Mathematics/Lessons/Integration/Definite Integrals
Pure · Integration · Year 1

Definite Integrals

Add two limits and an integral stops being a function and becomes a number: integrate, substitute top and bottom, subtract. The constant cancels itself out.

9 min read 3 worked examples

Integrals with limits

An indefinite integral gives you back a function, plus a $C$. A definite integral gives you a number. It carries two limits, a lower value $a$ and an upper value $b$:

$\displaystyle\int_a^b f(x)\,\mathrm{d}x.$

To evaluate one, integrate as usual, then substitute the upper limit and the lower limit into the result and subtract. Writing $F(x)$ for the integral of $f(x)$:

$\displaystyle\int_a^b f(x)\,\mathrm{d}x = \Big[F(x)\Big]_a^b = F(b) - F(a).$

The square brackets $\big[F(x)\big]_a^b$ are shorthand for "evaluate $F$ at $b$, evaluate it at $a$, and subtract".

A curve labelled y = f(x) lying above the x-axis. Two vertical lines drop from the curve to the axis, meeting it at points labelled a and b. The region enclosed by the curve, the axis and the two vertical lines is filled with diagonal hatching.
The limits a and b cut out a section of the curve. That region turns out to be what a definite integral measures — the next lesson makes it precise. For now, treat the integral as a number you calculate.
Why the constant of integration drops out

Suppose you kept the $+C$. You would compute $\big(F(b)+C\big) - \big(F(a)+C\big)$, and the two constants cancel. So $C$ makes no difference to a definite integral and you can leave it out from the start. The ambiguity that forced the $+C$ on indefinite integrals is subtracted away, which is why a definite integral comes out as one specific number.

The method: integrate, bracket, substitute, subtract

(1) Integrate $f(x)$, with no $+C$. (2) Write it in square brackets with the limits. (3) Substitute the upper limit, then the lower limit. (4) Subtract: upper minus lower. Keeping the two substitutions on separate lines is worth the extra second, because that is where sign errors happen.

Worked example 1

Evaluate a definite integral
Evaluate $\displaystyle\int_1^3 \left(3x^2 - 2x\right)\,\mathrm{d}x$.
MethodIntegrate, then take upper minus lower
1
Integrate, no $+C$ needed: $\displaystyle\int (3x^2 - 2x)\,\mathrm{d}x = x^3 - x^2$.
Term by term.
2
Write with the limits: $\Big[x^3 - x^2\Big]_1^3$.
Square brackets, limits on the right.
3
Upper limit $x = 3$: $27 - 9 = 18$.
Substitute $b = 3$.
4
Lower limit $x = 1$: $1 - 1 = 0$. Subtract: $18 - 0 = 18$.
$F(b) - F(a)$.
▪ solved
Leave the +C out

It would cancel anyway, so there is no reason to carry it. Set the two substitutions out clearly — $F(3) = 18$ and $F(1) = 0$ — before subtracting, rather than doing both in your head at once. The answer to a definite integral is always a plain number, never a function.

Worked example 2

A negative lower limit
Evaluate $\displaystyle\int_{-1}^{2} \left(4x^3 + 1\right)\,\mathrm{d}x$.
MethodIntegrate, substitute both limits, subtract
1
Integrate: $\displaystyle\int (4x^3 + 1)\,\mathrm{d}x = x^4 + x$.
Add one, divide.
2
Upper limit $x = 2$: $16 + 2 = 18$.
Substitute $b = 2$.
3
Lower limit $x = -1$: $(-1)^4 + (-1) = 1 - 1 = 0$.
Note $(-1)^4 = +1$.
4
Subtract: $18 - 0 = 18$.
$F(2) - F(-1)$.
▪ solved
Powers of negative numbers

Even powers come out positive and odd powers keep the sign: $(-1)^4 = +1$, but $(-1)^3 = -1$. Write each substitution out in full, brackets and all, before combining anything. And keep the order straight: it is upper minus lower, so if the lower value is negative you end up adding, which is easy to get the wrong way round.

Worked example 3

A root inside the integral
Evaluate $\displaystyle\int_1^4 \dfrac{6}{\sqrt{x}}\,\mathrm{d}x$.
MethodConvert to a power, integrate, then evaluate
1
Rewrite: $\dfrac{6}{\sqrt{x}} = 6x^{-1/2}$.
Index form.
2
Integrate: adding one to $-\tfrac12$ gives $\tfrac12$, and dividing by $\tfrac12$ doubles: $\dfrac{6x^{1/2}}{1/2} = 12\sqrt{x}$.
Add one, divide.
3
Upper $x = 4$: $12\sqrt4 = 24$. Lower $x = 1$: $12\sqrt1 = 12$.
Substitute both limits.
4
Subtract: $24 - 12 = 12$.
$F(4) - F(1)$.
▪ solved
The excluded power never came up

Integrating $6x^{-1/2}$ raised the power to $+\tfrac12$, well clear of the forbidden $n = -1$. The formula copes with $-\tfrac12$ without any trouble; it is only the exact power $x^{-1}$ that it cannot touch. Work the limits out exactly — most definite integrals are meant to give an exact value rather than a decimal.

Common mistake

Subtracting the wrong way round is the big one: it is always $F(b) - F(a)$, upper minus lower, and swapping them negates your answer. Watch the signs when a limit is negative, particularly with odd powers. And do not waste time on a $+C$ that cancels itself.

Check yourself

Evaluate $\displaystyle\int_0^2 \left(3x^2 + 4x\right)\,\mathrm{d}x$.
(Hint: integrate to $x^3 + 2x^2$, then take the value at 2 minus the value at 0.)

Method
1
Integrate: $\Big[x^3 + 2x^2\Big]_0^2$.
Add one, divide.
2
Upper $x = 2$: $8 + 8 = 16$.
Substitute $b = 2$.
3
Lower $x = 0$: $0 + 0 = 0$.
Substitute $a = 0$.
4
Subtract: $16 - 0 = 16$.
$F(2) - F(0)$.
▪ solved

A lower limit of $0$ often makes $F(a) = 0$, but write it out rather than assuming — a constant term in $F(x)$ would make it something else.

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Finding the Constant of Integration