MathsISC Class 12 · Integrals

Properties of Definite Integrals: The a + b − x Rule, Even and Odd Functions

Q: When should I use the property ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx?

Use it when the integrand has a structure that changes neatly under x → a + b − x: sin and cos swapping over [0, π/2], a factor x becoming π − x over [0, π], or terms like √x and √(a + b − x). Adding the two forms usually removes the difficult part.

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

Replace x by −x and simplify. If you get the same function, it is even; if you get its negative, it is odd. Many functions are neither, and then the property does not apply.

Q: Why do I need to split modulus integrals?

Because |g(x)| has a different formula on each side of the zeros of g. Splitting the interval at those zeros lets you remove the modulus correctly on each piece.

Q: Is the Fundamental Theorem of Calculus proved in ISC?

No. The CISCE 2027 syllabus states the Fundamental Theorem of Calculus without proof, and lists the properties of definite integrals that you must be able to use.

Q: How many marks do definite integral properties carry?

They appear across the paper: 1-mark Section A items, a 2-mark question (ISC 2024 Q5) and longer proofs such as the 6-mark ISC 2025 question proving the value of ∫ x/(1 + sin x) dx from π/4 to 3π/4.

Updated 15 October 2026 · Fact-checked

Properties let you evaluate definite integrals without a direct antiderivative. The most used is ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx: replace x by a + b − x, add the two forms, and simplify. Odd functions integrate to 0 over [−a, a]; even functions give twice the integral over [0, a].

Understand Properties of Definite Integrals

A definite integral $\int_{a}^{b} f (x) d x$ is a number: by the Fundamental Theorem of Calculus (stated without proof in the syllabus), it equals $F (b) - F (a)$ for any antiderivative $F$ . Some integrals have no easy antiderivative, yet their definite values are simple. Properties of definite integrals are the tools for those.

The basic properties (all listed in the CISCE syllabus):
1. $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t$ : the letter does not matter.
2. $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ , and so $\int_{a}^{a} f (x) d x = 0$ .
3. $\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$ for $a < c < b$ : split at any point, in particular where a modulus changes sign or a piecewise function changes formula.
4. $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ (often called the king property).
5. $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ , the special case $a = 0$ of property 4.
6. $\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ if $f (2 a - x) = f (x)$ , and $= 0$ if $f (2 a - x) = - f (x)$ .
7. $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ if $f$ is even ( $f (- x) = f (x)$ ), and $= 0$ if $f$ is odd ( $f (- x) = - f (x)$ ).

Why property 4 works. Substitute $x = a + b - t$ ; the limits swap and $d x = - d t$ , and the minus sign swaps them back. Geometrically, the graph is reflected so that the region runs from $b$ to $a$ instead, and its area is unchanged.

How to use property 4. Call the integral $I$ . Write $I$ again with $x$ replaced by $a + b - x$ . Then add the two expressions (sometimes subtract). The aim is that the awkward parts cancel or combine:

In $\int_{0}^{π /2} \frac{s i n x}{s i n x + c o s x} d x$ , the second form has $cos x$ on top, and adding gives $2 I = \int_{0}^{π /2} 1 d x$ .
In $\int_{0}^{π} \frac{x s i n x}{1 + c o s ^{2} x} d x$ , the factor $x$ becomes $π - x$ , and adding removes $x$ altogether, leaving $π$ times a simpler integral. ISC 2025 asked a 6-mark proof of this kind with limits $\frac{π}{4}$ to $\frac{3 π}{4}$ , and the examiners noted candidates who used the wrong property or shifted $π$ wrongly.
In $\int_{- a}^{a} \frac{d x}{1 + e ^{g (x)}}$ with $g$ odd, the two forms add to 1. ISC 2024 asked this type over $[0, 2 π]$ , and the examiners noted candidates who did not recognise that a property was needed.

Modulus functions. For $\int_{a}^{b} ∣ g (x) ∣ d x$ , find where $g (x) = 0$ , split the interval there (property 3), and drop the modulus with the right sign on each piece.

