Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: elementary integration (201-SN2-RE)

Q: What is the constant for if I am not given any initial condition?

Without a condition, the constant IS the answer: it says that all the parallel curves obtained are valid, since differentiation brings them all back to the same function. Leaving it out amounts to claiming that only one of them is right, which is false, and it costs half a mark on every integral.

Q: Why do we study antiderivatives in a differential calculus course?

Because the course framework for Differential Calculus, 201-SN2-RE asks you to compute elementary integrals, and the last week of the semester is devoted to them. So you meet them before the definite integral: here an antiderivative is just the table of derivatives read the other way, with no area and no limits of integration.

Q: How do I recognize that an integral is done by substitution?

Look in the integrand for an inner function and its derivative as a factor, up to a multiplicative constant. If the missing factor is a number, you adjust it and the integral falls into place. If a power of x is missing, you do not substitute: the integral changes nature and leaves the scope of this course.

Q: How do I find the constant when the data is not at time zero?

Integrate first and keep the constant visible in the formula. Then replace the time with the given value, for example three seconds, set the result equal to the measured number, and solve for C. The constant can be read directly only if the data is at time zero and the rest of the formula vanishes there.

Q: How do I check an antiderivative when I have no answer key?

Differentiate your own answer. If you get exactly the starting function back, the antiderivative is right, whatever it looks like. This check takes ten seconds and requires no theorem from the next course. It does not validate the constant: for that, substitute the numerical data from the problem into the final formula.

This sheet does not copy out the table of antiderivatives: you already have it twice over, since it is the semester's table of derivatives read from right to left. It deals with what decides the mark on the last differential calculus test of 201-SN2-RE (formerly 201-NYA), where almost all the marks lost come down to the constant $C$ and to a missing $u'$ that you try to manufacture. At this point of the course you have no definite integral, no area under a curve and no fundamental theorem: checking by differentiation is the only check available to you, and it is free.

It is written for CEGEP students in the Science program (Sciences de la nature). The corrected exercise set on the same chapter then puts every reflex to the test, from the reversed table all the way to problems with two constants.

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

At this point of the course, an antiderivative is nothing more than the table of derivatives read from right to left, and the only check available to you is to differentiate your answer. Everything that costs marks comes down to the constant $C$ , which differentiation erases and no computation can guess, or to a missing $u'$ that you try to manufacture.

This chapter is part of 201-SN2-RE, Differential Calculus

What is on this sheet

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

Topics covered

Antiderivative of a function
Constant of integration
Table of antiderivatives
Power rule for antiderivatives
Natural logarithm with absolute value
Reverse chain rule
Substitution rule
Mean Value Theorem
Initial condition
Checking by differentiation

The essentials

What you have at this point of the course

• DEFINITION: $F$ is an antiderivative of $f$ on an interval if $F'=f$ . The whole table of antiderivatives is the semester's table of derivatives, read from right to left.
• You do NOT yet have the definite integral, limits of integration, the area under a curve, Riemann sums or the fundamental theorem. None of these notions belongs on your test.
• So you have no numerical means of checking. The only check available is to differentiate your answer and get the integrand back: ten seconds, and it catches the great majority of errors.
• Only one rule distributes, LINEARITY: $\int\left(af+bg\right)dx=a\int f\,dx+b\int g\,dx$ for CONSTANT $a$ and $b$ . There is no product rule and no quotient rule.

Writing the check on your test, and not just doing it in your head, often earns half a mark: the marker sees that the answer was checked.

The constant $C$ : a family, not a function

• Differentiating ERASES information: $F$ and $F+7$ have the same derivative. Integrating must give it back, and nothing says which one: hence the $+C$ , which denotes a FAMILY of parallel curves.
• THEOREM: if $F'=G'$ on an INTERVAL, then $F-G$ is constant. It is a consequence of the Mean Value Theorem, seen in the middle of the semester, and the word interval carries all the weight.
• Practical consequence: two answers that look very different can both be right. $\frac{4}{3}x^{3}-6x^{2}+9x$ and $\frac{(2x-3)^{3}}{6}$ differ by $-4.5$ and have the same derivative $(2x-3)^{2}$ .
• An imposed condition, a point or a measurement, selects exactly ONE member of the family. Without a condition, the answer remains the family, and the $+C$ IS the answer.

