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Definite Integrals

Q: Do I need to write +c in a definite integral?

No — the constant always cancels out when you subtract F(a) from F(b), so it is unnecessary (and a waste of time) to include it.

Q: Can I split a definite integral into two parts?

Yes, for adjacent intervals: _a^b f(x),dx+ _b^c f(x),dx= _a^c f(x),dx, which can simplify some calculations or combine given results.

An indefinite integral gives a whole family of curves, all differing by a constant. Fix two specific $x$ -values as boundaries, though, and that ambiguous constant cancels out completely — leaving a single, exact number.

26 min Video by TLMaths Integration

The big picture

A $definite$ integral evaluates the antiderivative at an upper limit and a lower limit, then subtracts. Because BOTH evaluations would carry the same $+ c$ , it cancels out in the subtraction — which is exactly why a definite integral needs no constant of integration at all. This single number is the foundation for everything geometric that integration is used for: areas, volumes and totals accumulated over an interval.

A-Level Maths H3-03: Indefinite vs Definite IntegralsWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Evaluate a definite integral using the limits of integration
Understand why the constant of integration cancels and is not needed
Use definite integrals to evaluate exact numerical or algebraic results
Apply the additive property of integrals over adjacent intervals

Notation and evaluation

A $definite integral$ $\int_{a}^{b} f (x) d x$ has a lower limit $a$ and upper limit $b$ . Find the antiderivative $F (x)$ (no $+ c$ needed), then evaluate $F (b) - F (a)$ .

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

Upper limit value minus lower limit value.

1Antiderivative:

F (x) = \frac{x ^{3}}{3}

2Evaluate at the limits:

F (3) - F (1) = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3}

Answer

\frac{26}{3}

Tip — Always substitute the UPPER limit first, then subtract the result of substituting the LOWER limit — getting this order backwards flips the sign of your answer.

Why the constant cancels

If the antiderivative is written with a $+ c$ , evaluating $[F (x) + c]_{a}^{b}$ gives $(F (b) + c) - (F (a) + c)$ — the two $c$ terms are identical and cancel exactly, whatever value $c$ takes. This is precisely why a definite integral is a single, unambiguous number, unlike an indefinite integral.

1With

c

[x^{3} + c]_{0}^{2} = (8 + c) - (0 + c) = 8 + c - c

2The

c

terms cancel:

8

Answer

8

, regardless of the value of

c

Why this works

This cancellation is the whole reason definite integrals are the tool used for real physical quantities (areas, distances, total charge) — the answer cannot depend on an arbitrary unknown.

Splitting the interval

Definite integrals over adjacent intervals add together, which lets you split a complicated interval into simpler pieces, or combine results calculated separately.

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

Adjacent intervals combine additively.

1The intervals

[0, 2]

and

[2, 5]

are adjacent, so they add directly:

5 + 3

Answer

8

Think like an examiner

Show the antiderivative in square brackets with the limits BEFORE substituting — this is the expected method-mark step.
Substitute the upper limit first, then subtract the lower limit substitution — never the other way round.
You do not need $+ c$ in a definite integral calculation — including it wastes time since it cancels regardless.

Common misconceptions

Subtracting the limits in the wrong order (lower minus upper).

It is always UPPER limit result minus LOWER limit result:

F (b) - F (a)

Including and trying to solve for

+ c

in a definite integral.

The constant always cancels in a definite integral — there is no need to write it at all.

Forgetting the antiderivative must be evaluated (not just written down) at both limits.

You must actually substitute BOTH the upper and lower limit values into the antiderivative and subtract.

Definite integrals

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

Fundamental theorem of calculus.

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

Additive over adjacent intervals.

Stretch yourself

Evaluate $\int_{- 1}^{2} (3 x^{2} - 2 x) d x$ .

Hint — Find the antiderivative first, then substitute both limits carefully, remembering the sign of the lower (negative) limit.

Questions students ask

Do I need to write +c in a definite integral?

No — the constant always cancels out when you subtract

F (a)

from

F (b)

, so it is unnecessary (and a waste of time) to include it.

What order do I substitute the limits in?

Always substitute the UPPER limit into the antiderivative first, then subtract the result of substituting the LOWER limit:

F (b) - F (a)

Can I split a definite integral into two parts?

Yes, for adjacent intervals:

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

, which can simplify some calculations or combine given results.

Key takeaways

A definite integral $\int_{a}^{b} f (x) d x = F (b) - F (a)$ , no $+ c$ needed.
The constant of integration always cancels in the subtraction.
Substitute the upper limit first, then subtract the lower limit substitution.
Definite integrals over adjacent intervals add together.

How this fits the course

Build on

Integrating xⁿ

Area Under a Curve

Leads to

Area Under a Curve

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