MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: antiderivatives (MATH 140)

This sheet is not a summary of section 4.9 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on antiderivatives in MATH 140 at McGill University, and which precise gesture avoids each loss.

The chapter looks like the derivative table read backwards, and that is the danger: the table reads backwards line by line, but the RULES do not. Every value below is exact and done by hand, as on the final, and every trap comes with the one line of differentiation that would have caught it.

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The thread of the chapter

Antidifferentiation has no product, quotient or chain rule: REWRITE the function as a sum of table entries, read the table backwards, then CHECK by differentiating. And the answer is a family: one constant per antidifferentiation and per interval of the domain, fixed only by a condition.

This chapter is part of MATH 140, Calculus 1 (McGill)

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
General antiderivative
Arbitrary constant
Table of antiderivatives
Power rule for antiderivatives
Antiderivative of 1/x
Antiderivative of f(ax + b)
Initial value problem
Graph of an antiderivative
Motion from acceleration

The essentials

An antiderivative is a family, pinned by a condition

• $F$ is an antiderivative of $f$ on an interval $I$ when $F'(x) = f(x)$ for every $x$ in $I$ . The general antiderivative is $F(x) + C$ .
• Why $+C$ is everything: if $F' = G'$ on an INTERVAL, then $(F - G)' = 0$ and the Mean Value Theorem makes $F - G$ constant. On a domain with a gap, one constant per interval.
• The members of the family are vertical translates of each other: same slope at the same $x$ , never meeting. One point picks one member.
• Each antidifferentiation adds its own constant: from $f''$ you get $f(x) = G(x) + C_1 x + C_2$ , and you need two independent conditions.
• The check is always available: differentiate your answer and compare with $f$ . It takes one line, and it catches every missing factor.

Three antiderivatives of

f(x) = 2x - 2

: at

x = 2

their tangents are parallel, all of slope

f(2) = 2

. The curves differ only by a vertical shift.

Write $+ C$ on the very line where you antidifferentiate, and name the constants $C_1$ , $C_2$ as soon as there are two. A constant added at the end is a constant that was lost in between.

The table reads backwards, the rules do not

• Every line of the derivative table gives a line backwards: $\cos x \mapsto \sin x$ , $\sec^2 x \mapsto \tan x$ , $\frac{1}{1 + x^2} \mapsto \arctan x$ , $b^x \mapsto \frac{b^x}{\ln b}$ .
• Power rule backwards: $x^n \mapsto \frac{x^{n+1}}{n+1}$ for $n \ne -1$ only. For $n = -1$ : $\frac{1}{x} \mapsto \ln|x|$ .
• No product rule, no quotient rule, no chain rule backwards: expand, split over a single-term denominator, divide, or use an identity FIRST.
• One composition is allowed: a LINEAR inside. If $F' = f$ , then $\frac{1}{a}F(ax + b)$ is an antiderivative of $f(ax + b)$ , proved by the chain rule.
• A nonlinear inside, $e^{x^2}$ or $(x^2 + 1)^5$ , is out of reach of this chapter: expand if it is a polynomial, otherwise it is MATH 141.

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 read backwards, with the conditions that come with it

Read a line as: the function of the first column, under the condition of the second, has the antiderivative of the third (plus a constant on each interval). The red lines are rules that do not exist: they are the most frequent wrong lines on a MATH 140 final.

