MathsCBSE Class 12 · Integrals

Properties of Definite Integrals: a + b − x, Even–Odd and Modulus (Class 12)

Q: What is the king property of definite integrals?

It states that the integral of f(x) from a to b equals the integral of f(a + b − x) over the same limits. Writing the integral both ways and adding often cancels the difficult part.

Q: How do I integrate a modulus function?

Find where the expression inside the modulus is zero, split the interval at that point, and write the expression with a plus or minus sign on each piece before integrating.

Q: How do I test whether a function is odd or even?

Replace x by −x. If the function stays the same, it is even; if it becomes its negative, it is odd. Many functions are neither, but can be split into odd and even parts.

Q: Which NCERT section covers these properties?

Section 7.10 (Some Properties of Definite Integrals) of Chapter 7, Integrals, in NCERT Mathematics Part II for Class 12.

Q: Do I need to prove the properties in the exam?

Usually you only apply them, naming the property you use. Short proofs by substitution can be asked, so know how to prove the king property with t = a + b − x.

Updated 15 October 2026 · Fact-checked

The key properties: the integral of f(x) from a to b equals that of f(a + b − x); over [−a, a] an odd function gives 0 and an even one gives twice the integral over [0, a]; and an integral can be split at any point, which handles modulus functions.

Over −a to a, an even function's two halves match so double one half; an odd function's areas are equal and opposite so the integral is 0.

Illustration drawn for QuestionBuddy.

Understand Properties of Definite Integrals

Some definite integrals have no easy antiderivative, yet their value can be found by symmetry. The properties below (NCERT Section 7.10) are proved by substitution, and the board asks for them every year, often for 5 marks.

The basic properties.

$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t$ (the variable name does not matter).
$\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ , and $\int_{a}^{a} f (x) d x = 0$ .
$\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$ for any $c$ : split where a formula changes, as in modulus functions.
$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ , and in particular $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ . This is often called the king property.
$\int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x$ ; so it equals $2 \int_{0}^{a} f (x) d x$ if $f (2 a - x) = f (x)$ and $0$ if $f (2 a - x) = - f (x)$ .
$\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ if $f$ is even ( $f (- x) = f (x)$ ), and $0$ if $f$ is odd ( $f (- x) = - f (x)$ ).

The add-up trick. With the king property, write $I$ in its original form and in its transformed form, then add. For $I = \int_{0}^{π /2} \frac{s i n ^{4} x}{s i n ^{4} x + c o s ^{4} x} d x$ , replacing $x$ by $\frac{π}{2} - x$ gives $I = \int_{0}^{π /2} \frac{c o s ^{4} x}{c o s ^{4} x + s i n ^{4} x} d x$ ; adding, $2 I = \int_{0}^{π /2} 1 d x = \frac{π}{2}$ , so $I = \frac{π}{4}$ . When an $x$ multiplies a function symmetric about the midpoint, as in $\int_{0}^{π} x sin^{3} x d x$ , the same trick turns $x$ into the constant $\frac{π}{2}$ .

Even and odd parts. Over a symmetric interval, split the integrand into its odd pieces (which vanish) and even pieces (which double). $x^{7}$ , $sin x$ , $x ∣ x ∣$ and $x^{3} cos x$ are odd; $x^{2}$ , $cos x$ , $∣ x ∣$ are even.

Modulus functions. Find where the expression inside the modulus changes sign, split the interval there, and remove the modulus with the correct sign on each piece. For example $\int_{- 3}^{3} ∣ x + 1∣ d x = \int_{- 3}^{- 1} - (x + 1) d x + \int_{- 1}^{3} (x + 1) d x = 2 + 8 = 10$ .

Remember that $\int_{a}^{b} f (x) d x = F (b) - F (a)$ where $F^{'} = f$ (the Fundamental Theorem of Calculus, used without proof). Definite integral as the limit of a sum is not in the 2026-27 syllabus.

Properties of Definite Integrals: formulas to remember

King property: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$; Special case: $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ .
Splitting the interval: $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$; Split at points where $f$ changes formula.
Even and odd functions: $\int_{- a}^{a} f (x) d x = {2 \int_{0}^{a} f (x) d x, 0, f even f odd$; Check $f (- x)$ first.
The 2a property: $\int_{0}^{2 a} f (x) d x = {2 \int_{0}^{a} f (x) d x, 0, f (2 a - x) = f (x) f (2 a - x) = - f (x)$; Useful for $\int_{0}^{π}$ of sine powers.
Reversing limits: $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$; And $\int_{a}^{a} f = 0$ .
Fundamental Theorem of Calculus: $\int_{a}^{b} f (x) d x = F (b) - F (a), F^{'} = f$; Used without proof in the syllabus.

