What is an antiderivative?

A function F whose derivative is the given function f, that is one satisfying F′(x) = f(x). For f(x) = 2x an antiderivative is F(x) = x², because differentiating x² gives 2x. Looking for an antiderivative is integration — the operation inverse to differentiation.

Where does the constant C in an indefinite integral come from?

From the derivative of a constant being zero. If F is an antiderivative of f, then so are F + 5, F − 100 and every other vertical shift, because they all have the identical derivative. The result of an indefinite integral is therefore a whole family of functions, and C records that freedom.

How is an indefinite integral different from a definite one?

An indefinite integral is a FUNCTION (up to a constant) and answers the question "what is this the derivative of". A definite integral is a NUMBER: it carries limits of integration and is computed as F(b) − F(a). Geometrically that number measures the region between the graph and the axis over the given interval.

Why does the constant C not matter in a definite integral?

Because it cancels in the subtraction: (F(b) + C) − (F(a) + C) = F(b) − F(a). So when computing a definite integral you pick any antiderivative — most conveniently the one without a constant term — and forget about C entirely.

How are powers integrated?

The opposite way to how they are differentiated: the exponent GROWS by one and then you divide by that new exponent. Hence ∫x³ dx = x⁴/4 + C. The single exception is n = −1, where dividing by zero makes no sense — the integral of 1/x is a logarithm, which this lesson does not cover.

Does a definite integral always equal an area?

Only when the function is non-negative across the whole interval. Stretches below the axis enter the integral with a minus sign, so for f(x) = x − 2 from 0 to 4 the integral is 0 even though the region is not empty at all — its area is 4. In that case the area is computed piece by piece, taking absolute values.

How do I find the area when the graph crosses the axis?

First find the zeros inside the interval, then compute the integral separately on each piece and add the absolute values of the results. Adding the results without the modulus gives you the integral back, not the area — the commonest trap in area problems.

How can I check that I integrated correctly?

Differentiate the answer. If the derivative of the function you obtained equals the integrand, the integral is right — and that check is usually quicker than the integration itself. It is worth doing every time, because a slip in a coefficient shows up immediately.

Advanced level

The integral

Integration is differentiation run backwards: we look for the function whose derivative is the one we started with. An indefinite integral gives a formula up to a constant, a definite one gives a number — and that number is the area under the curve.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

The derivativeA derivative measures how fast a function changes at a point — it is the slope of the tangent to the graph. Meet the definition through the difference quotient and the six rules that compute it without taking a limit every time.

All formulas

Antiderivative
$F'(x) = f(x)$
F is an antiderivative of f when differentiating F gives f
Indefinite integral
$\int f(x)\,dx = F(x) + C$
the result is a whole FAMILY of functions, hence the constant C
Power rule
$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
the exponent grows by one, then divide by the new exponent
Newton–Leibniz formula
$\int_a^b f(x)\,dx = F(b) - F(a)$
the constant C cancels in the subtraction, so it never matters
Area under a curve
$f(x) \geq 0 \ \text{on} \ \langle a; b \rangle \implies A = \int_a^b f(x)\,dx$
the definite integral equals the area only above the axis
Area in general
$A = \int_a^b |f(x)|\,dx$
where the curve dips below the axis, the area is computed piece by piece
Trigonometric functions
$\int \sin x\,dx = -\cos x + C, \qquad \int \cos x\,dx = \sin x + C$
the minus sign changes sides compared with differentiation

Differentiation answers the question "how fast is this changing". Integration asks the reverse: given the rate of change, what was the quantity? It is the operation inverse to the derivative — and at the same time the way to measure regions bounded by a curve.

The antiderivative

An antiderivative of a function $f$ is a function $F$ whose derivative is $f$ :

F'(x) = f(x)

For $f(x) = 2x$ an antiderivative is $F(x) = x^2$ , because differentiating $x^2$ gives $2x$ . But so is $x^2 + 7$ , and so is $x^2 - 100$ — the derivative of a constant is zero, so shifting a graph vertically leaves the slope at every point untouched.

There are therefore infinitely many antiderivatives, and they differ by a constant alone.

The indefinite integral

We write it like this:

\int f(x)\,dx = F(x) + C

The sign $\int$ is an elongated "S" for sum, and $dx$ says which variable we are integrating over. The constant of integration $C$ is not decoration: without it the notation would name one function instead of a whole family.

The power rule is the exact reverse of the one from differentiation — the exponent grows by one and then we divide by the new exponent:

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \qquad (n \neq -1)

function	integral
$k$ (a constant)	$kx + C$
$x$	$\dfrac{x^2}{2} + C$
$x^2$	$\dfrac{x^3}{3} + C$
$\dfrac{k}{x^2}$	$-\dfrac{k}{x} + C$
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$

The restriction $n \neq -1$ is there because $n = -1$ would mean dividing by zero. The integral of $\tfrac{1}{x}$ is a logarithm — beyond the scope of this lesson.

A sum is integrated term by term, and a constant factor passes through an integral just as it passes through a derivative.

Compute the indefinite integral ∫ (6x² − 4x + 5) dx and check the result.

The definite integral and Newton–Leibniz

Once limits of integration are written on the integral sign, the result stops being a function and becomes a number:

\int_a^b f(x)\,dx = F(b) - F(a)

This is the Newton–Leibniz formula, one of the great theorems of mathematics: it ties differentiation to the measurement of areas. The computation has three steps — find an antiderivative, substitute the upper limit, subtract the value at the lower one.