Function	Condition	Antiderivative
$x^n$	$n \ne -1$	$\frac{x^{n+1}}{n+1}$
Example: $x^{-3} \mapsto -\frac{1}{2x^2}$ and $\sqrt x = x^{1/2} \mapsto \frac{2}{3}x^{3/2}$ .
$\frac{1}{x}$	$x \ne 0$	$\ln\|x\|$
Example: $\frac{5}{x} \mapsto 5\ln\|x\|$ ; at $x = -2$ , $(\ln\|x\|)' = -\frac{1}{2} = \frac{1}{x}$ .
$e^{kx}$	$k \ne 0$	$\frac{1}{k}e^{kx}$
Example: $e^{-2x} \mapsto -\frac{1}{2}e^{-2x}$ , and the check gives $-\frac{1}{2} \cdot (-2)e^{-2x} = e^{-2x}$ .
$b^x$	$b > 0$ , $b \ne 1$	$\frac{b^x}{\ln b}$
Example: $10^x \mapsto \frac{10^x}{\ln 10}$ , never $\frac{10^{x+1}}{x+1}$ .
$f(ax + b)$	$a \ne 0$	$\frac{1}{a}F(ax + b)$
Example: $(3x - 1)^4 \mapsto \frac{(3x - 1)^5}{15}$ and $\cos 2x \mapsto \frac{\sin 2x}{2}$ .
$f(x)g(x)$	any	$F(x)G(x)$ rule that does not exist
Example: $x \cdot x$ : the product $\frac{x^2}{2} \cdot \frac{x^2}{2} = \frac{x^4}{4}$ has derivative $x^3$ , not $x^2$ . What to do: Expand first: $x \cdot x = x^2 \mapsto \frac{x^3}{3}$ .
$\frac{f(x)}{g(x)}$	any	$\frac{F(x)}{G(x)}$ rule that does not exist
Example: $\frac{x^2}{x^2 + 1}$ is $1 - \frac{1}{x^2 + 1}$ , whose antiderivative is $x - \arctan x$ , not a quotient. What to do: Split over a single-term denominator, or make the numerator contain the denominator (long division).
$f(g(x))$	$g$ not linear	$\frac{F(g(x))}{g'(x)}$ rule that does not exist
Example: $\frac{(x^2 + 1)^3}{6x}$ has derivative $(x^2 + 1)^2 - \frac{(x^2 + 1)^3}{6x^2}$ , not $(x^2 + 1)^2$ . What to do: Expand the polynomial: $(x^2 + 1)^2 = x^4 + 2x^2 + 1 \mapsto \frac{x^5}{5} + \frac{2x^3}{3} + x$ .

The last three lines fail for the same reason: the derivative of a product or of a composition has more than one factor, and dividing by one of them only works when it is a CONSTANT.

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. Dropping the constant, then meeting the initial condition

the whole question

What not to write

“ $f'(x) = 3x^2 - 4x + 1$ , so $f(x) = x^3 - 2x^2 + x$ . But $f(2) = 2$ , not $5$ .”

What to write

“ $f(x) = x^3 - 2x^2 + x + C$ and $f(2) = 2 + C = 5$ , so $C = 3$ : $f(x) = x^3 - 2x^2 + x + 3$ .”

Why: Without $C$ you hold one member of the family, almost never the one the condition asks for, and nothing is left to adjust.

2. Applying the power rule to the exponent -1

1 to 2 marks

What not to write

“ $\frac{3}{x} = 3x^{-1} \mapsto 3 \cdot \frac{x^0}{0}$ ”, or “ $\frac{3}{x} \mapsto 3\ln x$ on $x \ne 0$ ”.

What to write

“ $\frac{3}{x} \mapsto 3\ln|x| + C_1$ for $x > 0$ and $3\ln|x| + C_2$ for $x < 0$ .”

Why: $n = -1$ is the one exponent the power rule excludes, and $\ln x$ alone is undefined on half of the domain of $\frac{1}{x}$ .

3. Multiplying the antiderivatives of the two factors

the whole question

What not to write

“ $(x^2 + 1)(2x - 3) \mapsto \left(\frac{x^3}{3} + x\right)(x^2 - 3x) + C$ .”

What to write

“ $(x^2 + 1)(2x - 3) = 2x^3 - 3x^2 + 2x - 3 \mapsto \frac{x^4}{2} - x^3 + x^2 - 3x + C$ .”

Why: The derivative of a product has two terms, so undoing a product cannot give one. Differentiating the wrong answer at $x = 1$ gives $-\frac{16}{3}$ instead of $-2$ .

4. Forgetting the 1/a of a linear inside function

1 mark, each time

What not to write

“ $\sin 3x \mapsto -\cos 3x + C$ .”

What to write

“ $\sin 3x \mapsto -\frac{1}{3}\cos 3x + C$ . Check: $-\frac{1}{3} \cdot (-\sin 3x) \cdot 3 = \sin 3x$ .”

Why: The chain rule puts the factor $3$ back when you differentiate; the antiderivative must carry $\frac{1}{3}$ in advance. The check finds it in one line.

5. Dividing by the derivative of a nonlinear inside

the whole question

What not to write

“ $(x^2 + 1)^2 \mapsto \frac{(x^2 + 1)^3}{3 \cdot 2x}$ , as for $(3x - 1)^4$ .”

What to write

“ $(x^2 + 1)^2 = x^4 + 2x^2 + 1 \mapsto \frac{x^5}{5} + \frac{2x^3}{3} + x + C$ .”

Why: Dividing by $a$ works because $a$ is a constant. $2x$ is not: the quotient rule differentiates it too, and an extra term appears.