Three antiderivatives of

f(x)=3\sqrt{x}

: the vertical gap is

8

both at

x=1

and at

x=4

, so the three curves have the same slope everywhere. Differentiation brings them all back to the same

f

: it ERASES

C

Substitution, the chain rule played backwards

• Look in the integrand for an inner function $u$ AND its derivative $u'$ as a factor, up to a multiplicative CONSTANT.
• Protocol in four steps: set $u$ , write $du=u'(x)\,dx$ , replace everything so that NO $x$ remains, integrate in $u$ , then go BACK to $x$ .
• THE LINE: a constant comes out of the integral, a function of $x$ never does. $\int x\sqrt{4-x^{2}}\,dx$ takes thirty seconds; $\int\sqrt{4-x^{2}}\,dx$ is not part of 201-SN2-RE.
• The $u$ that keep coming back: $\frac{1}{x}$ calls for $u=\ln x$ , $\frac{1}{1+x^{2}}$ calls for $u=\arctan x$ , $\cos x$ calls for $u=\sin x$ , and a denominator in $\sqrt{x}$ calls for $u=\sqrt{x}$ .

An answer left in terms of $u$ is not an answer: the marker does not know what $u$ is, and the question drops to zero even though the computation was right.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The table of derivatives, read from right to left

You do not memorize a second table. You read the semester's table the other way: the middle column is the derivative line that applies, the right-hand column is the antiderivative. A red cell flags a line that DOES NOT EXIST.

$f(x)$ to integrate	The derivative that produces it	$\int f(x)\,dx$
$x^{n}$ , $n\neq -1$	$\left(\frac{x^{n+1}}{n+1}\right)'=x^{n}$	$\frac{x^{n+1}}{n+1}+C$
Example: $\int\sqrt[3]{x^{2}}\,dx=\int x^{2/3}dx=\frac{3}{5}x^{5/3}+C$ , which equals $0.6$ at $x=1$ .
$x^{-1}$ , by the power rule	$\left(\frac{x^{0}}{0}\right)'$	$\frac{x^{0}}{0}$ power rule excluded
Example: $\int\frac{dx}{x^{5}}=-\frac{1}{4x^{4}}+C$ works, but $n=-1$ would give a division by $0$ . What to do: Switch to the logarithm line: $\int\frac{dx}{x}=\ln\left\|x\right\|+C$ , absolute value included.
$\frac{1}{x}$	$\left(\ln\left\|x\right\|\right)'=\frac{1}{x}$	$\ln\left\|x\right\|+C$
Example: $\int\frac{4}{x}dx=4\ln\left\|x\right\|+C$ , whereas $\int\frac{4}{x^{2}}dx=\frac{4}{x}+C$ .
$\sin x$	$\left(\cos x\right)'=-\sin x$	$-\cos x+C$
Example: $\int 4\sin x\,dx=-4\cos x+C$ , which equals $-4$ at $x=0$ .
$\cos x$	$\left(\sin x\right)'=\cos x$	$\sin x+C$
Example: $\int 2\cos x\,dx=2\sin x+C$ , which equals $2$ at $x=\frac{\pi}{2}$ .
$\sec^{2}x$	$\left(\tan x\right)'=\sec^{2}x$	$\tan x+C$
Example: $\int\sec^{2}(7x)\,dx=\frac{1}{7}\tan(7x)+C$ , which equals $\frac{1}{7}$ at $x=\frac{\pi}{28}$ .
$\frac{1}{1+x^{2}}$	$\left(\arctan x\right)'=\frac{1}{1+x^{2}}$	$\arctan x+C$
Example: $\int\frac{6}{1+x^{2}}dx=6\arctan x+C$ , which equals $\frac{3\pi}{2}\approx 4.71$ at $x=1$ .
$\frac{1}{\sqrt{1-x^{2}}}$	$\left(\arcsin x\right)'=\frac{1}{\sqrt{1-x^{2}}}$	$\arcsin x+C$
Example: $\int\frac{3\,dx}{\sqrt{1-x^{2}}}=3\arcsin x+C$ , which equals $\frac{\pi}{2}$ at $x=\frac{1}{2}$ .
$e^{x}$	$\left(e^{x}\right)'=e^{x}$	$e^{x}+C$
Example: $\int 5e^{x}dx=5e^{x}+C$ , which equals $5$ at $x=0$ .