How to solve Properties of Definite Integrals questions

A decision routine for definite integrals that resist direct integration.

1Symmetric limits $[- a, a]$ ? Check whether each part of the integrand is odd or even; drop the odd parts and double the even parts over $[0, a]$ .
2Modulus or piecewise integrand? Find the break points, split the interval there, and integrate each piece with the modulus removed.
3A ratio such as $\frac{f ( x )}{f ( x ) + f ( a + b - x )}$ , or a factor $x$ times a symmetric function? Use the king property, write the new form of $I$ , and add it to the original.
4Limits $0$ to $2 a$ with sine or cosine powers? Use the 2a property.
5Evaluate the simpler integral that remains, and divide by 2 if you added two forms of $I$ .
6Check the answer's size roughly (for example, a positive integrand must give a positive value).

Properties of Definite Integrals in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Distance versus displacement (splitting at a sign change)

Suppose a lift's velocity, in m/s, is $v (t) = t - 3$ for $0 \leq t \leq 6$ seconds (negative means going down).

Find the lift's displacement and the total distance it travels in these 6 seconds.

Try it first, then show the solutionSolution

Displacement $= \int_{0}^{6} (t - 3) d t = [\frac{t ^{2}}{2} - 3 t]_{0}^{6} = 18 - 18 = 0$ .
Distance $= \int_{0}^{6} ∣ t - 3∣ d t$ ; the sign changes at $t = 3$ .
$= \int_{0}^{3} (3 - t) d t + \int_{3}^{6} (t - 3) d t = \frac{9}{2} + \frac{9}{2} = 9$ m.

Answer: Displacement 0 m (it returns to the start); distance 9 m.

Memory hook: Displacement integrates v; distance integrates |v|, split where v changes sign.

Odd functions cancel over a symmetric interval

Imagine a pendulum bob whose velocity is $v (t) = cos t$ (in m/s) for $0 \leq t \leq π$ seconds.

Show that its displacement over this time is zero, and find the total distance it moves.

Try it first, then show the solutionSolution

Displacement $= \int_{0}^{π} cos t d t = [sin t]_{0}^{π} = 0$ . (Also, $cos (π - t) = - cos t$ , so by the 2a property the integral is 0.)
Distance $= \int_{0}^{π} ∣ cos t ∣ d t = \int_{0}^{π /2} cos t d t - \int_{π /2}^{π} cos t d t$ .
$= 1 - (0 - 1) = 2$ m.

Answer: Displacement 0; distance 2 m.

Memory hook: Forward and back cancel in the integral but not on the road.

Tricks to remember Properties of Definite Integrals

Short memory aids. Say each one aloud once, then use it on a question the same day.

The king flips x to a + b − x
Swap $x$ for (top + bottom − $x$ ) and the value stays the same. Then add the two versions of $I$ .
Use it when: Ratio integrals and integrals with a factor x.
Odd vanishes, even doubles
Over $[- a, a]$ : odd parts give 0, even parts give twice the half.
Use it when: Symmetric limits.
Modulus means split
Find where the inside is zero; on each side, write it with a + or − sign.
Use it when: Integrals of $∣ x - k ∣$ and similar.
x becomes half the sum
In $\int_{a}^{b} x g (x) d x$ with $g (a + b - x) = g (x)$ , the answer is $\frac{a + b}{2} \int_{a}^{b} g (x) d x$ .
Use it when: Integrals like $\int_{0}^{π} x sin^{3} x d x$ .

Properties of Definite Integrals: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Odd parts vanish · 2 marks · Very short answerBoard level

Evaluate $\int_{- 1}^{1} (x^{7} + sin x + 3 x^{2}) d x$ .

Try it first, then show the solutionSolution

$x^{7}$ and $sin x$ are odd, so their integrals over $[- 1, 1]$ are 0. (1 mark)
$3 x^{2}$ is even: $\int_{- 1}^{1} 3 x^{2} d x = 2 \int_{0}^{1} 3 x^{2} d x = 2 [x^{3}]_{0}^{1} = 2$ . (1 mark)

Answer: 2

Example 2: A modulus integral · 2 marks · Very short answerBoard level

Evaluate $\int_{- 3}^{3} ∣ x + 1∣ d x$ .