6. Forcing one constant across a gap in the domain

the whole question

What not to write

“ $f'(x) = \frac{1}{x^2}$ , $f(1) = 0$ , $f(-1) = 0$ : $f = -\frac{1}{x} + C$ gives $C = 1$ , then $f(-1) = 2$ . No solution.”

What to write

“ $f(x) = 1 - \frac{1}{x}$ for $x > 0$ and $f(x) = -1 - \frac{1}{x}$ for $x < 0$ : one constant per interval.”

Blue and orange have the same slope

\frac{1}{x^2}

at every

x \ne 0

, yet orange is blue moved up by

1

on the right and down by

2

on the left.

Why: The Mean Value Theorem only compares two antiderivatives on an INTERVAL. Across the hole at $0$ the constants are independent, and the figure shows two branches shifted by different amounts.

7. Using the same constant twice for a second derivative

2 to 3 marks, and a system that has no solution

What not to write

“ $f''(x) = 12x^2 + 4e^{2x}$ , so $f'(x) = 4x^3 + 2e^{2x} + C$ and $f(x) = x^4 + e^{2x} + C$ .”

What to write

“ $f'(x) = 4x^3 + 2e^{2x} + C_1$ , then $f(x) = x^4 + e^{2x} + C_1 x + C_2$ .”

Why: The antiderivative of the constant $C_1$ is $C_1 x$ , and the second antidifferentiation brings a new constant. With $f'(0) = 1$ , the wrong line gives $f'(0) = 2$ whatever $C$ is.

8. Reading the direction of F from the direction of f

2 marks on a sketch question

What not to write

“ $f$ is increasing on $[-3, 3]$ , so its antiderivative $F$ is increasing on $[-3, 3]$ .”

What to write

“ $F$ increases where $f > 0$ and decreases where $f < 0$ ; $f$ increasing means $F$ concave up.”

The orange line

f(x) = x

rises everywhere, but on

(-3, 0)

it is below the axis and

F(x) = \frac{x^2}{2} - 2

falls:

F

follows the sign of

f

Why: $F' = f$ : the SIGN of $f$ gives the direction of $F$ , the direction of $f$ gives the bending of $F$ . A corner of $f$ shows up as an inflection point of $F$ , never as a corner.

Which method to choose

Which gesture, by the FORM of the function

Look at the shape of f before reading any table: the form picks the rewriting

If a sum of powers, roots, or $\frac{c}{x^k}$ with $k \ne 1$ → rewrite as $x^n$ , then the power rule term by term

Example: $8x^3 - 6\sqrt x + \frac{3}{x^4} \mapsto 2x^4 - 4x^{3/2} - x^{-3}$
If $\frac{c}{x}$ , or $\frac{c}{ax + b}$ → $c\ln|x|$ , or $\frac{c}{a}\ln|ax + b|$ , one constant per side

Example: $\frac{1}{2x + 5} \mapsto \frac{1}{2}\ln|2x + 5|$
If a product of two polynomials or powers → expand, then term by term

Example: $(x^2 + 1)(2x - 3) = 2x^3 - 3x^2 + 2x - 3$
If a quotient with a single-term denominator → split into one fraction per term of the numerator

Example: $\frac{x^4 - 3x^2 + 6}{x^3} = x - \frac{3}{x} + \frac{6}{x^3}$
If a quotient of polynomials, numerator of degree at least the denominator → long division, or add and subtract to make the denominator appear

Example: $\frac{x + 3}{x + 1} = 1 + \frac{2}{x + 1} \mapsto x + 2\ln|x + 1|$
If a square, product or quotient of trigonometric functions → one identity: half-angle, Pythagoras, double angle, or a conjugate

Example: $\cos^2 x = \frac{1 + \cos 2x}{2} \mapsto \frac{x}{2} + \frac{\sin 2x}{4}$
If a table function of $ax + b$ → $\frac{1}{a}F(ax + b)$ , then differentiate to check

Example: $e^{5 - 2x} \mapsto -\frac{1}{2}e^{5 - 2x}$

If no branch applies, for instance $e^{x^2}$ or $x\cos(x^2)$ , the function is not an exam question of MATH 140: a proposed answer can still be CHECKED by differentiating, which is how $x\ln x - x$ is recognized as an antiderivative of $\ln x$ .

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.

