MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: the Fundamental Theorem of Calculus and net change (MATH 141)

Q: How do I differentiate an integral when the upper limit is a function of x?

Evaluate the integrand at the upper limit, then multiply by the derivative of that limit: this is the first part of the Fundamental Theorem combined with the chain rule. For example, the derivative of the integral from 1 to the square root of x of e to the t squared is e to the x, divided by two times the square root of x.

Q: Why is the integral of 1 over x squared from minus 1 to 1 not equal to minus 2?

Because the second part of the Fundamental Theorem requires the integrand to be continuous on the whole closed interval, and one over x squared is not even defined at zero. The computation minus 1 minus 1 still produces a number, but it is meaningless: a positive function cannot have a negative integral. The integral is improper, a topic of its own later in the course.

Q: What is the difference between displacement and distance travelled?

Displacement is the integral of the velocity, a net change in which motion to the left cancels motion to the right. Distance is the integral of the absolute value of the velocity. To compute it, find the times where the velocity is zero, integrate on each piece separately, and add the absolute values of the results.

Q: Can two different answers to the same indefinite integral both be correct?

Yes, if they differ by a constant on each interval of the domain. The antiderivatives of two sine x cosine x include sine squared x, minus cosine squared x, and minus one half cosine of 2x, which differ by constants. The safe test is always the same: differentiate your answer and compare it with the integrand.

Q: Do I need an antiderivative to use the first part of the Fundamental Theorem?

No. The first part gives the derivative of an accumulation function directly from the integrand, evaluated at the moving bound. That is why exam questions on it often use integrands like e to the t squared or sine of t squared, which have no elementary antiderivative: they force you to use the theorem instead of computing the integral.

This sheet is not a summary of sections 5.3 and 5.4 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on the Fundamental Theorem of Calculus in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every value below is exact and done by hand, as on the midterm, and every antiderivative quoted has been checked the way a marker checks it: by differentiating it.

The thread of the chapter

An integral is a SIGNED accumulation, linked to its rate in both directions: differentiating it reads the integrand AT THE MOVING BOUND, times the derivative of that bound; evaluating it gives $F(b) - F(a)$ only if $f$ is continuous on the whole interval, and that number is a NET change, never a total.

This chapter is part of MATH 141, Calculus 2 (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

Fundamental Theorem of Calculus, part 1
Fundamental Theorem of Calculus, part 2
Accumulation function
Antiderivative
Indefinite integral
Table of antiderivatives
Net Change Theorem
Displacement and distance travelled
Natural logarithm as an integral
Chain rule
Continuity on a closed interval

The essentials

Two halves, two directions, one hypothesis

• FTC 1 (differentiate an accumulation): if $f$ is continuous, $g(x) = \int_a^x f(t)\,dt$ has $g'(x) = f(x)$ . The area grows at the rate given by the HEIGHT of $f$ at the moving bound.
• FTC 2 (evaluate an integral): if $f$ is continuous on $[a, b]$ and $F' = f$ on $[a, b]$ , then $\int_a^b f(x)\,dx = F(b) - F(a)$ .
• The letter $t$ inside $\int_a^x f(t)\,dt$ is a dummy variable. The variable $x$ lives only in the bound, which is why the derivative is read AT the bound.
• Both halves need continuity. For FTC 2 it must hold on the WHOLE closed interval, endpoints included, and $F$ must be an antiderivative on the whole interval.
• An integral is SIGNED: parts below the axis subtract. $\int_a^b$ of a rate is a NET change, never a total, until the sign changes have been split off.

g(x)

is the shaded area from

a

x

. Moving

x

x + h

adds a strip of width

h

and height about

f(x)

: that is why

g'(x) = f(x)

Before writing anything, say which half you are using: a question with $x$ in a bound is FTC 1, a question with two numbers as bounds is FTC 2. That sentence is already a method mark.

Moving bounds: one rule covers every case

• $\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f(v(x))\,v'(x) - f(u(x))\,u'(x)$ , for $f$ continuous between the bounds.
• A CONSTANT bound contributes nothing: its derivative is $0$ . Both bounds constant: the integral is a number and its derivative is $0$ .
• A variable in the LOWER bound brings a minus sign: $\frac{d}{dx}\int_x^{\pi} \cos(t^2)\,dt = -\cos(x^2)$ .
• If $x$ also appears INSIDE the integrand, take it out first, then use the product rule.
• An integrand with no elementary antiderivative ( $e^{t^2}$ , $\sin(t^2)$ , $\sqrt{1 + t^3}$ ) is a signal: the question is about FTC 1.