Try it first, then show the solutionSolution

$∣ x + 1∣ = - (x + 1)$ for $x < - 1$ and $x + 1$ for $x \geq - 1$ . (½ mark)
$\int_{- 3}^{- 1} - (x + 1) d x = [- \frac{( x + 1 ) ^{2}}{2}]_{- 3}^{- 1} = 0 + 2 = 2$ . (½ mark)
$\int_{- 1}^{3} (x + 1) d x = [\frac{( x + 1 ) ^{2}}{2}]_{- 1}^{3} = 8$ . Total $= 10$ . (1 mark)

Answer: 10

Example 3: The king property on a general interval · 3 marks · Short answerHarder board level

Evaluate $\int_{2}^{5} \frac{x}{x + 7 - x} d x$ .

Try it first, then show the solutionSolution

Let $I$ be the integral. With $a + b = 7$ , replace $x$ by $7 - x$ : $I = \int_{2}^{5} \frac{7 - x}{7 - x + x} d x$ . (1 mark)
Adding the two forms: $2 I = \int_{2}^{5} \frac{x + 7 - x}{x + 7 - x} d x = \int_{2}^{5} 1 d x = 3$ . (1 mark)
So $I = \frac{3}{2}$ . (1 mark)

Answer: $\frac{3}{2}$

Example 4: Removing a factor x by symmetry · 3 marks · Short answerHarder board level

Evaluate $\int_{0}^{π} x sin^{3} x d x$ .

Try it first, then show the solutionSolution

Replace $x$ by $π - x$ : $I = \int_{0}^{π} (π - x) sin^{3} x d x$ , since $sin (π - x) = sin x$ . Adding: $2 I = π \int_{0}^{π} sin^{3} x d x$ . (1 mark)
$sin^{3} x = sin x (1 - cos^{2} x)$ ; put $t = cos x$ : $\int_{0}^{π} sin^{3} x d x = \int_{- 1}^{1} (1 - t^{2}) d t = \frac{4}{3}$ . (1 mark)
$2 I = \frac{4 π}{3}$ , so $I = \frac{2 π}{3}$ . (1 mark)

Answer: $\frac{2 π}{3}$

Example 5: King property plus the 2a rule · 5 marks · Long answerHOTS / competency

Evaluate $\int_{0}^{π} \frac{x}{1 + s i n ^{2} x} d x$ .

Try it first, then show the solutionSolution

By the king property, $I = \int_{0}^{π} \frac{π - x}{1 + s i n ^{2} x} d x$ ; adding, $2 I = π \int_{0}^{π} \frac{d x}{1 + s i n ^{2} x}$ . (1 mark)
Since $sin^{2} (π - x) = sin^{2} x$ , the 2a property gives $\int_{0}^{π} \frac{d x}{1 + s i n ^{2} x} = 2 \int_{0}^{π /2} \frac{d x}{1 + s i n ^{2} x}$ . (1 mark)
Divide top and bottom by $cos^{2} x$ : $\frac{s e c ^{2} x}{s e c ^{2} x + t a n ^{2} x} = \frac{s e c ^{2} x}{1 + 2 t a n ^{2} x}$ . Put $t = tan x$ (limits 0 to $\infty$ ): $\int_{0}^{\infty} \frac{d t}{1 + 2 t ^{2}} = \frac{1}{2} [tan^{- 1} (2 t)]_{0}^{\infty} = \frac{π}{2 2}$ . (1½ marks)
So $\int_{0}^{π} \frac{d x}{1 + s i n ^{2} x} = \frac{π}{2}$ . (½ mark)
$2 I = \frac{π ^{2}}{2}$ , hence $I = \frac{π ^{2}}{2 2}$ . (1 mark)

Answer: $\frac{π ^{2}}{2 2}$

Example 6: Case study: two shops on a road · 4 marks · Case-basedHOTS / competency