Solving an initial value problem with a second derivative

When to use it: Any question that gives $f''$ and two values, of $f$ , of $f'$ , or a tangent line, and asks for $f$

1 Antidifferentiate once and write the first constant: $f'(x) = G(x) + C_1$ .
2 If a condition on $f'$ is given, use it NOW to find $C_1$ .
3 Antidifferentiate again, remembering that $C_1$ becomes $C_1 x$ , and write the new constant $C_2$ .
4 Use the remaining condition or conditions; with two conditions on $f$ , solve the system in $C_1$ and $C_2$ , starting with the one at $x = 0$ .
5 Check: differentiate twice, and substitute each condition.

Concluding sentence

“Since $f''(x) = 12x - 6$ , $f'(x) = 6x^2 - 6x + C_1$ and $f(x) = 2x^3 - 3x^2 + C_1 x + C_2$ . From $f(0) = 2$ , $C_2 = 2$ ; from $f(1) = 3$ , $2 - 3 + C_1 + 2 = 3$ , so $C_1 = 2$ . Hence $f(x) = 2x^3 - 3x^2 + 2x + 2$ .”

The trap: Writing a single $C$ for both steps, or $C$ instead of $C_1 x$ : the system then has no solution, or the wrong one.

Marking: Typically 2 marks for each antidifferentiation with its constant, 1 for each condition used, 2 for the final function and its check.

Check before you hand in

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

Differentiate the answer

One line of differentiation, compared with the function you started from. A factor of 3 or 1/2 too many, or a wrong sign, shows at once.

$\frac{(3x - 1)^5}{5}$ differentiates to $3(3x - 1)^4$ : three times too much, the answer is $\frac{(3x - 1)^5}{15}$ .
Substitute every condition

Plug each given value back into your final function, and each slope into its derivative. A constant computed with a sign error fails here.

$f(x) = 2x^{3/2} - \frac{2}{5}x^{5/2} - \frac{16}{5}$ : $f(4) = 16 - \frac{64}{5} - \frac{16}{5} = 0$ , as required.
Count the constants

One constant per antidifferentiation, times the number of intervals of the domain. Then count the conditions: they must reach every constant, once.

$f'' = 6x$ with $f'(-1) = 3$ and $f'(1) = 3$ : two conditions, but both on $C_1$ , so $C_2$ stays free.
The physical sense of a motion answer

Times are positive, a downward velocity is negative on an upward axis, and energy gives the impact speed without the time.

Thrown up at $14.7$ m/s from $44.1$ m: $v^2 = 14.7^2 + 2 \cdot 9.8 \cdot 44.1 = 5 \cdot 14.7^2$ , so $|v| = 14.7\sqrt 5$ .

The typical problem, taken apart

A probe launched straight up on the Moon

On the Moon, where the acceleration of gravity is $1.6$ m/s $^2$ , a probe is launched straight up at $8$ m/s from a platform $2.4$ m above the ground. Find its maximum height, the time at which it lands, and its velocity at landing.

No calculator: give exact values, then an estimate.

Step 1

Axis upward, origin on the ground, $t$ in seconds from the launch: $a(t) = -1.6$ , $v(0) = 8$ , $s(0) = 2.4$ .

Why

The signs of every later answer come from this line. With the axis up, gravity is negative, and a final velocity found negative will mean downward.

Step 2

$v(t) = -1.6t + C_1$ and $v(0) = C_1 = 8$ : $v(t) = 8 - 1.6t$ . Then $s(t) = 8t - 0.8t^2 + C_2$ and $s(0) = C_2 = 2.4$ : $s(t) = 2.4 + 8t - 0.8t^2$ .

Why

Two antiderivatives, two constants, each found at once from its own initial value. Naming them shows the marker that you know what they are.

Step 3

Highest point: $v(t) = 0$ at $t = 5$ s, and $s(5) = 2.4 + 40 - 20 = 22.4$ m. It is a maximum since $v > 0$ before and $v < 0$ after.

Why

The top is where the velocity changes sign, not where it is small. The justification is the first derivative test on $s$ , with $s' = v$ .

Step 4

Landing: $s(t) = 0$ , that is $0.8t^2 - 8t - 2.4 = 0$ , or $t^2 - 10t - 3 = 0$ . So $t = 5 \pm \sqrt{28} = 5 \pm 2\sqrt 7$ ; the negative root is rejected: $t = 5 + 2\sqrt 7 \approx 10.3$ s.

Why

Dividing by $0.8$ first makes the equation solvable by hand. A root before the launch belongs to the parabola, not to the flight.

Step 5

$v(5 + 2\sqrt 7) = 8 - 8 - 3.2\sqrt 7 = -3.2\sqrt 7 \approx -8.5$ m/s. Check by energy: $v^2 = 8^2 + 2 \cdot 1.6 \cdot 2.4 = 71.68 = 3.2^2 \cdot 7$ .

