Mathematics Class 12 — Part…
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Definite Integrals and Properties

🎓 Class 12 Mathematics CBSE Theory Ch 7 — Integrals ⏱ ~25 min
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This MCQ module is based on: Definite Integrals and Properties

This mathematics assessment will be based on: Definite Integrals and Properties
Targeting Class 12 level in Calculus, with Advanced difficulty.

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7.7 Definite Integral

In previous sections, we have studied indefinite integrals and various methods of finding them. In this section, we shall study what is called the definite integral? of a function.

Definition
A definite integral is denoted by \(\int_a^b f(x)\,dx\), where \(a\) is called the lower limit of the integral and \(b\) is called the upper limit of the integral. The definite integral is introduced either as the limit of a sum or if it has an anti derivative F in the interval \([a, b]\), then its value is the difference between the values of F at the end points, i.e., \(F(b) - F(a)\).
X Y O y = f(x) x = a x = b A(x)
Fig 7.1 — The shaded region represents the area bounded by \(y = f(x)\), the x-axis, and the ordinates \(x = a\) and \(x = b\)

7.10 Some Properties of Definite Integrals

We state below some important properties of definite integrals. These will be useful in evaluating the definite integrals more easily.

Property P₀
\[\int_a^b f(x)\,dx = \int_a^b f(t)\,dt\] (The variable of integration is a dummy variable and can be changed.)
Property P₁
\[\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx\] (Interchanging limits changes the sign.) In particular, \(\int_a^a f(x)\,dx = 0\).
Property P₂
\[\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx\] (Splitting the interval at any point \(c\).)
Property P₃
\[\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx\]
Property P₄
\[\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx\] (This is a particular case of P₃.)
Property P₅
\[\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a - x)\,dx\]
Property P₆ (Even/Odd Functions)

(i) If \(f\) is an even function, i.e., \(f(-x) = f(x)\): \[\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx\]

(ii) If \(f\) is an odd function, i.e., \(f(-x) = -f(x)\): \[\int_{-a}^a f(x)\,dx = 0\]

Worked Examples

Example 28

Evaluate \(\int_{-1}^1 |x|\,dx\)

Solution
We note that \(|x| = x\) if \(x \geq 0\) and \(|x| = -x\) if \(x < 0\). Also \(|x|\) is an even function since \(|-x| = |x|\). \[\int_{-1}^1 |x|\,dx = 2\int_0^1 x\,dx = 2\left[\frac{x^2}{2}\right]_0^1 = 2 \cdot \frac{1}{2} = 1\]

Example 29

Evaluate \(\int_0^\pi \sin^2 x\,dx\)

Solution
We observe that \(\sin^2 x\) is an even function. Also, \(\sin^2(\pi - x) = \sin^2 x\). Therefore, by P₄: \[\int_0^\pi \sin^2 x\,dx = \int_0^\pi \sin^2(\pi - x)\,dx = \int_0^\pi \sin^2 x\,dx\] This confirms self-consistency. Using the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\): \[\int_0^\pi \frac{1 - \cos 2x}{2}\,dx = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_0^\pi = \frac{1}{2}\left[\pi - 0 - 0 + 0\right] = \frac{\pi}{2}\]

Example 30

Evaluate \(\int_0^\pi \log\sin x\,dx\)

Solution
Let \(I = \int_0^{\pi/2}\log\sin x\,dx\). By P₄: \[I = \int_0^{\pi/2}\log\sin\left(\frac{\pi}{2} - x\right)\,dx = \int_0^{\pi/2}\log\cos x\,dx\] Adding the two values of I: \[2I = \int_0^{\pi/2}[\log\sin x + \log\cos x]\,dx = \int_0^{\pi/2}\log(\sin x\cos x)\,dx\] \[= \int_0^{\pi/2}\log\frac{\sin 2x}{2}\,dx = \int_0^{\pi/2}\log\sin 2x\,dx - \int_0^{\pi/2}\log 2\,dx\] Put \(2x = t\) in the first integral. Then \(dx = \frac{dt}{2}\), when \(x = 0\), \(t = 0\) and when \(x = \frac{\pi}{2}\), \(t = \pi\). \[2I = \frac{1}{2}\int_0^\pi \log\sin t\,dt - \frac{\pi}{2}\log 2\] By P₅, since \(\sin(\pi - t) = \sin t\): \(\int_0^\pi \log\sin t\,dt = 2\int_0^{\pi/2}\log\sin t\,dt = 2I\). \[2I = \frac{1}{2}(2I) - \frac{\pi}{2}\log 2 = I - \frac{\pi}{2}\log 2\] \[I = -\frac{\pi}{2}\log 2\]
Interactive: Definite Integral Evaluator
Evaluate \(\int_a^b x^n\,dx\) for given values of \(a\), \(b\), and \(n\)
Enter values and click Evaluate.