Even and odd. Before working hard on an integral over $[- a, a]$ , test $f (- x)$ . Products such as $x^{3} cos x$ or $sin^{5} x$ are odd, so their integral is 0 at once.

Properties of Definite Integrals: formulas to remember

Fundamental theorem: $\int_{a}^{b} f (x) d x = F (b) - F (a), F^{'} = f$; Stated without proof in the syllabus.
Reversing limits: $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$; So ∫ₐᵃ f(x) dx = 0.
Splitting: $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$; Split where a modulus or piecewise rule changes.
King property: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$; With a = 0: ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx.
Double interval: $\int_{0}^{2 a} f (x) d x = {2 \int_{0}^{a} f (x) d x, 0, f (2 a - x) = f (x) f (2 a - x) = - f (x)$; Symmetry about the middle of [0, 2a].
Even and odd: $\int_{- a}^{a} f (x) d x = {2 \int_{0}^{a} f (x) d x, 0, f even f odd$; Check f(−x) first.
A standard result: $\int_{0}^{π /2} lo g sin x d x = \int_{0}^{π /2} lo g cos x d x = - \frac{π}{2} lo g 2$; Proved with properties 5 and 6.

How to solve Properties of Definite Integrals questions

When a definite integral looks hard to integrate directly, run through this checklist.

1If the limits are $- a$ and $a$ , test $f (- x)$ . Odd: the answer is 0. Even: write $2 \int_{0}^{a}$ .
2If there is a modulus, find where the expression inside is zero, split the interval there, and remove the modulus with the correct sign on each part.
3Otherwise call the integral $I$ and write it again with $x$ replaced by $a + b - x$ , simplifying trigonometric terms (for example $sin (π - x) = sin x$ , $sin (\frac{π}{2} - x) = cos x$ ).
4Add the two forms of $I$ (or subtract, if that cancels more). Look for the numerator to match the denominator, or for a variable factor $x$ to disappear.
5Evaluate the simpler integral that remains, and divide by 2.
6If the remaining integral is symmetric about the middle of the interval, use property 6 to halve the range.
7Check the sign and size of the answer: a positive integrand must give a positive result.

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.

Odd functions integrate to zero

Suppose a city's temperature on a particular day differs from its daily mean by $d (t) = 4 sin \frac{π t}{12}$ °C, where $t$ is measured in hours from noon, from $t = - 12$ (midnight before) to $t = 12$ (midnight after).

Show without integrating directly that $\int_{- 12}^{12} d (t) d t = 0$ , and explain what this means.

Try it first, then show the solutionSolution

$d (- t) = 4 sin (- \frac{π t}{12}) = - d (t)$ , so $d$ is odd.
By the even-odd property, $\int_{- 12}^{12} d (t) d t = 0$ .
The warm afternoon hours exactly balance the cool morning hours, so the average of the deviation over the day is zero.

Answer: The integral is 0: the temperature above the mean in the afternoon balances the temperature below it in the morning.

Memory hook: Odd function, symmetric limits: the two halves cancel.

Splitting at a modulus

Imagine a toy car moving on a straight track with velocity $v (t) = 2 - t$ m/s for $0 \leq t \leq 4$ seconds: forwards at first, then backwards.

Find its displacement $\int_{0}^{4} v d t$ and the total distance it travels $\int_{0}^{4} ∣ v ∣ d t$ .

Try it first, then show the solutionSolution

Displacement: $\int_{0}^{4} (2 - t) d t = [2 t - \frac{t ^{2}}{2}]_{0}^{4} = 8 - 8 = 0$ .
Distance: $v = 0$ at $t = 2$ , so split: $\int_{0}^{2} (2 - t) d t + \int_{2}^{4} (t - 2) d t$ .
Each part equals $2$ , so the distance is $4$ m.

Answer: Displacement 0 m (it returns to the start); distance travelled 4 m.

Memory hook: For distance, split where velocity changes sign and add the pieces as positives.