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 derivative of an integral, by where x sits

Read a line as: for this expression, apply this rule, and the derivative is in the last column. The red line is the rule students invent.

Expression	Rule	Derivative
$\int_0^x \sqrt{1 + t^3}\,dt$	FTC 1	$\sqrt{1 + x^3}$
Example: At $x = 2$ : $\sqrt{9} = 3$ . Not $\sqrt{1 + x^3} - 1$ : the constant bound adds nothing.
$\int_x^{\pi} \cos(t^2)\,dt$	swap the bounds	$-\cos(x^2)$
Example: At $x = 0$ : $-\cos 0 = -1$ .
$\int_1^{\sqrt{x}} e^{t^2}\,dt$	chain rule	$\frac{e^{x}}{2\sqrt{x}}$
Example: At $x = 1$ : $\frac{e}{2}$ , about $1.36$ .
$\int_x^{x^2} \sqrt{1 + t^4}\,dt$	split at a constant	$2x\sqrt{1 + x^8} - \sqrt{1 + x^4}$
Example: At $x = 1$ : $2\sqrt{2} - \sqrt{2} = \sqrt{2}$ .
$\int_2^5 \sqrt{1 + t^4}\,dt$	a number	$0$
Example: The integral is about $39.1$ , a constant; its derivative is $0$ .
$\int_0^x (x - t)\cos t\,dt$	take $x$ out, product rule	$\int_0^x \cos t\,dt = \sin x$
Example: Then $G'' = \cos x$ , $G(0) = G'(0) = 0$ , so $G(x) = 1 - \cos x$ and $G(\pi) = 2$ .
$\int_x^{x^2} f(t)\,dt$	f at the two bounds	$f(x^2) - f(x)$ no such rule
Example: For $f(t) = t$ : the integral is $\frac{x^4 - x^2}{2}$ , whose derivative is $2x^3 - x$ , not $x^2 - x$ . What to do: Multiply each value by the derivative of its bound: $f(x^2) \cdot 2x - f(x) \cdot 1$ .

Every blue line is the same rule, $f(v)\,v' - f(u)\,u'$ , read in a particular case. If you cannot say which $u$ and $v$ you are using, you are about to write the red line.

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 derivative of the bound

half the marks of the question, every time the bound is not simply x

What not to write

“ $\frac{d}{dx}\int_1^{\sqrt{x}} e^{t^2}\,dt = e^{x}$ .”

What to write

“By FTC 1 and the chain rule, $\frac{d}{dx}\int_1^{\sqrt{x}} e^{t^2}\,dt = e^{(\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{x}}{2\sqrt{x}}$ .”

Why: The accumulation is a composition $G(u(x))$ with $G(u) = \int_1^u e^{t^2}\,dt$ and $u = \sqrt{x}$ . FTC 1 differentiates $G$ , the chain rule owes the factor $u'(x)$ . Test on a case you can compute: $\int_1^{x^2} \frac{dt}{t} = 2\ln|x|$ has derivative $\frac{2}{x} = \frac{1}{x^2} \cdot 2x$ , factor included.

2. Subtracting the value at a constant lower bound

1 mark, and the whole question when it is used for a tangent line

What not to write

“ $\frac{d}{dx}\int_0^x \sqrt{1 + t^3}\,dt = \sqrt{1 + x^3} - \sqrt{1 + 0^3} = \sqrt{1 + x^3} - 1$ .”

What to write

“ $\frac{d}{dx}\int_0^x \sqrt{1 + t^3}\,dt = \sqrt{1 + x^3}$ : the lower bound is a constant and contributes nothing.”

Why: FTC 1 is not an evaluation. Evaluating the INTEGRAND at the bounds, $f(b) - f(a)$ , computes nothing at all; the lower bound only fixes where the accumulation starts, a constant that disappears when differentiating.

3. Evaluating F(b) minus F(a) across a discontinuity

all the marks of the question, plus a negative area for a positive function

What not to write

“ $\int_{-1}^{1} \frac{dx}{x^2} = \Big[-\frac{1}{x}\Big]_{-1}^{1} = -1 - 1 = -2$ .”