No line handles $\tan^{2}x$ , $\sin^{2}x$ or a product: these integrands are REWRITTEN first, with an identity or by expanding.

Is the $u'$ there? Three decisions and one exit

Faced with a product or a quotient, a single question settles it: is one piece, up to a constant, the derivative of the other? The red cell is the exit from the chapter, not an answer.

Integrand	Candidate $u$ and its $u'$	Decision and result
$8x\left(x^{2}+3\right)^{3}$	$u=x^{2}+3$ , $u'=2x$	$\left(x^{2}+3\right)^{4}+C$
Example: $8x=4\cdot 2x$ , so the integral equals $\left(x^{2}+3\right)^{4}+C$ , that is $256$ at $x=1$ with $C=0$ .
$x\sqrt{4-x^{2}}$	$u=4-x^{2}$ , $u'=-2x$	$-\frac{1}{3}\left(4-x^{2}\right)^{3/2}+C$
Example: $x\,dx=-\frac{1}{2}du$ , and the antiderivative equals $-\frac{8}{3}\approx -2.67$ at $x=0$ .
$\sqrt{4-x^{2}}$	$u=4-x^{2}$ , but no $x$	stop not in 201-SN2-RE
Example: Differentiating the candidate $-\frac{1}{3}\left(4-x^{2}\right)^{3/2}$ gives $x\sqrt{4-x^{2}}$ , with an extra factor $x$ . What to do: Without $u'$ as a factor, you do not substitute. This integral calls for a trigonometric substitution, a technique of the integral calculus course that follows.
$\frac{x+1}{x^{2}}$	$u=x^{2}$ , $u'=2x$ missing	rewrite: $\ln\left\|x\right\|-\frac{1}{x}+C$
Example: $\frac{x+1}{x^{2}}=\frac{1}{x}+\frac{1}{x^{2}}$ , and the antiderivative equals $-1$ at $x=1$ with $C=0$ .
$\frac{x+1}{x^{2}+2x+5}$	$u=x^{2}+2x+5$ , $u'=2(x+1)$	$\frac{1}{2}\ln\left(x^{2}+2x+5\right)+C$
Example: $(x+1)dx=\frac{1}{2}du$ , and the antiderivative equals $\frac{1}{2}\ln 5\approx 0.80$ at $x=0$ .

The last two lines have the SAME numerator $x+1$ : it is the denominator that decides, not the look of the fraction.

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. Forgetting the constant of integration

half a mark per integral, so 2 to 3 marks on a whole test

What not to write

“ $\int\left(6x^{2}-8x^{3}+5\right)dx=2x^{3}-2x^{4}+5x$ .”

What to write

“ $\int\left(6x^{2}-8x^{3}+5\right)dx=2x^{3}-2x^{4}+5x+C$ .”

Why: The result of an indefinite integral is not a function but a FAMILY of functions. Without the $+C$ , the answer denotes a single curve and therefore claims something false.

2. Setting the constant to zero because the initial value is zero

the whole problem, 5 marks

What not to write

“ $x(0)=0$ , so $C_{2}=0$ and $x(t)=8(t+1)^{3/2}+6t$ .”

What to write

“ $x(0)=8\cdot 1^{3/2}+0+C_{2}=0$ , so $C_{2}=-8$ and $x(t)=8(t+1)^{3/2}+6t-8$ .”

With

C_{2}=0

, the curve starts at

x(0)=8

(red): the condition

x(0)=0

is not met. Only

C_{2}=-8

(green) makes the position pass through the origin.

Why: The constant is not the initial value: it is what must be ADDED so that the formula gives back the initial value. Here the term $8(t+1)^{3/2}$ already equals $8$ at $t=0$ .

3. Manufacturing the missing factor by pulling it out of the integral

the whole question, and often the next question as well

What not to write

“ $\int\sqrt{4-x^{2}}\,dx=\frac{1}{x}\int x\sqrt{4-x^{2}}\,dx=-\frac{1}{3x}\left(4-x^{2}\right)^{3/2}+C$ .”

