Properties of Definite Integrals

A complete lesson on the algebra and geometry of definite integrals: signed area, changing limits, splitting intervals, linearity, constant multiples and symmetry.

IB MathematicsCalculusOnline textbookWorked examplesRadford Mathematics

Integration pathway

Use this mini-chapter as a sequence. Start with the meaning and properties of definite integrals, then apply them to area questions.

Chapter mapIntegration overviewSee how the integration pages connect as one textbook chapter. Current lessonProperties of definite integralsSigned area, reversing limits, splitting intervals, linearity and symmetry. Next applicationArea under a curveUse definite integrals to calculate area between one curve and the x-axis. ExtensionArea between two curvesUse upper minus lower and split intervals when the upper curve changes. Numerical methodTrapezoidal ruleApproximate areas numerically when exact integration is not required or not possible. Next planned pageKinematics with integrationUse integrals to connect acceleration, velocity and displacement.

1. What does a definite integral measure?

A definite integral is written with a lower limit and an upper limit:

\[\int_a^b f(x)\,dx\]

Geometrically, it measures signed area between the graph of \(y=f(x)\) and the \(x\)-axis from \(x=a\) to \(x=b\).

Area above the \(x\)-axis contributes positively.
Area below the \(x\)-axis contributes negatively.
If the question asks for total area, you must make all regions positive, usually by splitting at the \(x\)-intercepts.

The definite integral adds signed area. Positive and negative contributions can cancel, so a definite integral is not always the same as total geometric area.

2. The essential properties at a glance

The following properties are used constantly in IB Mathematics. They are not separate tricks; they all come from the same idea of adding signed areas over intervals.

Zero-width interval

\[\int_a^a f(x)\,dx=0\]

There is no horizontal width, so there is no area.

Reversing limits

\[\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\]

Changing direction changes the sign.

Splitting an interval

\[\int_a^c f(x)\,dx=\int_a^b f(x)\,dx+\int_b^c f(x)\,dx\]

The area from \(a\) to \(c\) can be split at any intermediate point \(b\).

Adding functions

\[\int_a^b (f(x)+g(x))\,dx=\int_a^b f(x)\,dx+\int_a^b g(x)\,dx\]

The integral of a sum is the sum of the integrals.

Subtracting functions

\[\int_a^b (f(x)-g(x))\,dx=\int_a^b f(x)\,dx-\int_a^b g(x)\,dx\]

This is the algebraic foundation for area between two curves.

Constant multiple

\[\int_a^b kf(x)\,dx=k\int_a^b f(x)\,dx\]

Multiplying all heights by \(k\) multiplies the signed area by \(k\).

Dummy variable

\[\int_a^b f(x)\,dx=\int_a^b f(t)\,dt\]

The variable name does not matter, provided the limits stay the same.

Order comparison

\[f(x)\geq g(x)\Rightarrow \int_a^b f(x)\,dx\geq \int_a^b g(x)\,dx\]

For \(a

3. Reversing limits changes the sign

The limits \(a\) and \(b\) do more than mark the ends of the interval. They also specify the direction in which the interval is travelled. Going from \(a\) to \(b\) gives the opposite sign to going from \(b\) to \(a\).

\[\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\]

The shaded region is the same, but the orientation is reversed. This is why the sign changes.

Example. If \(\int_2^7 f(x)\,dx=11\), then

\[\int_7^2 f(x)\,dx=-11.\]

4. Splitting an interval

If \(b\) lies between \(a\) and \(c\), the area from \(a\) to \(c\) is the area from \(a\) to \(b\), plus the area from \(b\) to \(c\):

\[\int_a^c f(x)\,dx=\int_a^b f(x)\,dx+\int_b^c f(x)\,dx.\]

Splitting at \(x=b\) does not change the total signed area. It simply writes the same accumulation as two pieces.

This property is especially important when a graph crosses the \(x\)-axis, or when the upper curve changes in an area-between-curves problem.