Exercise 7.9 (Using Properties)

1. \(\int_0^1 \frac{x}{x^2+1}\,dx\)
Put \(x^2 + 1 = t\), \(2x\,dx = dt\). When \(x=0, t=1\); when \(x=1, t=2\). \[\frac{1}{2}\int_1^2 \frac{dt}{t} = \frac{1}{2}[\log t]_1^2 = \frac{1}{2}\log 2\]
2. \(\int_0^{\pi/2} \sqrt{\sin x}\cos x\,dx\)
Put \(\sin x = t\), \(\cos x\,dx = dt\). When \(x = 0, t = 0\); when \(x = \pi/2, t = 1\). \[\int_0^1 \sqrt{t}\,dt = \left[\frac{2}{3}t^{3/2}\right]_0^1 = \frac{2}{3}\]
3. \(\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2}\right)\,dx\)
Put \(x = \tan\theta\), \(dx = \sec^2\theta\,d\theta\). When \(x = 0, \theta = 0\); when \(x = 1, \theta = \pi/4\). \(\sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta\). \[\int_0^{\pi/4} 2\theta\sec^2\theta\,d\theta\] By IBP: \(= 2[\theta\tan\theta]_0^{\pi/4} - 2\int_0^{\pi/4}\tan\theta\,d\theta = 2\cdot\frac{\pi}{4} - 2[\log\sec\theta]_0^{\pi/4} = \frac{\pi}{2} - 2\log\sqrt{2} = \frac{\pi}{2} - \log 2\).
4. \(\int_0^2 x\sqrt{x+2}\,dx\) (Put \(x + 2 = t^2\))
Put \(x + 2 = t^2\), \(dx = 2t\,dt\), \(x = t^2 - 2\). When \(x = 0, t = \sqrt{2}\); when \(x = 2, t = 2\). \[\int_{\sqrt{2}}^2 (t^2 - 2)\cdot t \cdot 2t\,dt = 2\int_{\sqrt{2}}^2 (t^4 - 2t^2)\,dt = 2\left[\frac{t^5}{5} - \frac{2t^3}{3}\right]_{\sqrt{2}}^2\] \[= 2\left[\left(\frac{32}{5} - \frac{16}{3}\right) - \left(\frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3}\right)\right] = 2\left[\frac{16}{15} + \frac{8\sqrt{2}}{15}\right] = \frac{16(2 + \sqrt{2})}{15}\]
5. \(\int_0^{\pi/2} \frac{\sin x}{1+\cos^2 x}\,dx\)
Put \(\cos x = t\), \(-\sin x\,dx = dt\). When \(x = 0, t = 1\); when \(x = \pi/2, t = 0\). \[-\int_1^0 \frac{dt}{1+t^2} = \int_0^1 \frac{dt}{1+t^2} = [\tan^{-1}t]_0^1 = \frac{\pi}{4}\]
6. \(\int_0^2 \frac{dx}{x+4-x^2}\)
Complete the square: \(4+x-x^2 = \frac{17}{4} - (x - \frac{1}{2})^2\). Put \(t = x - \frac{1}{2}\): \[\int_{-1/2}^{3/2} \frac{dt}{\frac{17}{4} - t^2} = \frac{1}{2\cdot\frac{\sqrt{17}}{2}}\log\left|\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\right|\Bigg|_{-1/2}^{3/2} = \frac{1}{\sqrt{17}}\log\left|\frac{\sqrt{17}+3}{\sqrt{17}-3}\right| - \frac{1}{\sqrt{17}}\log\left|\frac{\sqrt{17}-1}{\sqrt{17}+1}\right|\]

Using properties of definite integrals, evaluate:

7. \(\int_0^{\pi/2} \cos^2 x\,dx\)
\(\cos^2 x = \frac{1+\cos 2x}{2}\). \(\int_0^{\pi/2}\frac{1+\cos 2x}{2}\,dx = \frac{1}{2}\left[x + \frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{1}{2}\left[\frac{\pi}{2} + 0\right] = \frac{\pi}{4}\).
8. \(\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx\)
Let \(I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx\). By P₄: \[I = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx\] Adding: \(2I = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}\). Therefore \(I = \frac{\pi}{4}\).
9. Evaluate \(\int_{-5}^5 |x+2|\,dx\)
\(|x+2| = -(x+2)\) when \(x < -2\) and \(= (x+2)\) when \(x \geq -2\). Split at \(x = -2\): \[\int_{-5}^{-2}[-(x+2)]\,dx + \int_{-2}^5(x+2)\,dx = \left[-\frac{x^2}{2}-2x\right]_{-5}^{-2} + \left[\frac{x^2}{2}+2x\right]_{-2}^5\] \[= [(-2+4) - (-\frac{25}{2}+10)] + [(\frac{25}{2}+10) - (2-4)] = 2 - (-\frac{5}{2}) + \frac{45}{2} + 2 = \frac{9}{2} + \frac{45}{2} = 29\]
Activity: Property P₄ Magic
L4 Analyse
Predict: Can you evaluate these integrals without computing any anti derivative, just using properties?
  1. Evaluate \(\int_0^{\pi/2}\frac{\sin^3 x}{\sin^3 x + \cos^3 x}\,dx\) using P₄.
  2. Evaluate \(\int_0^1 \log\left(\frac{1}{x} - 1\right)\,dx\) using P₄.
  3. For each, let \(I\) be the integral. Apply P₄ to get a second expression for \(I\), then add to simplify.