Suppose two shops stand at the 2 km and 4 km marks of a 6 km straight road. For a house at position $x$ km ( $0 \leq x \leq 6$ ), the total walking distance to both shops is $D (x) = ∣ x - 2∣ + ∣ x - 4∣$ .
(i) Write $D (x)$ without modulus signs on $[0, 2]$ , $[2, 4]$ and $[4, 6]$ . (1 mark)
(ii) Evaluate $\int_{0}^{6} ∣ x - 2∣ d x$ . (1 mark)
(iii) Evaluate $\int_{0}^{6} D (x) d x$ . (1 mark)
(iv) Find the average value of $D$ over the road. (1 mark)

Try it first, then show the solutionSolution

(i) $D = 6 - 2 x$ on $[0, 2]$ ; $D = 2$ on $[2, 4]$ ; $D = 2 x - 6$ on $[4, 6]$ . (1 mark)
(ii) $\int_{0}^{2} (2 - x) d x + \int_{2}^{6} (x - 2) d x = 2 + 8 = 10$ . (1 mark)
(iii) $\int_{0}^{2} (6 - 2 x) d x + \int_{2}^{4} 2 d x + \int_{4}^{6} (2 x - 6) d x = 8 + 4 + 8 = 20$ . (1 mark)
(iv) Average $= \frac{1}{6} \int_{0}^{6} D (x) d x = \frac{20}{6} = \frac{10}{3}$ km. (1 mark)

Answer: (i) $6 - 2 x$ , $2$ , $2 x - 6$ (ii) 10 (iii) 20 (iv) $\frac{10}{3}$ km.

Where students lose marks in Properties of Definite Integrals

The slips examiners see most often, and what to write instead.

Mistake: Treating a modulus as ordinary brackets
The sign inside changes, so the formula changes.
Wrong working
$\int_{- 3}^{3} ∣ x + 1∣ d x = [\frac{( x + 1 ) ^{2}}{2}]_{- 3}^{3} = 8 - 2 = 6$ .
Right working
Split at $- 1$ : $2 + 8 = 10$ .
Do this: Always find the zero of the expression inside the modulus and split there.
Usually costs: Most of the marks.
Mistake: Applying the king property with the wrong a + b
On $[2, 5]$ the replacement is $7 - x$ , not $5 - x$ .
Wrong working
$\int_{2}^{5} f (x) d x = \int_{2}^{5} f (5 - x) d x$ .
Right working
$\int_{2}^{5} f (x) d x = \int_{2}^{5} f (7 - x) d x$ .
Do this: Write 'replace $x$ by $a + b - x$ ' with the actual numbers.
Usually costs: The method marks.
Mistake: Forgetting to divide by 2 after adding
Adding two equal forms gives $2 I$ .
Wrong working
$I + I = \frac{π}{2}$ , so $I = \frac{π}{2}$ .
Right working
$2 I = \frac{π}{2}$ , so $I = \frac{π}{4}$ .
Do this: Write ' $2 I = \dots$ ' explicitly.
Usually costs: The final mark.
Mistake: Calling a function odd without checking
$x^{2} sin x$ is odd, but $x sin x$ is even.
Wrong working
$\int_{- π}^{π} x sin x d x = 0$ since x is odd.
Right working
$(- x) sin (- x) = x sin x$ : even, so the integral is $2 \int_{0}^{π} x sin x d x = 2 π$ .
Do this: Compute $f (- x)$ for the whole integrand.
Usually costs: The whole question.

Exam tips

Definite-integral property questions carried the most marks in this chapter across 2023 to 2026 (including 5-mark questions in 2024); name each property you use.
Write $I$ twice (original and transformed) and add them in a clearly labelled line ' $2 I = \dots$ '.
For modulus questions, sketch a quick number line showing where the inside changes sign.
Check odd and even parts term by term; a single even term changes the answer.
Keep exact answers such as $\frac{π ^{2}}{2 2}$ ; rationalising is optional.