5. Linearity: sums, differences and constant multiples

Integration is linear. This means we can distribute the integral across addition and subtraction, and we can move constant factors outside the integral.

\[\int_a^b (\alpha f(x)+\beta g(x))\,dx=\alpha\int_a^b f(x)\,dx+\beta\int_a^b g(x)\,dx\]

Here \(\alpha\) and \(\beta\) are constants.

Multiplying a function by \(2\) doubles every vertical height, so it also doubles the signed area over the same interval.

Example. Suppose \(\int_0^3 f(x)\,dx=4\) and \(\int_0^3 g(x)\,dx=-2\). Then

\[\int_0^3 (3f(x)-2g(x))\,dx=3(4)-2(-2)=16.\]

6. Signed area is not always total area

A common mistake is to calculate one definite integral and assume it is the total area. That only works when the function stays on one side of the \(x\)-axis.

Important. If a graph crosses the \(x\)-axis, split the interval at the crossing point. Regions below the axis must be made positive if the question asks for total area.

\[A_{\mathrm{total}}=\int_a^b |f(x)|\,dx\]

In exact calculations, this usually means splitting the integral and using the positive expression for the height on each part.

For \(y=x-1\), the graph is below the \(x\)-axis on \([0,1]\) and above it on \([1,3]\). The signed integral subtracts the left triangle, but total area adds both triangles.

7. Symmetry properties

When the interval is symmetric, such as \([-a,a]\), even and odd functions can simplify definite integrals.

Odd function

\[f(-x)=-f(x)\Rightarrow \int_{-a}^{a} f(x)\,dx=0\]

The negative and positive signed areas cancel.

Even function

\[f(-x)=f(x)\Rightarrow \int_{-a}^{a} f(x)\,dx=2\int_0^a f(x)\,dx\]

The left and right halves have equal area.

Odd symmetry gives cancellation. Even symmetry lets us double one half of the integral.

8. Worked example 1: using interval splitting

Suppose \(\int_1^4 f(x)\,dx=7\) and \(\int_4^6 f(x)\,dx=-2\). Find \(\int_6^1 f(x)\,dx\).

First combine the intervals from \(1\) to \(6\):

\[\int_1^6 f(x)\,dx=\int_1^4 f(x)\,dx+\int_4^6 f(x)\,dx=7+(-2)=5.\]

The required integral has the limits reversed:

\[\int_6^1 f(x)\,dx=-\int_1^6 f(x)\,dx=-5.\]

9. Worked example 2: using linearity

Given \(\int_0^2 f(x)\,dx=3\) and \(\int_0^2 g(x)\,dx=-1\), find

\[\int_0^2 (2f(x)-3g(x))\,dx.\]

Use linearity:

\[\int_0^2 (2f(x)-3g(x))\,dx=2\int_0^2 f(x)\,dx-3\int_0^2 g(x)\,dx.\]

\[=2(3)-3(-1)=6+3=9.\]

10. Worked example 3: signed area versus total area

For \(f(x)=x-1\), find the signed integral and the total area between the graph and the \(x\)-axis from \(x=0\) to \(x=3\).

The signed integral is

\[\int_0^3 (x-1)\,dx=\left[\frac{x^2}{2}-x\right]_0^3=\frac{9}{2}-3=\frac{3}{2}.\]

For total area, split at \(x=1\), where the graph crosses the \(x\)-axis:

\[A_{\mathrm{total}}=\int_0^1 (1-x)\,dx+\int_1^3 (x-1)\,dx.\]

\[=\frac{1}{2}+2=\frac{5}{2}.\]

The two answers are different because the first triangle lies below the \(x\)-axis and is counted negatively by the definite integral.

The signed integral is \(\frac{3}{2}\), while the total area is \(\frac{5}{2}\).

