}=0+1=1$

What to write

$\int_{1}^{+\infty}\frac{dx}{x^{2}}=\lim_{t\to+\infty}\left[-\frac{1}{x}\right]_{1}^{t}=\lim_{t\to+\infty}\left(1-\frac{1}{t}\right)=1$

Why: $\infty$ is not a number: you do not substitute it into an antiderivative. The variable limit $t$ and the passage to the limit are exactly what the marking scheme grades, and they only cost one line.

4. Handling the two halves of a doubly improper integral together

the whole question

What not to write

$\int_{-\infty}^{+\infty}x\,dx=\lim_{t\to+\infty}\int_{-t}^{t}x\,dx=\lim_{t\to+\infty}0=0$

What to write

Split: $\int_{-\infty}^{0}x\,dx$ diverges, so the whole integral DIVERGES, whatever the symmetry.

From

-R

R

, the red and the green cancel; from

-R

2R

+6

is left over for

R=2

. The “value” depends on how you go to infinity: each half diverges.

Why: The definition requires two INDEPENDENT limits, not a single symmetric one. The symmetric computation gives $0$ , but it would just as well give $+\infty$ with the limits $-t$ and $2t$ : an expression whose value depends on the path chosen is not defined.

5. Using the Comparison Test the wrong way round

the whole convergence question

What not to write

“ $\frac{1}{x^{2}+1}\le\frac{1}{x^{2}}$ and $\int\frac{dx}{x^{2}}$ converges, so $\int\frac{dx}{x^{2}+1}$ diverges.”

What to write

“Bounded above by a convergent one, so CONVERGENT. Bounding from above passes convergence downward and divergence upward.”

\frac{1}{x^{2}+1}

stays BELOW

\frac{1}{x^{2}}

, whose area on

\left[1,+\infty\right)

is finite: so its area is finite too. Being smaller than a convergent one makes it converge.

Why: The direction is the only content of the test: a function smaller than an integrable one is integrable, a function larger than a non-integrable one is not. Stating it with the words “smaller” and “larger” rather than with symbols keeps you from reversing it under pressure.

6. Mixing up the two p-tests

the whole question

What not to write

“ $\int_{0}^{1}\frac{dx}{x^{2}}$ converges, since $2>1$ .”

What to write

“At $0$ , the test is REVERSED: $\int_{0}^{1}\frac{dx}{x^{p}}$ converges if $p<1$ . Here $p=2$ , so it DIVERGES.”

Why: The two tests are reversed because the danger is not in the same place: at infinity, the function must decrease FAST, so $p$ large; at $0$ , it must blow up SLOWLY, so $p$ small. Remembering the reason rather than the two inequalities keeps you from mixing them up.

7. Writing a $+C$ in a definite integral

half a mark, and a confusion about what kind of object each one is

What not to write

$\int_{0}^{1}x\,dx=\frac{x^{2}}{2}+C\Big|_{0}^{1}=\frac{1}{2}+C$

What to write

$\int_{0}^{1}x\,dx=\left[\frac{x^{2}}{2}\right]_{0}^{1}=\frac{1}{2}$

Why: A definite integral is a NUMBER. The constant disappears in the subtraction $F(b)-F(a)$ , which is precisely why the choice of antiderivative does not matter. Writing a $+C$ in a definite integral signals confusion about what is being computed.

8. Substituting without changing the limits of integration

the whole question

What not to write

$\int_{0}^{2}x\,e^{x^{2}}\,dx=\frac{1}{2}\int_{0}^{2}e^{u}\,du$

What to write

$u=x^{2}$ gives $u=0$ for $x=0$ and $u=4$ for $x=2$ , so $\frac{1}{2}\int_{0}^{4}e^{u}\,du=\frac{e^{4}-1}{2}$

Why: The limits belong to the variable, not to the integral: changing the variable forces you to translate them. There are two valid routes, changing the limits or going back to $x$ before evaluating, but never a mix of the two, which gives a number that is wrong by a large factor.