What to write

“The factor $x$ is missing and it DEPENDS on $x$ : it does not come out of the integral. Without $u'$ as a factor, you do not substitute, and $\int\sqrt{4-x^{2}}\,dx$ is not part of this course.”

Two neighbouring integrands, two worlds. On the one from the origin, the factor

x

u'

up to a constant: one substitution. The upper curve,

\sqrt{4-x^{2}}

, is a circular arc that 201-SN2-RE cannot integrate.

Why: Only a CONSTANT passes through the integral sign. A $\frac{1}{x}$ taken out and put back in changes the function being integrated, and differentiating the wrong answer shows it in one line.

4. Forgetting the coefficient the chain rule brings out

3 marks out of 10

What not to write

“ $\int(3x+2)^{5}dx=\frac{(3x+2)^{6}}{6}+C$ , by the power rule.”

What to write

“ $u=3x+2$ gives $du=3\,dx$ , so $\int(3x+2)^{5}dx=\frac{(3x+2)^{6}}{18}+C$ , the denominator being $6\times 3$ .”

Why: When you differentiate the wrong answer, the chain rule brings out a $3$ : you get $3(3x+2)^{5}$ , that is three times the integrand. It is the most frequent error of the chapter and the easiest to catch.

5. Lowering the exponent instead of raising it

2 marks, and the check by differentiation shows it in five seconds

What not to write

“ $\int\frac{dx}{x^{5}}=\int x^{-5}dx=\frac{x^{-6}}{-6}+C$ .”

What to write

“ $\int x^{-5}dx=\frac{x^{-4}}{-4}+C=-\frac{1}{4x^{4}}+C$ : the exponent INCREASES by $1$ , even when it stays negative.”

Why: Differentiation lowers the exponent, an antiderivative raises it: these are two opposite moves, and one test in three applies the move from earlier in the semester out of habit.

6. Applying the logarithm line to every quotient

3 marks out of 10

What not to write

“ $\int\frac{dx}{x^{5}}=\ln\left|x^{5}\right|+C$ , since there is a quotient.”

What to write

“The logarithm line handles ONLY the exponent $-1$ . Here $\int\frac{dx}{x^{5}}=-\frac{1}{4x^{4}}+C$ , and you refute the wrong answer by differentiating: $\left(\ln\left|x^{5}\right|\right)'=\frac{5}{x}$ .”

Why: The logarithm is not the antiderivative of quotients, it is the antiderivative of $\frac{1}{x}$ . The refutation does not even require knowing the right answer, only differentiating the wrong one.

7. Losing the absolute value, and forgetting the word interval

1 rigour mark, systematically deducted

What not to write

“ $\int\frac{dx}{x}=\ln x+C$ on $\mathbb{R}\setminus\{0\}$ .”

What to write

“ $\int\frac{dx}{x}=\ln\left|x\right|+C$ , and on an INTERVAL only: $\mathbb{R}\setminus\{0\}$ is the union of two intervals, so it needs two constants.”

Both branches have derivative

\frac{1}{x}

, and yet they differ by different constants: one constant PER interval, because

\mathbb{R}\setminus\{0\}

has two of them.

Why: Without the absolute value the formula is wrong for $x<0$ , whereas $\left(\ln(-x)\right)'=\frac{-1}{-x}=\frac{1}{x}$ is perfectly well defined there. And the theorem on antiderivatives holds only on an interval.

8. Reversing sine and cosine without the sign

half the marks of the question

What not to write

“ $\int 4\sin x\,dx=4\cos x+C$ , since sine gives cosine.”

What to write

“ $\int 4\sin x\,dx=-4\cos x+C$ , since $\left(\cos x\right)'=-\sin x$ . The table is asymmetric: the minus sign appears on one side and not on the other.”

\sin x

is positive between

0

and

\pi

, so its antiderivative must RISE there: it is

-\cos x

, which starts at

-1

and increases. The candidate

\cos x

, dashed, goes down: the graph rules it out.

Why: An answer of the opposite sign is spotted by evaluating at one point: $-4\cos x$ equals $-4$ at $x=0$ . A single number separates the right answer from its opposite.