What to write

“ $\frac{1}{x^2}$ is not continuous at $0 \in [-1, 1]$ , so FTC 2 does not apply: the integral is improper.”

The antiderivative

-\frac{1}{x}

has two separate branches:

F(-1) = 1

and

F(1) = -1

are not on the same piece, so their difference measures no area.

Why: FTC 2 requires continuity on the WHOLE closed interval. A positive integrand cannot have a negative integral; the $-2$ only compares two points on two separate branches of $-\frac{1}{x}$ . Scan every denominator for zeros inside $[a, b]$ before evaluating.

4. Using an antiderivative that jumps inside the interval

the whole question, and the sign of the answer

What not to write

“ $F(x) = -\arctan\frac{1}{x}$ has $F'(x) = \frac{1}{1 + x^2}$ , so $\int_{-1}^{1} \frac{dx}{1 + x^2} = F(1) - F(-1) = -\frac{\pi}{2}$ .”

What to write

“ $\arctan x$ is an antiderivative on all of $\mathbb{R}$ : $\int_{-1}^{1} \frac{dx}{1 + x^2} = \arctan 1 - \arctan(-1) = \frac{\pi}{2}$ .”

Why: Here the integrand is continuous but $F$ is not defined at $0$ and jumps by $\pi$ there: it is an antiderivative on each half separately, with two different constants. FTC 2 needs ONE antiderivative valid on the whole interval.

5. Reporting the displacement as the distance travelled

2 to 3 marks, the most tested sentence of the chapter

What not to write

“ $v(t) = t^2 - 2t - 3$ , so the distance on $[0, 4]$ is $\left|\int_0^4 v(t)\,dt\right| = \frac{20}{3}$ m.”

What to write

“ $v = (t - 3)(t + 1)$ changes sign at $t = 3$ : distance $= -\int_0^3 v + \int_3^4 v = 9 + \frac{7}{3} = \frac{34}{3}$ m.”

Why: The Net Change Theorem integrates a signed rate and returns a signed, NET change. The distance is $\int |v|$ , and the absolute value is removed only by splitting at the zeros of $v$ . Taking the absolute value at the end cancels nothing: the lobes have already been subtracted.

6. Inventing a quotient rule for integrals

the whole question: nothing after the first line is marked

What not to write

“ $\int_1^4 \frac{x^2 + 1}{\sqrt{x}}\,dx = \left[\frac{x^3/3 + x}{\frac{2}{3}x^{3/2}}\right]_1^4$ .”

What to write

“ $\frac{x^2 + 1}{\sqrt{x}} = x^{3/2} + x^{-1/2}$ , so the integral is $\Big[\frac{2}{5}x^{5/2} + 2x^{1/2}\Big]_1^4 = \frac{84}{5} - \frac{12}{5} = \frac{72}{5}$ .”

Why: The table has no product rule and no quotient rule. Rewrite by algebra (divide term by term, expand, use an identity) until every term is a table entry; then check by differentiating.

7. Writing ln x where ln |x| is needed

1 to 2 marks

What not to write

“ $\int_{-e}^{-1} \frac{dx}{x} = \ln(-1) - \ln(-e)$ , which is undefined, so the integral does not exist.”

What to write

“ $\frac{1}{x}$ is continuous on $[-e, -1]$ and $\ln|x|$ is an antiderivative there: $\ln 1 - \ln e = -1$ .”

Why: On $(-\infty, 0)$ , $\ln|x| = \ln(-x)$ has derivative $\frac{-1}{-x} = \frac{1}{x}$ . The absolute value is what makes the table valid for negative $x$ ; the negative answer matches the negative integrand.

8. Leaving x inside the integrand when differentiating

the whole question

What not to write

“ $G(x) = \int_0^x (x - t)\cos t\,dt$ , so by FTC 1, $G'(x) = (x - x)\cos x = 0$ .”

What to write

“ $G(x) = x\int_0^x \cos t\,dt - \int_0^x t\cos t\,dt$ , so $G'(x) = \int_0^x \cos t\,dt + x\cos x - x\cos x = \sin x$ .”

Why: FTC 1 applies to $\int_a^x f(t)\,dt$ where the integrand depends on $t$ only. An $x$ inside is a constant for the integration, so it comes out in front, and then the product rule applies. The false answer would make $G$ constant, yet $G(\pi) = 2 \neq G(0) = 0$ .