Which method to choose

Is this integral improper, and how should it be handled

Before any computation, look at the LIMITS OF INTEGRATION and at the DOMAIN of the integrand on the interval. That glance, and that glance alone, decides the method.

If the integrand is continuous on the whole closed interval and the limits are finite → ordinary integral: $F(b)-F(a)$ directly

Example: $\int_{0}^{1}x\,dx=\frac{1}{2}$
If a limit of integration is infinite → replace it by a variable limit $t$ and take the limit

Example: $\int_{1}^{+\infty}\frac{dx}{x^{2}}=\lim_{t\to+\infty}\left(1-\frac{1}{t}\right)=1$
If the integrand blows up at an ENDPOINT of the interval → variable limit tending to that endpoint, from the correct side

Example: $\int_{0}^{1}\frac{dx}{\sqrt{x}}=\lim_{t\to 0^{+}}\left(2-2\sqrt{t}\right)=2$
If the integrand blows up INSIDE the interval → SPLIT at the singular point and handle the two pieces separately; both must converge

Example: $\int_{-1}^{1}\frac{dx}{x^{2}}$ diverges, each half diverging
If both limits of integration are infinite → split at any point, two INDEPENDENT limits

Example: $\int_{-\infty}^{+\infty}\frac{dx}{1+x^{2}}=\frac{\pi}{2}+\frac{\pi}{2}=\pi$

An integrand containing a denominator, a root or a logarithm always deserves this glance: interior singularities are invisible in the way the integral is written.

Converges or diverges, without computing

When the question only asks whether the integral converges, compare with a reference integral rather than computing.

If the integrand behaves like $\frac{1}{x^{p}}$ at $+\infty$ → converges if and only if $p>1$

Example: $\frac{1}{x^{2}+1}$ behaves like $\frac{1}{x^{2}}$ , so it converges
If the integrand behaves like $\frac{1}{x^{p}}$ at $0$ → converges if and only if $p<1$

Example: $\frac{1}{\sqrt{x}}$ has $p=\frac{1}{2}<1$ , so it converges on $\left(0, 1\right]$
If the integrand is positive and BOUNDED ABOVE by a function whose integral converges → the integral converges

Example: $e^{-x^{2}}\le e^{-x}$ for $x\ge 1$ , so $\int_{1}^{+\infty}e^{-x^{2}}$ converges
If the integrand is positive and BOUNDED BELOW by a function whose integral diverges → the integral diverges

Example: $\frac{1}{x+1}\ge\frac{1}{2x}$ for $x\ge 1$ , so the integral diverges

The Comparison Test requires POSITIVE functions. For a function that changes sign, first bound its absolute value.

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 improper integral, written up for the marker

When to use it: As soon as a limit of integration is infinite or the integrand is not defined on the whole closed interval.

1 Spot and NAME the difficulty: “The integrand is not defined at $0$ ; the integral is improper at that endpoint.”
2 Write the limit with a variable limit of integration: “ $\int_{0}^{1}f=\lim_{t\to 0^{+}}\int_{t}^{1}f$ .”
3 Compute the ordinary integral from $t$ to $1$ , keeping $t$ as a letter.
4 Take the limit, and say whether it is finite.
5 Conclude with the word CONVERGES or DIVERGES, and the value if there is one.

Concluding sentence

Since the limit is finite, the integral converges and equals $2$ .

The trap: Substituting $0$ or $+\infty$ directly into the antiderivative. These symbols are not numbers, and an answer that writes $\left[-\frac{1}{x}\right]_{1}^{+\infty}$ loses the mark even when the final value is correct.

Marking: Usually 1 mark for identifying why the integral is improper, 1 mark for the limit written out, 1 mark for the computation, 1 mark for the conclusion naming convergence or divergence.

Check before you hand in

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

Check the sign of the result

The integral of a positive function over an interval oriented the right way is positive. A negative result signals either swapped limits or a singularity crossed.

$\int_{-1}^{1}\frac{dx}{x^{2}}=-2$ is impossible: the integrand is positive everywhere, the singularity at $0$ has been ignored.
Check continuity on the whole closed interval

Before writing $F(b)-F(a)$ , look for zeros of the denominator, negative radicands and zero arguments of a logarithm INSIDE the interval, endpoints included.