The constant $C$ is irrelevant here, because it cancels in the subtraction:

(F(b) + C) - (F(a) + C) = F(b) - F(a)

Compute the definite integral of (3x² − 4x + 1) from 1 to 3.

The area under a curve

Where do areas come into it? Slice the region under a graph into narrow vertical strips. Each is almost a rectangle of area $f(x) \cdot \Delta x$ , and the narrower the strips, the smaller the error. Passing to the limit — exactly the limit we met with the derivative — turns the sum of rectangle areas into an integral.

The region under f(x) = 2x from 0 to 3. Here the answer can be checked with geometry: it is a triangle of base 3 and height 6.

For a function that is non-negative across the interval, the definite integral simply is the area:

f(x) \geq 0 \ \text{on} \ \langle a; b \rangle \implies A = \int_a^b f(x)\,dx

Check it against the drawing above. The area of a triangle of base $3$ and height $6$ is $\tfrac{1}{2} \cdot 3 \cdot 6 = 9$ , and the integral:

\int_0^3 2x\,dx = \big[x^2\big]_0^3 = 9 - 0 = 9

They agree — and that is the whole power of the method: for a straight line geometry already knew the answer, but the integral computes the region under any curve, including those no formula in a book of shapes covers.

Find the area of the region bounded by f(x) = x², the x-axis and the lines x = 0 and x = 2.

When the curve dips below the axis

A definite integral measures area with a sign: stretches below the axis enter it with a minus. That is why "integral = area" only holds for non-negative functions.

The two shaded triangles have the identical area of 2, but they lie on opposite sides of the axis. The integral subtracts them and returns 0; the area of the region is 4.

Compute both numbers. The antiderivative is $F(x) = \tfrac{x^2}{2} - 2x$ , so:

\int_0^4 (x - 2)\,dx = (8 - 8) - 0 = 0

The area, on the other hand, is taken piece by piece with absolute values:

A = \left| \int_0^2 (x - 2)\,dx \right| + \left| \int_2^4 (x - 2)\,dx \right| = |-2| + |2| = 4

In general:

A = \int_a^b |f(x)|\,dx

The recipe is simple: find the zeros inside the interval, integrate over each piece separately, and add the absolute values of the results.

Two questions, two answers

It pays to keep three easily confused things apart:

the indefinite integral — a function (a family of functions), the answer to "what is this the derivative of";
the definite integral — a number, the result of the subtraction $F(b) - F(a)$ , possibly negative;
the area — a non-negative number, equal to the integral only when the curve stays above the axis.

That last distinction is the commonest source of wrong answers in area problems.

Exercises

The three kinds of question match the three parts of this lesson. For the indefinite integral, type an antiderivative — the constant C may be written out or left off, because the answer is graded up to a constant. For the definite integral the answer is a number, and it may well come out negative. For the area under a graph the answer is always non-negative, and in the harder questions the curve crosses the axis inside the interval — then the area is computed piece by piece.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

∫ (2x + 6) dx =

Common mistakes

Dropping the constant C — the result of an indefinite integral is a family of functions, so $+\,C$ belongs to the answer.
Integrating as if differentiating — in an integral the exponent grows, and then you divide by the new exponent.
Equating the integral with the area — below the axis the integral is negative and an area never is.
Adding the pieces without the modulus — that gives the integral back, not the area.
Subtracting in the wrong order — it is $F(b) - F(a)$ : the lower limit is taken away from the upper one.
Carrying C through a definite integral — it cancels anyway and only clutters the computation.

Formula card

Topic: The integral

Antiderivative
$F'(x) = f(x)$
F is an antiderivative of f when differentiating F gives f
Indefinite integral
$\int f(x)\,dx = F(x) + C$
the result is a whole FAMILY of functions, hence the constant C
Power rule
$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
the exponent grows by one, then divide by the new exponent
Newton–Leibniz formula
$\int_a^b f(x)\,dx = F(b) - F(a)$
the constant C cancels in the subtraction, so it never matters
Area under a curve
$f(x) \geq 0 \ \text{on} \ \langle a; b \rangle \implies A = \int_a^b f(x)\,dx$
the definite integral equals the area only above the axis
Area in general
$A = \int_a^b |f(x)|\,dx$
where the curve dips below the axis, the area is computed piece by piece
Trigonometric functions
$\int \sin x\,dx = -\cos x + C, \qquad \int \cos x\,dx = \sin x + C$
the minus sign changes sides compared with differentiation

The shaded region under f(x) = 2x on ⟨0; 3⟩. It is a triangle of base 3 and height 6, so its area is 9 — exactly what the definite integral returns.

The region between f(x) = x − 2 and the axis on ⟨0; 4⟩. Two triangles of area 2 sit on opposite sides of the axis: the integral is 0, the area is 4.

Frequently asked questions

A luminous parabola and a straight line crossing above a faint coordinate grid, the vertex flaring like a star, against a cosmic nebula in violet and magenta

Functions and analysis

Why a function's graph can never double back

The whole definition of a function comes down to one word: uniqueness. One argument, one value — and not a single one more. That seemingly minor condition decides why a circle is not the graph of a function, why a line and a parabola describe so much of the world, and why four datasets with identical summary statistics can look nothing alike.

12 min