Which method to choose

Which half of the theorem, by the FORM of the question

Look at where x sits, and at what is asked, before writing any antiderivative

If x only in the upper bound, lower bound constant → FTC 1: the integrand evaluated at x, nothing else

Example: $\frac{d}{dx}\int_0^x \sqrt{1 + t^3}\,dt = \sqrt{1 + x^3}$
If the bound is a function of x → integrand at that function, times its derivative

Example: $\frac{d}{dx}\int_1^{x^2} \sqrt{1 + t^3}\,dt = 2x\sqrt{1 + x^6}$
If x in both bounds → split at a constant where f is continuous, apply the rule to each piece

Example: $\int_x^{x^2} = \int_0^{x^2} - \int_0^x$
If x also inside the integrand → take x out of the integral, then product rule and FTC 1

Example: $\int_0^x (x - t)\cos t\,dt = x\int_0^x \cos t\,dt - \int_0^x t\cos t\,dt$
If two numbers as bounds, a value is asked → FTC 2: check continuity on [a, b], rewrite by algebra, table, then F(b) minus F(a)

Example: $\int_1^2 \frac{(x - 1)^2}{x}\,dx = \ln 2 - \frac{1}{2}$
If a rate is given, an amount or a distance is asked → Net Change Theorem; for a distance or a total, split at the sign changes of the rate first

Example: $v = t^2 - 2t - 3$ on $[0, 4]$ : displacement $-\frac{20}{3}$ , distance $\frac{34}{3}$
If a limit of an integral with a variable bound → check the 0 over 0 form, then L'Hôpital with FTC 1 on the numerator

Example: $\lim_{x \to 0} \frac{1}{x^3}\int_0^x \sin(t^2)\,dt = \frac{1}{3}$

If the integrand has no elementary antiderivative, FTC 2 is impossible, so the question is necessarily one of the FTC 1 branches.

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.

Evaluating a definite integral with FTC 2

When to use it: Any question that says evaluate $\int_a^b f(x)\,dx$ with numbers as bounds

1 State the continuity of the integrand on the closed interval $[a, b]$ , naming any denominator or root and why it causes no trouble there.
2 Rewrite the integrand, by algebra only, as a sum of table entries: divide term by term, expand, use $\sec^2 x = 1 + \tan^2 x$ .
3 Write an antiderivative $F$ and check it by differentiating, at least mentally.
4 Evaluate with brackets, $\Big[F(x)\Big]_a^b = F(b) - F(a)$ , keeping exact values.
5 Check the sign and the size against the integrand: positive integrand and $a < b$ give a positive result, between $m(b - a)$ and $M(b - a)$ .

Concluding sentence

“Since $f$ is continuous on $[1, 4]$ , FTC 2 applies: $\int_1^4 \frac{x^2 + 1}{\sqrt{x}}\,dx = \Big[\frac{2}{5}x^{5/2} + 2x^{1/2}\Big]_1^4 = \frac{72}{5}$ .”

The trap: Skipping step 1 is invisible when the answer is right and fatal when it is not: it is the step that catches $\frac{1}{x^2}$ on $[-1, 1]$ and $\sec^2 x$ on $[0, \pi]$ .

Marking: Typically 1 mark for the rewriting, 1 for the antiderivative, 1 for the evaluation. The continuity sentence is what protects all three.

Differentiating an integral whose bounds move

When to use it: Any question that asks for $\frac{d}{dx}$ of an integral with $x$ in a bound

1 Check that the integrand depends on $t$ only; if $x$ appears inside, take it out first.
2 If both bounds move, split at a constant where the integrand is continuous.
3 For each moving bound, write the integrand evaluated AT the bound, times the derivative of the bound, with a minus sign for a lower bound.
4 Simplify only what is legal on the stated domain, such as $\sqrt{\cos^2 x} = \cos x$ when $\cos x > 0$ .

Concluding sentence

“By FTC 1 and the chain rule, $h'(x) = \sqrt{1 + (x^2)^4} \cdot 2x - \sqrt{1 + x^4} = 2x\sqrt{1 + x^8} - \sqrt{1 + x^4}$ .”

The trap: Evaluating the integrand at the bound but forgetting its derivative, or replacing $t$ by $x$ instead of by the bound.