$\int_{1}^{3}\frac{dx}{x-2}$ contains $2$ : improper, to be split, and divergent in the end.
Compare with a reference integral

Before computing, predict convergence or divergence with the p-test. If the computation ends with a finite value while the test predicts divergence, one of the two is wrong.

$\int_{1}^{+\infty}\frac{dx}{\sqrt{x}}$ has $p=\frac{1}{2}\le 1$ , so it diverges: a finite answer needs to be checked again.
Check a value with a rough bound

Bound the integrand above and below by constants on the interval, and check that the result falls between the two products obtained.

On $\left[0, 1\right]$ , $\frac{1}{1+x^{2}}$ lies between $\frac{1}{2}$ and $1$ , so the integral lies between $0.5$ and $1$ : the value $\frac{\pi}{4}\approx 0.785$ ✓.
Check the limits after a change of variable

After a substitution, recompute the two new limits by substituting the old ones into the change of variable formula, and check that they are in the right order.

$u=x^{2}$ maps $\left[0, 2\right]$ onto $\left[0, 4\right]$ ; keeping $\left[0, 2\right]$ gives $\frac{e^{2}-1}{2}\approx 3.19$ instead of $26.8$ .

The typical problem, taken apart

A doubly improper integral, handled by the rules

Determine whether $\displaystyle\int_{0}^{+\infty}\frac{dx}{\sqrt{x}\,(1+x)}$ converges, and compute it if it does.

The integrand blows up at

0

\frac{1}{\sqrt{x}}

and decreases at infinity like

\frac{1}{x^{3/2}}

: two improper ends, two limits, hence the split of the interval.

Step 1

Spotting: the integrand is not defined at $0$ , because of $\sqrt{x}$ in the denominator, and the upper limit is infinite. So the integral is DOUBLY improper.

Why

Name both difficulties in the very first line. A solution that sees only one handles a single passage to the limit and loses half the rigour marks.

Step 2

Split at any point, for example $1$ : $\int_{0}^{+\infty}=\int_{0}^{1}+\int_{1}^{+\infty}$ , and require BOTH to converge.

Why

The split point is arbitrary and needs no justification; what matters is that each piece contains only one difficulty, at only one of its endpoints.

Step 3

Behaviour near $0$ : $\frac{1}{\sqrt{x}(1+x)}\sim\frac{1}{\sqrt{x}}$ , with power $p=\frac{1}{2}<1$ , so the first piece CONVERGES. Behaviour at $+\infty$ : $\frac{1}{\sqrt{x}(1+x)}\sim\frac{1}{x^{3/2}}$ , with power $p=\frac{3}{2}>1$ , so the second one CONVERGES too.

Why

The p-test answers both questions before any computation. Noticing that the two conditions are reversed, $p<1$ at $0$ and $p>1$ at infinity, is exactly what the chapter tests.

Step 4

Computation: let $u=\sqrt{x}$ , so $x=u^{2}$ and $dx=2u\,du$ . The integral becomes $\int\frac{2u\,du}{u\left(1+u^{2}\right)}=2\int\frac{du}{1+u^{2}}=2\arctan u+C$ .

Why

The substitution $u=\sqrt{x}$ removes the root AND the singularity: it is the natural change of variable whenever a root sits in the denominator.

Step 5

Limits: $x=0$ gives $u=0$ , and $x\to+\infty$ gives $u\to+\infty$ . So $\int_{0}^{+\infty}=\lim_{T\to+\infty}\left[2\arctan u\right]_{0}^{T}=2\cdot\frac{\pi}{2}-0=\pi$ .

Why

The limits are translated into the new variable, the infinite one included. The passage to the limit stays written, even when the antiderivative is bounded.

Step 6

Check: $\arctan$ is bounded by $\frac{\pi}{2}$ , so the result cannot exceed $\pi$ ; and a numerical estimate on $\left[0.0001, 10000\right]$ gives about $3.10$ , consistent with $\pi\approx 3.1416$ ✓.

