Integrals explained well for Calculus 1

Integrals are the chapter where many Calculus 1 (Analisi I) students realise that high school was not enough. Derivatives were “apply the rule and go”; integrals are not: you have to think, recognise the right form, choose the technique. It is a change of mindset that frightens, but once the underlying idea is understood integrals become much more manageable.

In this article I start from what an integral really is (because without the geometric intuition you study by rote and get stuck at the first non-standard exercise), go through the calculation techniques you use 90% of the time in Calculus 1, and close with the mistakes I see most often when I prepare first-year engineering, physics or maths students.

The integral as the area under the curve

The definite integral of a function $f (x)$ between $a$ and $b$ is written

\int_{a}^{b} f (x) d x

and it is the (signed) area enclosed between the graph of $f$ , the $x$ -axis and the vertical lines $x = a$ and $x = b$ . This is the geometric idea, and the whole rest of the chapter rests on it.

Formally the area is built as the limit of Riemann sums: you divide the interval $[a, b]$ into $n$ small pieces, on each piece you approximate the area with a thin rectangle of base $Δ x$ and height $f (x_{i})$ , add up all the rectangles and let $n$ tend to infinity. When the limit exists and does not depend on the subdivision, the function is called integrable and that limit is the integral.

You will never be asked to calculate an integral “by hand” with Riemann sums (except perhaps in the first tutorial). But it is worth having seen the construction at least once, because if tomorrow you are asked “why is an integral additive over the interval” or “why, if you shift the function by a constant, does the integral increase by the constant times $(b - a)$ ”, the answer is obvious if you have the area in your head, cryptic if you have only formulas.

The fundamental theorem: where everything holds together

The fundamental theorem of integral calculus is the bridge between integrals and derivatives. I state it in its operational version:

If $F$ is an antiderivative of $f$ (that is, $F^{'} (x) = f (x)$ ), then:

\int_{a}^{b} f (x) d x = F (b) - F (a)

This theorem turns the calculation of the area into a problem of “finding a function whose derivative is $f$ ”. Hence the classic division into two types of integral:

Indefinite integral $\int f (x) d x$ : the set of all antiderivatives of $f$ , that is $F (x) + c$ .
Definite integral $\int_{a}^{b} f (x) d x$ : the number $F (b) - F (a)$ , that is the signed area.

The $+ c$ in the indefinite integral is not a decoration: if you forget it in the exam, you risk losing half a mark (and in some courses it is considered a serious mistake). Always put it in.

The standard antiderivatives to know by heart

Before talking about techniques, you must have in your head a table of basic antiderivatives. They are the inverse of the elementary derivatives you already know:

Integral	Note
$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c$	valid for $n \neq = - 1$
$\int \frac{1}{x} d x = ln ∣ x ∣ + c$	the absolute value is important
$\int e^{x} d x = e^{x} + c$
$\int a^{x} d x = \frac{a ^{x}}{l n a} + c$
$\int sin x d x = - cos x + c$	the minus sign is where very many go wrong
$\int cos x d x = sin x + c$
$\int \frac{1}{1 + x ^{2}} d x = arctan x + c$
$\int \frac{1}{1 - x ^{2}} d x = arcsin x + c$

These eight rows are your dictionary. Almost all Calculus 1 integrals reduce to one of these forms, after applying a technique.

The three techniques you really need

In Calculus 1, for 95% of exercises, you only need to master three techniques: integration by substitution, integration by parts, and integrals of rational functions (with partial fractions). Let’s take them one by one.

1. Substitution

The idea: if inside the integral there is a composite function $f (g (x))$ multiplied by $g^{'} (x)$ , you set $t = g (x)$ and the integral becomes

\int f (t) d t

It is the inverse of the chain rule.

Classic example:

\int 2 x cos (x^{2}) d x

I set $t = x^{2}$ , so $d t = 2 x d x$ , and the integral becomes

\int cos t d t = sin t + c = sin (x^{2}) + c

The typical mistake is forgetting to “go back to $x$ ” at the end when the integral is indefinite, or not updating the limits when it is definite (if you change variable, you must also change the limits: this saves you from going back to $x$ ).

2. Integration by parts

From the product rule for derivatives we get:

\int u (x) \cdot v^{'} (x) d x = u (x) \cdot v (x) - \int u^{'} (x) \cdot v (x) d x

You use it when you have a product of functions and one of the two “simplifies” when differentiated (typically a polynomial or a logarithm). A practical rule for choosing which is $u$ and which is $v^{'}$ : follow the acronym LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) — the first in the list that appears in the integral is $u$ .

Example:

\int x e^{x} d x

Algebraic and Exponential: $u = x$ (algebraic), $v^{'} = e^{x}$ . So $u^{'} = 1$ , $v = e^{x}$ . Result:

x e^{x} - \int e^{x} d x = x e^{x} - e^{x} + c = e^{x} (x - 1) + c

A less obvious example:

\int ln x d x

There seems to be no product, but you can write $ln x \cdot 1$ . Then $u = ln x$ , $v^{'} = 1$ , $u^{'} = 1/ x$ , $v = x$ . Result:

x ln x - \int 1 d x = x ln x - x + c

3. Rational functions

An integral of the type

\int \frac{P ( x )}{Q ( x )} d x

where $P$ and $Q$ are polynomials. Three steps:

If the degree of $P$ is $\geq$ the degree of $Q$ , first do polynomial division.
Factorise $Q (x)$ (real roots, irreducible quadratic factors).
Decompose $P / Q$ into partial fractions and integrate each one.