Marking: Usually half the marks for the integrand at the right place, half for the chain rule factor.

Check before you hand in

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

Differentiate your antiderivative

Every indefinite integral and every F used in FTC 2 is checked by one derivative. It catches invented product and quotient rules, a power rule applied to an exponential, and a lost constant factor.

$\frac{d}{dx}(\tan x - x) = \sec^2 x - 1 = \tan^2 x$ : consistent. $\frac{d}{dx}\frac{\tan^3 x}{3} = \tan^2 x\sec^2 x$ : not.
The sign must match the picture

A positive integrand over an interval from left to right gives a positive integral. A negative result for a positive function means FTC 2 was used where it does not apply.

$\int_{-1}^{1} \frac{dx}{1 + x^2}$ must be positive: $\frac{\pi}{2}$ , never $-\frac{\pi}{2}$ .
Bracket the value by the extremes of f

If m is at most f and f is at most M on the interval, the integral lies between m times the length and M times the length. Five seconds, and it catches a wrong exact value.

$\int_0^{1/2} \frac{4\,dx}{\sqrt{1 - x^2}}$ lies between $2$ and $2.31$ : $\frac{2\pi}{3} \approx 2.09$ fits, $\frac{4\pi}{3}$ does not.
Distance at least the displacement

The distance travelled can never be smaller than the absolute value of the displacement, with equality only if the velocity never changes sign.

Distance $\frac{34}{3}$ m against displacement $-\frac{20}{3}$ m: $\frac{34}{3} \ge \frac{20}{3}$ .

The typical problem, taken apart

A population that falls, then recovers: net change against the sum of changes

The fish population of a lake changes at the rate $P'(t) = 60t^2 - 300t + 240$ fish per year, for $0 \le t \le 6$ (years), and $P(0) = 1000$ .

Find $P(t)$ , the net change over the six years, the smallest and the largest population with the times they occur, and the sum of all the gains and losses.

The rate is positive, then negative on

(1, 4)

, then positive again: three lobes of signed areas

+110

-270

and

+520

Step 1

By the Net Change Theorem, $P(t) = P(0) + \int_0^t P'(s)\,ds = 1000 + \Big[20s^3 - 150s^2 + 240s\Big]_0^t = 1000 + 20t^3 - 150t^2 + 240t$ .

Why

The integral of the rate gives the change; the initial value gives the level. Using the dummy letter $s$ keeps the bound $t$ apart from the variable of integration.

Step 2

Net change over six years: $P(6) - P(0) = 20(216) - 150(36) + 240(6) = 4320 - 5400 + 1440 = 360$ fish.

Why

One FTC 2 evaluation, no need to know the sign of the rate: a NET change subtracts the losses automatically.

Step 3

$P'(t) = 60(t^2 - 5t + 4) = 60(t - 1)(t - 4)$ : positive on $[0, 1)$ , negative on $(1, 4)$ , positive on $(4, 6]$ . So $P(1) = 1110$ is a local maximum and $P(4) = 840$ a local minimum.

Why

The population turns where the RATE changes sign, exactly as an accumulation function turns where its integrand does. Factoring the rate is the whole study of variations.

Step 4

Candidates: $P(0) = 1000$ , $P(1) = 1110$ , $P(4) = 840$ , $P(6) = 1360$ . Smallest population $840$ at $t = 4$ , largest $1360$ at $t = 6$ .

Why

On a closed interval the extremes are among the turning points AND the endpoints. The local maximum $1110$ is not the largest value: forgetting $t = 6$ loses the mark.

Step 5

Gains and losses: $+110$ on $[0, 1]$ , $-270$ on $[1, 4]$ , $+520$ on $[4, 6]$ . Their sum in absolute value is $110 + 270 + 520 = 900$ , while the net change is $110 - 270 + 520 = 360$ .

Why

A total of changes needs the split at the sign changes, like a distance; the net change does not. Both numbers are correct answers to two different questions.

The conclusion, written out

“ $P(t) = 1000 + 20t^3 - 150t^2 + 240t$ . The net change is $+360$ fish; the population is smallest ( $840$ ) at $t = 4$ and largest ( $1360$ ) at $t = 6$ ; the gains and losses add up to $900$ fish.”

The classic mistake on this problem: Giving $360$ as the sum of all gains and losses, or taking $1110$ as the largest population because it is where $P' = 0$ with a sign change from positive to negative.