9. Handing in an answer still written in u

1 mark, sometimes the whole question

What not to write

“ $\int\frac{\cos x}{2+\sin x}dx=\ln\left|u\right|+C$ .”

What to write

“With $u=2+\sin x$ , you get $\ln\left|u\right|+C$ , so $\int\frac{\cos x}{2+\sin x}dx=\ln\left(2+\sin x\right)+C$ , with no absolute value since $2+\sin x\geq 1$ .”

Why: The variable $u$ is a computing tool, not a variable of the problem. An answer in $u$ defines no function of $x$ , and the marker does not have to reconstruct the substitution.

10. Inventing a product rule for integrals

the whole question

What not to write

“ $\int f(x)g(x)\,dx=\left(\int f(x)dx\right)\left(\int g(x)dx\right)$ , just like the derivative of a product.”

What to write

“There is no product rule. Faced with a product, you expand, divide, or recognize a substitution. With $f=g=x$ : on the left $\frac{x^{3}}{3}$ , which equals $9$ at $x=3$ ; on the right $\frac{x^{4}}{4}$ , which equals $20.25$ .”

Why: The rule invoked does not even exist for differentiation, where $\left(fg\right)'=f'g+fg'$ . Two numbers are enough to refute it, and a numerical counterexample is worth more than a protest.

Which method to choose

Faced with an integral: read, rewrite or substitute

You do not choose based on the topic of the chapter, but on the FORM of the integrand. Three decisions, and one exit.

If a sum of powers of $x$ , possibly disguised as roots or quotients → rewrite as $x^{n}$ , then read the table term by term

Example: $\int\sqrt[3]{x^{2}}\,dx=\int x^{2/3}dx=\frac{3}{5}x^{5/3}+C$

The only exponent that resists is $-1$ , handled by the logarithm.
If a product or a quotient in which one piece is, up to a constant, the derivative of the other → substitute: set $u$ equal to the piece whose derivative is available

Example: $\int\frac{x+1}{x^{2}+2x+5}dx=\frac{1}{2}\ln\left(x^{2}+2x+5\right)+C$

$u$ is chosen according to whose derivative is THERE, never according to what looks complicated.
If a power or a transcendental function of a binomial $ax+b$ → chain rule backwards: integrate as if the inside were $x$ , then divide by $a$

Example: $\int(3x+2)^{5}dx=\frac{(3x+2)^{6}}{18}+C$ and $\int\sec^{2}(7x)dx=\frac{1}{7}\tan(7x)+C$

Works ONLY if the inside has degree 1, where $u'=a$ is constant.
If a square that can be expanded, a quotient of equal degrees, a power of tangent → rewrite: expand, divide, or apply an identity

Example: $\int\tan^{2}x\,dx=\int\left(\sec^{2}x-1\right)dx=\tan x-x+C$

A quotient whose numerator has a degree greater than or equal to that of the denominator is divided first.
If no factor looks like the derivative of another and nothing can be rewritten → stop: the integral is not part of 201-SN2-RE

Example: $\int\sqrt{4-x^{2}}\,dx$ , $\int\frac{dx}{\sqrt{x^{2}+9}}$ , $\int xe^{x}dx$

These three call for techniques of the integral calculus course that follows.

If no branch applies, reread the problem: at this level, an exam integral always falls into one of the first four. Integration by parts, partial fractions and trigonometric substitutions are not yet available to you.

How many constants, and where to fix them

In a motion problem, count the constants BEFORE computing: that count tells you whether the problem can be solved.

If the problem gives $y'$ and a value of $y$ → one integration, one constant, one condition: the problem is determined

Example: $y'=3\sqrt{x}$ with $y(4)=10$ gives $y=2x^{3/2}-6$
If the problem gives $a(t)$ and asks for the position → two integrations, two constants, so TWO conditions are needed

Example: $a(t)=\frac{6}{\sqrt{t+1}}$ , $v(3)=30$ and $x(0)=0$ give $x(t)=8(t+1)^{3/2}+6t-8$
If the condition is not given at time $0$ → integrate first, then replace $t$ with the given value and solve for $C$

Example: $v(3)=30$ gives $12\sqrt{4}+C_{1}=30$ , so $C_{1}=6$

The constant can be read at a glance only if the condition is at $t=0$ AND the rest vanishes there.
If a condition is missing → conclude that the family remains indexed, and SAY so rather than setting the constant to zero

Example: With $x(0)=0$ alone, $x(t)=8(t+1)^{3/2}+C_{1}t-8$ : the initial velocity remains unknown

A wrong constant leaves the check by differentiation perfectly green. That is why it goes unnoticed, and why checking the constants is a SEPARATE check.