Why

The negative sign says downward, as it should. The energy check does not use the landing time, so it catches an error made in the previous step.

The conclusion, written out

“The probe reaches $22.4$ m above the ground at $t = 5$ s, lands at $t = 5 + 2\sqrt 7 \approx 10.3$ s, with velocity $-3.2\sqrt 7 \approx -8.5$ m/s, that is about $8.5$ m/s downward.”

The classic mistake on this problem: Taking $s(0) = 0$ at the platform and then solving $s(t) = 0$ for the landing: that time is when the probe passes back at the platform, $t = 10$ s, not when it lands.

Learn by heart

• $F' = f$ on an interval: every antiderivative is $F + C$ . One constant per interval of the domain.
• $x^n \mapsto \frac{x^{n+1}}{n+1}$ for $n \ne -1$ ; $\frac{1}{x} \mapsto \ln|x|$ ; $b^x \mapsto \frac{b^x}{\ln b}$ .
• No product, quotient or chain rule backwards: expand, split, divide or use an identity FIRST.
• Linear inside only: $f(ax + b) \mapsto \frac{1}{a}F(ax + b)$ .
• From $f''$ : $f = G + C_1 x + C_2$ , two constants, two conditions that reach both.
• $F$ increases where $f > 0$ ; $F$ is concave up where $f$ increases.
• Motion: $v' = a$ , $s' = v$ ; the constants are $v(0)$ and $s(0)$ , or continuity at a junction.
• Always finish by differentiating your answer.

Frequently asked questions

What is the difference between an antiderivative and the general antiderivative?

An antiderivative of f is one function whose derivative is f. The general antiderivative is the whole family obtained by adding an arbitrary constant, because two antiderivatives on the same interval always differ by a constant. On a domain made of several intervals, such as all x except zero, there is one independent constant on each interval.

Why is the antiderivative of 1 over x the natural log of the absolute value of x?

The power rule divides by n plus one, which is zero when n is minus one, so it cannot be used. For positive x the derivative of ln x is one over x. For negative x, the chain rule gives the derivative of ln of minus x as one over x as well. The absolute value covers both sides, with a separate constant on each.

How do I find the antiderivative of a product in MATH 140?

There is no product rule for antiderivatives, and the product of the two antiderivatives is wrong. Expand the product into a sum of powers, or use a trigonometric identity to turn it into a sum, and then antidifferentiate term by term. Check the result by differentiating it: the check takes one line and catches the error every time.

How do I solve an initial value problem with a second derivative?

Antidifferentiate once and add a first constant, then antidifferentiate again: the first constant becomes that constant times x, and a second constant appears. Use the two given conditions to find the two constants, starting with any condition at x equal to zero. Finish by differentiating twice and substituting each condition.

Is an antiderivative the same as an indefinite integral?

They describe the same family of functions. MATH 140 calls it the general antiderivative, and MATH 141 gives it that second name, with its own notation, once the Fundamental Theorem of Calculus has been introduced. The techniques of this chapter, the table read backwards and the constant fixed by a condition, carry over unchanged.

Practise it

Corrected exercises: Antiderivatives, MATH 140 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

Previous sheet Optimization problems Next sheet MATH 141: Riemann sums

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-antiderivatives. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: antiderivatives (MATH 140)

The essentials

An antiderivative is a family, pinned by a condition

The table reads backwards, the rules do not

The rules in table form

The table read backwards, with the conditions that come with it

The mistakes that cost marks

1. Dropping the constant, then meeting the initial condition

2. Applying the power rule to the exponent -1

3. Multiplying the antiderivatives of the two factors

4. Forgetting the 1/a of a linear inside function

5. Dividing by the derivative of a nonlinear inside

6. Forcing one constant across a gap in the domain

7. Using the same constant twice for a second derivative

8. Reading the direction of F from the direction of f

Which method to choose

Which gesture, by the FORM of the function

How the answer is expected to be written

Solving an initial value problem with a second derivative

Check before you hand in

The typical problem, taken apart

A probe launched straight up on the Moon

Learn by heart

Frequently asked questions

What is the difference between an antiderivative and the general antiderivative?

Why is the antiderivative of 1 over x the natural log of the absolute value of x?

How do I find the antiderivative of a product in MATH 140?

How do I solve an initial value problem with a second derivative?

Is an antiderivative the same as an indefinite integral?

Corrected exercises: Antiderivatives, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?