Learn by heart

• FTC 1: $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$ , for $f$ continuous.
• Moving bounds: $\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f(v)\,v' - f(u)\,u'$ . Constant bounds: derivative $0$ .
• FTC 2: $\int_a^b f = F(b) - F(a)$ , ONLY if $f$ is continuous on all of $[a, b]$ and $F' = f$ on all of $[a, b]$ .
• No product rule, no quotient rule: rewrite by algebra, then use the table.
• $\int \frac{dx}{x} = \ln|x| + C$ , one constant per interval. $\int b^x\,dx = \frac{b^x}{\ln b} + C$ . $\int \frac{dx}{1 + x^2} = \arctan x + C$ .
• Displacement $\int v$ ; distance $\int |v|$ , split at the zeros of $v$ .
• $\ln x = \int_1^x \frac{dt}{t}$ , negative for $0 < x < 1$ .
• The maximum of an accumulation is where the rate changes sign; the maximum of the rate is an inflection point.

Frequently asked questions

How do I differentiate an integral when the upper limit is a function of x?

Evaluate the integrand at the upper limit, then multiply by the derivative of that limit: this is the first part of the Fundamental Theorem combined with the chain rule. For example, the derivative of the integral from 1 to the square root of x of e to the t squared is e to the x, divided by two times the square root of x.

Why is the integral of 1 over x squared from minus 1 to 1 not equal to minus 2?

Because the second part of the Fundamental Theorem requires the integrand to be continuous on the whole closed interval, and one over x squared is not even defined at zero. The computation minus 1 minus 1 still produces a number, but it is meaningless: a positive function cannot have a negative integral. The integral is improper, a topic of its own later in the course.

What is the difference between displacement and distance travelled?

Displacement is the integral of the velocity, a net change in which motion to the left cancels motion to the right. Distance is the integral of the absolute value of the velocity. To compute it, find the times where the velocity is zero, integrate on each piece separately, and add the absolute values of the results.

Can two different answers to the same indefinite integral both be correct?

Yes, if they differ by a constant on each interval of the domain. The antiderivatives of two sine x cosine x include sine squared x, minus cosine squared x, and minus one half cosine of 2x, which differ by constants. The safe test is always the same: differentiate your answer and compare it with the integrand.

Do I need an antiderivative to use the first part of the Fundamental Theorem?

No. The first part gives the derivative of an accumulation function directly from the integrand, evaluated at the moving bound. That is why exam questions on it often use integrands like e to the t squared or sine of t squared, which have no elementary antiderivative: they force you to use the theorem instead of computing the integral.

Practise it

Corrected exercises: The Fundamental Theorem of Calculus, MATH 141 at McGill

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

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Riemann sums and the definite integral Next sheet The substitution rule

Looking for a MATH 141 tutor in Montreal?

Get in touch for a first session. The Fundamental Theorem is the hinge of Calculus 2: every technique of integration that follows is a way of finding the F that FTC 2 needs.

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Revision sheet: the Fundamental Theorem of Calculus and net change (MATH 141)

The essentials

Two halves, two directions, one hypothesis

Moving bounds: one rule covers every case

The rules in table form

The derivative of an integral, by where x sits

The mistakes that cost marks

1. Forgetting the derivative of the bound

2. Subtracting the value at a constant lower bound

3. Evaluating F(b) minus F(a) across a discontinuity

4. Using an antiderivative that jumps inside the interval

5. Reporting the displacement as the distance travelled

6. Inventing a quotient rule for integrals

7. Writing ln x where ln |x| is needed

8. Leaving x inside the integrand when differentiating

Which method to choose

Which half of the theorem, by the FORM of the question

How the answer is expected to be written

Evaluating a definite integral with FTC 2

Differentiating an integral whose bounds move

Check before you hand in

The typical problem, taken apart

A population that falls, then recovers: net change against the sum of changes

Learn by heart

Frequently asked questions

How do I differentiate an integral when the upper limit is a function of x?

Why is the integral of 1 over x squared from minus 1 to 1 not equal to minus 2?

What is the difference between displacement and distance travelled?

Can two different answers to the same indefinite integral both be correct?

Do I need an antiderivative to use the first part of the Fundamental Theorem?

Corrected exercises: The Fundamental Theorem of Calculus, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?