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.

Computing an antiderivative, and checking it on the test

When to use it: As soon as the problem says “compute $\int\ldots dx$ ” or “find an antiderivative of”.

1 Rewrite the integrand as lines of the table: $\frac{1}{x^{5}}=x^{-5}$ , $\sqrt[3]{x^{2}}=x^{2/3}$ , $\frac{\left(\sqrt{x}+1\right)^{2}}{x}=1+2x^{-1/2}+\frac{1}{x}$ . No integration until this line is written.
2 Decide: table directly, rewriting or substitution. If you substitute, announce “let $u=\ldots$ , so $du=\ldots dx$ ”: the marker is looking for this line.
3 Integrate term by term by linearity, taking out the constants and only the constants.
4 Go back to $x$ if there was a substitution, then write $+C$ at the end of the line, not in your head.
5 Write the check: “ $\left(F(x)\right)'=\ldots$ , which is indeed the integrand.” It costs one line and it pays.

Concluding sentence

“So $\int f(x)\,dx=F(x)+C$ , and we check that $F'(x)=f(x)$ .”

The trap: Skipping the first step and attacking a product or a quotient directly: that is where rules that do not exist are born.

Marking: Out of 4 marks: 1 for the rewriting or the announced $u$ , 2 for the antiderivative, 0.5 for the $+C$ , 0.5 for the written check.

Working back from an acceleration to a position

When to use it: The problem gives $a(t)$ or $v(t)$ and asks for the velocity or the position, with one or two numerical values.

1 Count the integrations BEFORE computing, and write it down: from $a$ to $v$ , one constant; from $a$ to $x$ , two. You need as many conditions as constants.
2 Integrate once and write $v(t)=\ldots+C_{1}$ , with the constant visible.
3 Fix $C_{1}$ by substituting the velocity data at the instant it is given, which is not necessarily $t=0$ .
4 Integrate the COMPLETE formula for $v$ , constant included, to get $x(t)=\ldots+C_{2}$ .
5 Fix $C_{2}$ with the position data, first computing what the rest equals at that instant: a zero initial position does not give a zero constant.
6 Check by differentiating twice, then substitute EACH piece of data from the problem into the final formula.

Concluding sentence

“Two integrations call for two conditions: $v(3)=30$ fixes $C_{1}=6$ and $x(0)=0$ fixes $C_{2}=-8$ , so $x(t)=8(t+1)^{3/2}+6t-8$ .”

The trap: Integrating $v$ while leaving $C_{1}$ aside: the term $6t$ disappears and the position is wrong everywhere, while the differentiation of $v$ stays green.

Marking: Out of 10 marks: 3 per integration, 2 per constant correctly fixed.

Check before you hand in

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

Differentiate your answer, always

It is the only check this course has: no definite integral, no numerical value to compare with. Differentiate the antiderivative you found and compare it with the integrand in the problem, symbol by symbol.

$\left(\frac{(3x+2)^{6}}{6}\right)'=3(3x+2)^{5}$ : three times too much, the denominator should have been $18$ .
Scan your test for every $+C$

Before handing in, reread ONLY the equal signs of your indefinite integrals and check that each one ends with a $+C$ . Ten seconds for the whole test.

Three integrals without $+C$ often cost $1.5$ marks, more than any computational error in the chapter.
Substitute the condition back, separately

Differentiation validates the antiderivative but stays GREEN with a wrong constant. Put the instant given in the problem back into your final formula, and see whether it gives back the data.

With $C_{1}=0$ you get $v(3)=24$ whereas the radar said $30$ : the error shows without knowing the right answer.
Evaluate at one point to settle a sign

A sign error on a transcendental function turns the answer into its opposite. Evaluate at $x=0$ or $x=\frac{\pi}{2}$ with $C=0$ , and compare with the expected sign.