Example:

\int \frac{1}{x ^{2} - 1} d x

I factorise: $x^{2} - 1 = (x - 1) (x + 1)$ . I decompose:

\frac{1}{( x - 1 ) ( x + 1 )} = \frac{A}{x - 1} + \frac{B}{x + 1}

Solving: $A = 1/2$ , $B = - 1/2$ . I integrate:

\frac{1}{2} ln ∣ x - 1∣ - \frac{1}{2} ln ∣ x + 1∣ + c

Partial fractions are the point where many get lost in Calculus 1, but it is only a matter of practice: once you have seen ten examples, the decomposition comes automatically.

Improper integrals: when the interval is infinite

In Calculus 1 the improper integral is also introduced, that is, the case where the interval of integration is infinite or the function blows up inside the interval. The definition is a limit:

\int_{1}^{+ \infty} f (x) d x = b \to + \infty lim \int_{1}^{b} f (x) d x

If the limit exists and is finite, the integral converges. Otherwise it diverges.

Key example:

\int_{1}^{+ \infty} \frac{1}{x ^{α}} d x

converges if and only if $α > 1$ . Memorise this result because it is the reference for the comparison tests of convergence. Typical mistake: confusing convergence at infinity with convergence at a singular point.

\int_{0}^{1} \frac{1}{x ^{α}} d x

converges if and only if $α < 1$ — exactly the opposite.

Typical mistakes (and how to avoid them)

After dozens of students prepared for Calculus 1, the mistakes on integrals are almost always these five:

Forgetting the $+ c$ in the indefinite integral. It seems trivial, but in many courses it is penalised. Put it in as the last thing, always.
Getting the sign of the antiderivative of $sin x$ and $cos x$ wrong. The antiderivative of $sin x$ is $- cos x$ , not $cos x$ . Memorise the pair: sine “gains a minus”, cosine does not.
Not updating the limits in a definite substitution. If you substitute $t = g (x)$ in a definite integral, the limits become $g (a)$ and $g (b)$ . Very many substitute the variable but leave the original limits, and the result is completely wrong.
Confusing substitution and by parts. Faced with
$\int x sin (x^{2}) d x$
the temptation is “by parts because there is a product”. But here the substitution $t = x^{2}$ collapses the integral in three lines. Rule: if in the product you see a function and its “almost-derivative”, it is substitution; if the two functions are independent, it is by parts.
Skipping the absolute value in $ln ∣ x ∣$ . The antiderivative of $1/ x$ is $ln ∣ x ∣$ , not $ln x$ . Without the absolute value you are not defined for $x < 0$ , and in many integrals (for example those of rational functions) the function lives on the negatives too.

How much time it takes to sort them out

A realistic estimate, based on the students I have followed in Calculus 1 at the Politecnico, Bicocca and Sapienza:

If you start from zero but can differentiate well: 6-8 hours of targeted lessons to reach exam level, plus 15-20 hours of exercises on your own.
If you have already done some exercises but do not recognise the right technique: 3-4 hours to sort out recognising the technique, then lots of exercises.
If you have an exam in a week and have noticed that the chapter is uncovered: 2-3 intensive lessons on the core (substitution, by parts, partial fractions) and a practice run on past years’ papers.

In the last two years 97% of the students I have prepared passed the exam at the first attempt. The critical point is almost never the calculation: it is recognising “this is a substitution, this is by parts, this is a partial fraction”. That can be trained only by doing guided exercises.

How we work in lessons

Typically I start from your professor’s exercises (tutorial sheet, last year’s papers, the professor in class). We see where you get stuck: whether the problem is the right technique (recognition), the calculation (algebra), or the very idea (what it means to integrate). From there we build the missing piece, not a standard programme.

If you want to understand how I set up the method before an exam, also have a look at how to study maths without stress and at derivatives explained well — integrals rest on derivatives, and if the latter are shaky the first chapter ends up a disaster.

For the part that comes before, limits explained well (antiderivatives of rational functions and improper integrals take up indeterminate forms and standard limits again) and, if your course covers them, complex numbers explained well: they are needed for the antiderivatives of rational functions with complex roots.

Want a hand with Calculus 1?

If you are struggling with integrals and the exam is getting close, a half-hour free introductory call is enough for us to work out what is really missing: often it is not “everything”, it is one specific piece (substitution, partial fractions, improper integrals) that blocks the rest.

In the maths lessons we work on your real exercises and on your professor’s papers, not on a generic syllabus. See the available packages (single lesson or 10-hour package) or write to me on WhatsApp to organise the first lesson.

Want to test yourself? Take the integrals review quiz: 15 questions, with solutions explained step by step.

Integrals explained well: from geometry to Calculus 1