Tricks to remember Properties of Definite Integrals

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

Flip and add
Replace x by a + b − x, then add: the hard part usually cancels.
Use it when: Every king-property question.
Odd vanishes, even doubles
Over [−a, a]: odd functions give 0, even functions give twice the half.
Use it when: One-line answers in Section A.
Kill the x
If x multiplies a function of sin x or cos x, the king property turns x into (a + b) − x and removes it.
Use it when: ∫₀^π x f(sin x) dx questions.
Zero of the inside
For |g(x)|, split exactly where g(x) = 0.
Use it when: Modulus integrals.

Properties of Definite Integrals: solved examples

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

Example 1: An odd integrand · 1 mark · MCQBoard level

$\int_{- 1}^{1} x^{3} cos x d x$ equals:
(a) 0
(b) 1
(c) $2 cos 1$
(d) $2 sin 1$

Try it first, then show the solutionSolution

$f (- x) = (- x)^{3} cos (- x) = - x^{3} cos x = - f (x)$ : odd, so the integral over $[- 1, 1]$ is 0.

Answer: (a) 0.

Example 2: The classic sin over sin plus cos · 2 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

Let $I$ be the integral. By $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ : $I = \int_{0}^{π /2} \frac{c o s x}{c o s x + s i n x} d x$ . (1 mark)
Adding: $2 I = \int_{0}^{π /2} 1 d x = \frac{π}{2}$ , so $I = \frac{π}{4}$ . (1 mark)

Answer: $\frac{π}{4}$ .

Example 3: A modulus integral · 2 marks · Short answerHarder board level

Evaluate $\int_{0}^{4} ∣ x - 2∣ d x$ .

Try it first, then show the solutionSolution

Split at $x = 2$ : $\int_{0}^{2} (2 - x) d x + \int_{2}^{4} (x - 2) d x$ . (1 mark)
$= [2 x - \frac{x ^{2}}{2}]_{0}^{2} + [\frac{x ^{2}}{2} - 2 x]_{2}^{4} = 2 + 2 = 4$ . (1 mark)

Answer: 4.

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

Evaluate $\int_{0}^{π} x sin x cos^{2} x d x$ .

Try it first, then show the solutionSolution

Replace $x$ by $π - x$ ; since $sin (π - x) = sin x$ and $cos^{2} (π - x) = cos^{2} x$ : $I = \int_{0}^{π} (π - x) sin x cos^{2} x d x$ . (1 mark)
Add: $2 I = π \int_{0}^{π} sin x cos^{2} x d x$ . Put $t = cos x$ , $d t = - sin x d x$ : $= π \int_{- 1}^{1} t^{2} d t$ . (1 mark)
$= π [\frac{t ^{3}}{3}]_{- 1}^{1} = \frac{2 π}{3}$ , so $I = \frac{π}{3}$ . (1 mark)

Answer: $\frac{π}{3}$ .

Example 5: The king property with general limits · 3 marks · Short answerHarder board level

Evaluate $\int_{2}^{8} \frac{x}{x + 10 - x} d x$ .

Try it first, then show the solutionSolution

Here $a + b = 10$ . Replace $x$ by $10 - x$ : $I = \int_{2}^{8} \frac{10 - x}{10 - x + x} d x$ . (1 mark)
Add: $2 I = \int_{2}^{8} 1 d x = 6$ . (1 mark)
$I = 3$ . (1 mark)

Answer: 3.

Example 6: Log sin x from 0 to π/2 · 5 marks · Long answerHOTS / competency

Prove that $\int_{0}^{π /2} lo g sin x d x = - \frac{π}{2} lo g 2$ .

Try it first, then show the solutionSolution

Let $I = \int_{0}^{π /2} lo g sin x d x$ . By property 5, $I = \int_{0}^{π /2} lo g cos x d x$ . (1 mark)
Add: $2 I = \int_{0}^{π /2} lo g (sin x cos x) d x = \int_{0}^{π /2} lo g \frac{s i n 2 x}{2} d x = \int_{0}^{π /2} lo g sin 2 x d x - \frac{π}{2} lo g 2$ . (1 mark)
In $\int_{0}^{π /2} lo g sin 2 x d x$ put $t = 2 x$ : $= \frac{1}{2} \int_{0}^{π} lo g sin t d t$ . (1 mark)
Since $sin (π - t) = sin t$ , property 6 gives $\int_{0}^{π} lo g sin t d t = 2 \int_{0}^{π /2} lo g sin t d t = 2 I$ . So that integral is $\frac{1}{2} \cdot 2 I = I$ . (1 mark)
$2 I = I - \frac{π}{2} lo g 2$ , so $I = - \frac{π}{2} lo g 2$ . (1 mark)

Answer: Proved.