$-4\cos x$ equals $-4$ at $x=0$ ; if your answer equals $+4$ there, you read the sine line backwards.
Has the exponent gone UP by one?

Differentiation lowers the exponent, an antiderivative raises it. On each power in your answer, compare the exponent with the one in the problem: it must be one more, even when it stays negative.

From $x^{-5}$ you go to $x^{-4}$ , never to $x^{-6}$ : $-4$ is indeed greater than $-5$ .

The typical problem, taken apart

A cart, two integrations, two constants

A cart leaves a platform and accelerates along a horizontal rail. A sensor gives its acceleration $a(t)=\frac{6}{\sqrt{t+1}}$ in metres per second squared, for $t\geq 0$ in seconds.

The radar on the platform measured $v(3)=30$ metres per second, and the origin of positions is taken at the platform, so $x(0)=0$ . Find $x(t)$ , then the position at $t=3$ seconds.

The data in the problem: the acceleration decreases but stays positive, so the cart speeds up less and less without ever slowing down. This graph says NOTHING about the velocity, hence the measurement

v(3)=30

Step 1

$a(t)=6(t+1)^{-1/2}$ , so $v(t)=\int 6(t+1)^{-1/2}dt=6\cdot\frac{(t+1)^{1/2}}{1/2}+C_{1}=12\sqrt{t+1}+C_{1}$ .

Why

Rewrite as a power BEFORE integrating, otherwise the power rule cannot be recognized under the radical. The inside $t+1$ has derivative $1$ : no coefficient to offset, which would not be the case with $2t+1$ .

Step 2

The data is $v(3)=30$ : $12\sqrt{4}+C_{1}=24+C_{1}=30$ , so $C_{1}=6$ and $v(t)=12\sqrt{t+1}+6$ .

Why

The constant is fixed AFTER integrating, never before: $C_{1}$ does not exist until the antiderivative is written. And the data is not at $t=0$ , so you replace $t$ with $3$ in the formula, you do not read the constant at a glance.

Step 3

$x(t)=\int\left(12(t+1)^{1/2}+6\right)dt=12\cdot\frac{(t+1)^{3/2}}{3/2}+6t+C_{2}=8(t+1)^{3/2}+6t+C_{2}$ .

Why

Second integration, second constant, and the COMPLETE formula for $v$ goes into the integral: it is $C_{1}=6$ that produces the term $6t$ . Forgetting it makes the position wrong everywhere without breaking anything in the check of $v$ .

Step 4

$x(0)=0$ gives $8\cdot 1^{3/2}+0+C_{2}=0$ , so $C_{2}=-8$ and $x(t)=8(t+1)^{3/2}+6t-8$ .

Why

A zero initial position does NOT mean a zero constant: the term $8(t+1)^{3/2}$ already equals $8$ at $t=0$ , and the constant is there to cancel it. This is the line that earns the most marks in the problem.

Step 5

$x(3)=8\cdot 4^{3/2}+6\cdot 3-8=8\cdot 8+18-8=74$ metres.

Why

Answer the question asked, with its unit. The order of magnitude can be checked along the way: a velocity rising from $18$ to $30$ metres per second over three seconds puts the position around $70$ metres.

Step 6

Check: $x'(t)=8\cdot\frac{3}{2}\sqrt{t+1}+6=12\sqrt{t+1}+6=v(t)$ , then $v'(t)=12\cdot\frac{1}{2\sqrt{t+1}}=\frac{6}{\sqrt{t+1}}=a(t)$ . Conditions: $x(0)=8+0-8=0$ and $v(3)=24+6=30$ .

Why

The two checks are INDEPENDENT. The double differentiation validates the antiderivatives and would stay green with wrong constants; it is the substituted conditions that validate the constants.

The conclusion, written out

“The position of the cart is $x(t)=8(t+1)^{3/2}+6t-8$ metres, and $x(3)=74$ metres. The two differentiations give back $v$ and then $a$ , and both conditions $x(0)=0$ and $v(3)=30$ are satisfied.”