11. Worked example 4: using symmetry

Evaluate

\[\int_{-2}^{2}(x^3+3x^2)\,dx.\]

Split the integrand into an odd part and an even part:

\[x^3+3x^2=(x^3)+(3x^2).\]

The function \(x^3\) is odd, so its integral from \(-2\) to \(2\) is \(0\). The function \(3x^2\) is even, so

\[\int_{-2}^{2}(x^3+3x^2)\,dx=0+2\int_0^2 3x^2\,dx.\]

\[=2[x^3]_0^2=2(8)=16.\]

The odd part cancels over \([-2,2]\). The even part can be calculated by doubling the right half.

12. IB-style checklist

Check the order of the limits before doing any calculation.
Remember that reversing the limits changes the sign.
Split the integral when the interval naturally breaks into pieces.
Use linearity to separate sums, differences and constant multiples.
Distinguish signed area from total area.
For total area, split at points where the graph crosses the \(x\)-axis.
Look for even or odd symmetry on intervals of the form \([-a,a]\).
Check whether your final answer should be positive, negative or zero.

Connected integration skills

These properties are the algebraic foundation for area questions. Use area under a curve to practise signed area with one graph, then use area between two curves to apply interval splitting and upper-minus-lower reasoning. Use the trapezoidal rule when the task asks for a numerical approximation from function values or a table.

For the full sequence, use the Integration overview page.

13. Related Radford Mathematics videos

Definite integrals: introduction

Evaluate definite integrals on TI-Nspire

Area under a curve

Area between two curves

14. Practice with solutions

What is \(\int_5^5 f(x)\,dx\)?
Show solution
The interval has zero width.
\[\int_5^5 f(x)\,dx=0.\]
If \(\int_1^5 f(x)\,dx=12\) and \(\int_1^2 f(x)\,dx=3\), find \(\int_2^5 f(x)\,dx\).
Show solution
\[\int_1^5 f(x)\,dx=\int_1^2 f(x)\,dx+\int_2^5 f(x)\,dx.\]
\[12=3+\int_2^5 f(x)\,dx\]
So \(\int_2^5 f(x)\,dx=9\).
If \(\int_0^4 f(x)\,dx=7\), find \(\int_4^0 2f(x)\,dx\).
Show solution
\[\int_4^0 2f(x)\,dx=-\int_0^4 2f(x)\,dx=-2\int_0^4 f(x)\,dx=-14.\]
Evaluate \(\int_{-3}^{3} x^5\,dx\) using symmetry.
Show solution
The function \(x^5\) is odd, so the integral over \([-3,3]\) is zero.
\[\int_{-3}^{3}x^5\,dx=0.\]
Evaluate \(\int_{-2}^{2}(x^4+2)\,dx\) using symmetry.
Show solution
The function \(x^4+2\) is even.
\[\int_{-2}^{2}(x^4+2)\,dx=2\int_0^2(x^4+2)\,dx.\]
\[=2\left[\frac{x^5}{5}+2x\right]_0^2=2\left(\frac{32}{5}+4\right)=\frac{104}{5}.\]
Find the total area between \(y=x-2\) and the \(x\)-axis from \(x=0\) to \(x=5\).
Show solution
The graph crosses the \(x\)-axis at \(x=2\). Split the interval:
\[A_{\mathrm{total}}=\int_0^2(2-x)\,dx+\int_2^5(x-2)\,dx.\]
\[=2+\frac{9}{2}=\frac{13}{2}.\]
Given \(\int_a^b f(x)\,dx=5\) and \(\int_a^b g(x)\,dx=8\), find \(\int_a^b(4f(x)-g(x))\,dx\).
Show solution
\[\int_a^b(4f(x)-g(x))\,dx=4\int_a^b f(x)\,dx-\int_a^b g(x)\,dx=4(5)-8=12.\]
Explain why \(\int_{-1}^{1}x\,dx=0\) does not mean the graph encloses no area with the \(x\)-axis.
Show solution
The positive area on \([0,1]\) cancels the negative signed area on \([-1,0]\). The total geometric area is not zero.
\[\int_{-1}^{1}|x|\,dx=2\int_0^1x\,dx=1.\]

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