Where students lose marks in Properties of Definite Integrals

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

Mistake: Not recognising that a property is needed
The ISC 2024 and 2025 examiners both reported candidates who tried to integrate directly when a property was required.
Do this: If the antiderivative is not obvious and the limits are symmetric or "nice" (0 to π/2, 0 to π, −a to a), try the king property or the even-odd test first.
Usually costs: The whole question.
Mistake: Applying a + b − x wrongly
In ISC 2025 candidates shifted π wrongly with limits π/4 to 3π/4; the replacement must be a + b − x = π − x there.
Wrong working
Limits $\frac{π}{4}$ to $\frac{3 π}{4}$ : replace $x$ by $\frac{3 π}{4} - x$ .
Right working
Replace $x$ by $\frac{π}{4} + \frac{3 π}{4} - x = π - x$ .
Do this: Always compute a + b first and write it down.
Usually costs: 3 to 4 marks.
Mistake: Ignoring the sign inside a modulus
|g(x)| equals −g(x) where g is negative; integrating g directly can give zero or a negative answer.
Do this: Find the zeros of g in the interval, split there, and attach the right sign to each piece.
Usually costs: 1 to 2 marks.
Mistake: Wrong even-odd test
Students judge by appearance; x³ cos x looks mixed but is odd.
Do this: Substitute −x and simplify: f(−x) = f(x) means even, f(−x) = −f(x) means odd.
Usually costs: 1 mark.

Exam tips

State the property you use, in symbols, before applying it; examiners give a mark for it.
After replacing $x$ by $a + b - x$ , simplify every trigonometric term before adding.
Number the two forms ① and ② and write "① + ②" so the step is clear.
For modulus integrals, write the split explicitly with each piece's sign.
Remember that $\int_{a}^{b} f (x) d x$ is a number; there is no $+ C$ .