Why

Two independent checks, one theoretical by bounding and the other numerical. Together, they confirm both the convergence and the value.

The conclusion, written out

The integral is doubly improper; it converges at $0$ because $p=\frac{1}{2}<1$ and at $+\infty$ because $p=\frac{3}{2}>1$ , and its value is $\pi$ .

The classic mistake on this problem: Handling both improper ends with a single limit, $\lim_{t\to 0,\,T\to+\infty}$ with $t$ and $T$ tied together. The definition requires two INDEPENDENT limits, and that is what prevents illusory cancellations between the two pieces.

Learn by heart

• $\int_{a}^{b}f=F(b)-F(a)$ requires CONTINUITY on the whole of $\left[a,b\right]$ , endpoints included.
• Definite integral: a NUMBER, no $+C$ . Indefinite integral: a family of functions.
• Integral and area differ as soon as $f$ changes sign: the area splits at the zeros.
• Improper means LIMIT: variable limit of integration written out, then the passage to the limit.
• INTERIOR singularity: split, and require BOTH pieces to converge.
• $\int_{1}^{+\infty}\frac{dx}{x^{p}}$ converges if $p>1$ ; $\int_{0}^{1}\frac{dx}{x^{p}}$ converges if $p<1$ .
• Comparison: smaller than a convergent one, it converges; larger than a divergent one, it diverges.
• Change of variable in a definite integral: translate the LIMITS.

Frequently asked questions

What is the difference between a definite integral and an area?

The integral counts what lies below the x-axis as negative, while the area counts everything as positive. The two agree only if the function keeps the same sign. For an area, split the interval at the zeros of the function and change the sign of the negative pieces.

What is an improper integral?

It is an integral with an infinite limit of integration, or whose integrand is undefined somewhere on the interval. It is defined as a limit: replace the troublesome endpoint with a variable, compute the ordinary integral, then let that variable tend to the endpoint. If the limit is finite, the integral converges.

How do you spot a singularity hidden inside the interval?

Look for zeros of the denominator, negative radicands and zero arguments of a logarithm, over the whole interval and not only at the endpoints. The classic example is the integral of one over x squared from minus one to one, where the singularity at zero is invisible in the way the integral is written.

When does an integral of one over x to the power p converge?

At infinity it converges if p is strictly greater than one, because the function must decrease fast enough. Near zero it is the other way round: it converges if p is strictly less than one, because the function must blow up slowly enough. The case p equal to one diverges in both situations.

Which way round is the Comparison Test used?

A positive function smaller than a function whose integral converges also has a convergent integral. A positive function larger than a function whose integral diverges also has a divergent integral. Both functions must be positive, otherwise you first have to bound the absolute value.

Practise it

Corrected exercises: Definite and improper integrals, 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

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Revision sheet: the definite integral and improper integrals, 201-SN3-RE (201-NYB)

The essentials

What the Fundamental Theorem says, and what it requires

The three ways to be improper

The p-test, to know by heart

The mistakes that cost marks

1. Confusing the integral with the area

2. Applying the Fundamental Theorem across a singularity

3. Computing an improper integral without writing the limit

4. Handling the two halves of a doubly improper integral together

5. Using the Comparison Test the wrong way round

6. Mixing up the two p-tests

7. Writing a +C+C+C in a definite integral

8. Substituting without changing the limits of integration

Which method to choose

Is this integral improper, and how should it be handled

Converges or diverges, without computing

How the answer is expected to be written

An improper integral, written up for the marker

Check before you hand in

The typical problem, taken apart

A doubly improper integral, handled by the rules

Learn by heart

Frequently asked questions

What is the difference between a definite integral and an area?

What is an improper integral?

How do you spot a singularity hidden inside the interval?

When does an integral of one over x to the power p converge?

Which way round is the Comparison Test used?

Corrected exercises: Definite and improper integrals, 201-SN3-RE (201-NYB)

See also

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

7. Writing a $+C$ in a definite integral