The classic mistake on this problem: Setting $C_{1}=C_{2}=0$ “since the cart starts from the platform”: you then find $x(3)=64$ metres instead of $74$ , and above all $v(3)=24$ , which contradicts the radar. Ten metres off, and the contradiction shows without knowing the right answer.

Learn by heart

• An antiderivative is the table of derivatives read from right to left: $F'=f$ , and nothing else.
• $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for $n\neq -1$ ; the case $n=-1$ gives $\ln\left|x\right|+C$ , absolute value INCLUDED.
• The exponent goes UP by $1$ : $x^{-5}$ gives $x^{-4}$ , never $x^{-6}$ .
• No product rule, no quotient rule. Faced with a product, rewrite or substitute.
• Substitution: $u$ AND $u'$ as a factor, up to a CONSTANT. A missing constant can be offset, a missing $x$ never.
• Two antiderivatives of the same function differ by a constant ON AN INTERVAL: a consequence of the Mean Value Theorem.
• Each integration adds a constant: from $a$ to $x$ , two constants, so two conditions.
• Free check: differentiating the answer must give back the integrand, and substituting the data must give back the data.

Frequently asked questions

What is the constant for if I am not given any initial condition?

Without a condition, the constant IS the answer: it says that all the parallel curves obtained are valid, since differentiation brings them all back to the same function. Leaving it out amounts to claiming that only one of them is right, which is false, and it costs half a mark on every integral.

Why do we study antiderivatives in a differential calculus course?

Because the course framework for Differential Calculus, 201-SN2-RE asks you to compute elementary integrals, and the last week of the semester is devoted to them. So you meet them before the definite integral: here an antiderivative is just the table of derivatives read the other way, with no area and no limits of integration.

How do I recognize that an integral is done by substitution?

Look in the integrand for an inner function and its derivative as a factor, up to a multiplicative constant. If the missing factor is a number, you adjust it and the integral falls into place. If a power of x is missing, you do not substitute: the integral changes nature and leaves the scope of this course.

How do I find the constant when the data is not at time zero?

Integrate first and keep the constant visible in the formula. Then replace the time with the given value, for example three seconds, set the result equal to the measured number, and solve for C. The constant can be read directly only if the data is at time zero and the rest of the formula vanishes there.

How do I check an antiderivative when I have no answer key?

Differentiate your own answer. If you get exactly the starting function back, the antiderivative is right, whatever it looks like. This check takes ten seconds and requires no theorem from the next course. It does not validate the constant: for that, substitute the numerical data from the problem into the final formula.

Practise it

Corrected exercises: Elementary integration, 201-SN2-RE

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.

12 corrected exercises
120 points
150 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 Applied optimization

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc1-elementary-integration. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: elementary integration (201-SN2-RE)

The essentials

What you have at this point of the course

The constant CCC: a family, not a function

Substitution, the chain rule played backwards

The rules in table form

The table of derivatives, read from right to left

Is the u′u'u′ there? Three decisions and one exit

The mistakes that cost marks

1. Forgetting the constant of integration

2. Setting the constant to zero because the initial value is zero

3. Manufacturing the missing factor by pulling it out of the integral

4. Forgetting the coefficient the chain rule brings out

5. Lowering the exponent instead of raising it

6. Applying the logarithm line to every quotient

7. Losing the absolute value, and forgetting the word interval

8. Reversing sine and cosine without the sign

9. Handing in an answer still written in u

10. Inventing a product rule for integrals

Which method to choose

Faced with an integral: read, rewrite or substitute

How many constants, and where to fix them

How the answer is expected to be written

Computing an antiderivative, and checking it on the test

Working back from an acceleration to a position

Check before you hand in

The typical problem, taken apart

A cart, two integrations, two constants

Learn by heart

Frequently asked questions

What is the constant for if I am not given any initial condition?

Why do we study antiderivatives in a differential calculus course?

How do I recognize that an integral is done by substitution?

How do I find the constant when the data is not at time zero?

How do I check an antiderivative when I have no answer key?

Corrected exercises: Elementary integration, 201-SN2-RE

See also

Looking for a 201-SN2-RE tutor in Montreal?

The constant $C$ : a family, not a function

Is the $u'$ there? Three decisions and one exit