Practice questions on Properties of Definite Integrals

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $f$ is an odd function, then $\int_{- a}^{a} f (x) d x$ equals:
(a) $2 \int_{0}^{a} f (x) d x$
(b) 0
(c) $\int_{0}^{a} f (x) d x$
(d) $2 a$
Hint
The areas on both sides cancel.
Answer
(b) 0.
Q2 · 2 marks · Very short answerBoard level
Evaluate $\int_{0}^{4} ∣ x - 1∣ d x$ .
Hint
Split at $x = 1$ .
Answer
$\int_{0}^{1} (1 - x) d x + \int_{1}^{4} (x - 1) d x = \frac{1}{2} + \frac{9}{2} = 5$ .
Q3 · 2 marks · Very short answerBoard level
Using properties, evaluate $\int_{0}^{π /2} cos^{2} x d x$ .
Hint
By the king property, it equals $\int_{0}^{π /2} sin^{2} x d x$ .
Answer
$I = \int_{0}^{π /2} cos^{2} x d x = \int_{0}^{π /2} sin^{2} x d x$ . Adding: $2 I = \int_{0}^{π /2} 1 d x = \frac{π}{2}$ , so $I = \frac{π}{4}$ .
Q4 · 3 marks · Short answerHarder board level
Evaluate $\int_{0}^{π /2} \frac{s i n ^{4} x}{s i n ^{4} x + c o s ^{4} x} d x$ .
Hint
Replace $x$ by $\frac{π}{2} - x$ and add.
Answer
The transformed integral has $cos^{4} x$ on top; adding gives $2 I = \frac{π}{2}$ , so $I = \frac{π}{4}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\int_{- 1}^{1} x ∣ x ∣ d x = 0$ .
Reason (R): $x ∣ x ∣$ is an odd function.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compute $(- x) ∣ - x ∣$ .
Answer
(a). $(- x) ∣ - x ∣ = - x ∣ x ∣$ , so the function is odd and its integral over $[- 1, 1]$ is 0.
Q6 · 3 marks · Short answerHarder board level
Evaluate $\int_{0}^{1} x (1 - x)^{5} d x$ using a property.
Hint
Replace $x$ by $1 - x$ .
Answer
$\int_{0}^{1} x (1 - x)^{5} d x = \int_{0}^{1} (1 - x) x^{5} d x = \int_{0}^{1} (x^{5} - x^{6}) d x = \frac{1}{6} - \frac{1}{7} = \frac{1}{42}$ .
Q7 · 5 marks · Long answerHOTS / competency
Evaluate $\int_{- π /2}^{π /2} \frac{s i n ^{2} x}{1 + e ^{x}} d x$ .
Hint
Use $x \to - x$ (the king property on $[- a, a]$ ) and note $\frac{1}{1 + e ^{x}} + \frac{1}{1 + e ^{- x}} = 1$ .
Answer
Replacing $x$ by $- x$ : $I = \int_{- π /2}^{π /2} \frac{s i n ^{2} x}{1 + e ^{- x}} d x = \int \frac{e ^{x} s i n ^{2} x}{e ^{x} + 1} d x$ . Adding: $2 I = \int_{- π /2}^{π /2} sin^{2} x d x = 2 \int_{0}^{π /2} sin^{2} x d x = \frac{π}{2}$ . Hence $I = \frac{π}{4}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the king property.Show answer
$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ .
2.What is $\int_{- a}^{a} f (x) d x$ when f is even?Show answer
$2 \int_{0}^{a} f (x) d x$ .
3.Evaluate $\int_{- 2}^{2} x^{3} d x$ .Show answer
0 (odd function).
4.Where do you split $\int_{0}^{5} ∣ x - 3∣ d x$ ?Show answer
At $x = 3$ ; the value is $\frac{9}{2} + 2 = \frac{13}{2}$ .
5.What is $\int_{0}^{π /2} \frac{c o s ^{n} x}{s i n ^{n} x + c o s ^{n} x} d x$ ?Show answer
$\frac{π}{4}$ .
6.Is the limit-of-a-sum definition in the 2026-27 syllabus?Show answer
No.

Properties of Definite Integrals: frequently asked questions

What is the king property of definite integrals?

It states that the integral of f(x) from a to b equals the integral of f(a + b − x) over the same limits. Writing the integral both ways and adding often cancels the difficult part.

How do I integrate a modulus function?

Find where the expression inside the modulus is zero, split the interval at that point, and write the expression with a plus or minus sign on each piece before integrating.

How do I test whether a function is odd or even?

Replace x by −x. If the function stays the same, it is even; if it becomes its negative, it is odd. Many functions are neither, but can be split into odd and even parts.

Which NCERT section covers these properties?

Section 7.10 (Some Properties of Definite Integrals) of Chapter 7, Integrals, in NCERT Mathematics Part II for Class 12.

Do I need to prove the properties in the exam?

Usually you only apply them, naming the property you use. Short proofs by substitution can be asked, so know how to prove the king property with t = a + b − x.