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
$\int_{- π /2}^{π /2} sin^{5} x d x$ equals:
(a) 0
(b) 1
(c) $\frac{8}{15}$
(d) $\frac{16}{15}$
Hint
Is $sin^{5} x$ odd or even?
Answer
(a) 0, since $sin^{5} x$ is odd.
Q2 · 2 marks · Short answerBoard level
Using properties, evaluate $\int_{0}^{π /2} cos^{2} x d x$ .
Hint
Show it equals $\int_{0}^{π /2} sin^{2} x d x$ , then add.
Answer
By property 5, $I = \int_{0}^{π /2} sin^{2} x d x$ . Adding, $2 I = \int_{0}^{π /2} 1 d x = \frac{π}{2}$ , so $I = \frac{π}{4}$ .
Q3 · 3 marks · Short answerHarder board level
Evaluate $\int_{0}^{1} x (1 - x)^{5} d x$ .
Hint
Replace $x$ by $1 - x$ .
Answer
$I = \int_{0}^{1} (1 - x) x^{5} d x = \int_{0}^{1} (x^{5} - x^{6}) d x = \frac{1}{6} - \frac{1}{7} = \frac{1}{42}$ .
Q4 · 3 marks · Short answerHarder board level
Evaluate $\int_{- 1}^{2} ∣ x^{3} - x ∣ d x$ .
Hint
$x^{3} - x = x (x - 1) (x + 1)$ changes sign at $- 1, 0, 1$ .
Answer
On $[- 1, 0]$ : $\int (x^{3} - x) = \frac{1}{4}$ . On $[0, 1]$ : $\int (x - x^{3}) = \frac{1}{4}$ . On $[1, 2]$ : $\int (x^{3} - x) = \frac{9}{4}$ . Total $\frac{11}{4}$ .
Q5 · 2 marks · Short answerHarder board level
Evaluate $\int_{- π /4}^{π /4} \frac{d x}{1 + e ^{t a n x}}$ .
Hint
Replace $x$ by $- x$ and add.
Answer
Replacing $x$ by $- x$ : $I = \int \frac{e ^{t a n x}}{e ^{t a n x} + 1} d x$ . Adding, $2 I = \int_{- π /4}^{π /4} 1 d x = \frac{π}{2}$ , so $I = \frac{π}{4}$ .
Q6 · 3 marks · Short answerHOTS / competency
Evaluate $\int_{0}^{π} \frac{x}{1 + s i n x} d x$ .
Hint
After the king property, multiply by $\frac{1 - s i n x}{1 - s i n x}$ .
Answer
$2 I = π \int_{0}^{π} \frac{d x}{1 + s i n x}$ . Since $sin (π - x) = sin x$ , property 6 gives $2 \int_{0}^{π /2} \frac{d x}{1 + s i n x}$ , and property 5 turns this into $2 \int_{0}^{π /2} \frac{d x}{1 + c o s x} = 2 \int_{0}^{π /2} \frac{1}{2} sec^{2} \frac{x}{2} d x = 2 [tan \frac{x}{2}]_{0}^{π /2} = 2$ . So $2 I = 2 π$ and $I = π$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a solar panel on a clear day produces power $P (t) = sin \frac{π t}{12}$ kW, where $t$ is the time in hours after 6 a.m., for $0 \leq t \leq 12$ .
(i) Show that $P (12 - t) = P (t)$ . (1 mark)
(ii) Find $\int_{0}^{6} P (t) d t$ . (1 mark)
(iii) Use a property to find the total energy for the day. (1 mark)
(iv) Find the average power over the 12 hours. (1 mark)
Hint
Use $\int_{0}^{2 a} f = 2 \int_{0}^{a} f$ .
Answer
(i) $sin (π - \frac{π t}{12}) = sin \frac{π t}{12}$ . (ii) $[- \frac{12}{π} cos \frac{π t}{12}]_{0}^{6} = \frac{12}{π}$ . (iii) $2 \times \frac{12}{π} = \frac{24}{π} \approx 7.64$ kWh. (iv) $\frac{24/ π}{12} = \frac{2}{π} \approx 0.64$ kW.

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 $\int_{- a}^{a} f (x) d x$ if $f$ is odd?Show answer
0.
3.Evaluate $\int_{0}^{π /2} \frac{s i n x}{s i n x + c o s x} d x$ .Show answer
$\frac{π}{4}$ .
4.Evaluate $\int_{- 2}^{2} x^{3} d x$ .Show answer
0.
5.Evaluate $\int_{- 1}^{1} ∣ x ∣ d x$ .Show answer
1.
6.Write the value of $\int_{0}^{π /2} lo g cos x d x$ .Show answer
$- \frac{π}{2} lo g 2$ .

Properties of Definite Integrals: frequently asked questions

When should I use the property ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx?

Use it when the integrand has a structure that changes neatly under x → a + b − x: sin and cos swapping over [0, π/2], a factor x becoming π − x over [0, π], or terms like √x and √(a + b − x). Adding the two forms usually removes the difficult part.

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

Replace x by −x and simplify. If you get the same function, it is even; if you get its negative, it is odd. Many functions are neither, and then the property does not apply.

Why do I need to split modulus integrals?

Because |g(x)| has a different formula on each side of the zeros of g. Splitting the interval at those zeros lets you remove the modulus correctly on each piece.

Is the Fundamental Theorem of Calculus proved in ISC?

No. The CISCE 2027 syllabus states the Fundamental Theorem of Calculus without proof, and lists the properties of definite integrals that you must be able to use.

How many marks do definite integral properties carry?

They appear across the paper: 1-mark Section A items, a 2-mark question (ISC 2024 Q5) and longer proofs such as the 6-mark ISC 2025 question proving the value of ∫ x/(1 + sin x) dx from π/4 to 3π/4.