(1) Let \(I = \int_0^{\pi/2}\frac{\sin^3 x}{\sin^3 x + \cos^3 x}\,dx\). By P₄: \(I = \int_0^{\pi/2}\frac{\cos^3 x}{\cos^3 x + \sin^3 x}\,dx\). Adding: \(2I = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}\). So \(I = \frac{\pi}{4}\).

(2) Let \(I = \int_0^1 \log\frac{1-x}{x}\,dx\). By P₄: \(I = \int_0^1 \log\frac{x}{1-x}\,dx = -I\). So \(2I = 0\), \(I = 0\).

Competency-Based Questions

Scenario: A data scientist models the probability density function (PDF) of a random variable as \(f(x) = \frac{3}{2}x^2\) for \(-1 \leq x \leq 1\) and \(f(x) = 0\) otherwise. For a valid PDF, \(\int_{-\infty}^{\infty} f(x)\,dx = 1\).
Q1. Verify that \(f(x)\) is a valid PDF by evaluating \(\int_{-1}^1 \frac{3}{2}x^2\,dx\).
L3 Apply
  • (a) 0
  • (b) 1
  • (c) 3/2
  • (d) 2
Answer: (b) 1. Since \(x^2\) is an even function: \(\frac{3}{2}\int_{-1}^1 x^2\,dx = \frac{3}{2}\cdot 2\int_0^1 x^2\,dx = 3\cdot\frac{1}{3} = 1\). Valid PDF confirmed.
Q2. Using Property P₆, explain why \(\int_{-1}^1 x \cdot f(x)\,dx = 0\) (the expected value is zero).
L4 Analyse
Answer: The integrand is \(x \cdot \frac{3}{2}x^2 = \frac{3}{2}x^3\). Since \(x^3\) is an odd function (\((-x)^3 = -x^3\)), by Property P₆(ii): \(\int_{-1}^1 \frac{3}{2}x^3\,dx = 0\). The symmetry of the odd function about the origin causes positive and negative areas to cancel exactly.
Q3. Compute the variance \(E(X^2) = \int_{-1}^1 x^2\cdot f(x)\,dx\) and determine whether this distribution has more or less spread than a uniform distribution on \([-1, 1]\) (which has variance \(\frac{1}{3}\)).
L5 Evaluate
Answer: \(E(X^2) = \int_{-1}^1 x^2 \cdot \frac{3}{2}x^2\,dx = \frac{3}{2}\int_{-1}^1 x^4\,dx = \frac{3}{2}\cdot 2\int_0^1 x^4\,dx = 3\cdot\frac{1}{5} = \frac{3}{5}\). Since the mean is 0, variance = \(\frac{3}{5}\). This is greater than \(\frac{1}{3}\) (uniform), meaning this distribution has more spread — the density concentrates more weight near the tails (\(x = \pm 1\)).
Q4. If we changed the PDF to \(g(x) = c|x|\) on \([-1, 1]\), what value of \(c\) makes it a valid PDF? Use the even function property.
L6 Create
Answer: Since \(|x|\) is even: \(\int_{-1}^1 c|x|\,dx = 2c\int_0^1 x\,dx = 2c\cdot\frac{1}{2} = c\). For a valid PDF: \(c = 1\). So \(g(x) = |x|\) for \(x \in [-1, 1]\).

Assertion–Reason Questions

Assertion (A): \(\int_{-2}^2 x^3\,dx = 0\).
Reason (R): If \(f(x)\) is an odd function, then \(\int_{-a}^a f(x)\,dx = 0\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Both true. \(x^3\) is odd since \((-x)^3 = -x^3\). By R (P₆), the integral over symmetric limits is zero.
Assertion (A): \(\int_0^{\pi/2}\frac{\sin x}{\sin x + \cos x}\,dx = \frac{\pi}{4}\).
Reason (R): \(\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx\) (Property P₄).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Let \(I = \int_0^{\pi/2}\frac{\sin x}{\sin x + \cos x}\,dx\). By P₄: \(I = \int_0^{\pi/2}\frac{\cos x}{\cos x + \sin x}\,dx\). Adding: \(2I = \frac{\pi}{2}\), so \(I = \frac{\pi}{4}\). R is the property that makes this derivation possible.
Assertion (A): \(\int_a^b f(x)\,dx = \int_a^b f(t)\,dt\).
Reason (R): The value of a definite integral depends on the variable of integration.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (c) — A is true (Property P₀: the variable of integration is a dummy variable). R is false: the value of a definite integral does NOT depend on the variable of integration; it is determined entirely by the function and the limits.

Frequently Asked Questions

What is a definite integral?

A definite integral from a to b of f(x)dx represents the signed area between the curve, x-axis, and vertical lines x = a and x = b. It gives a numerical value, not a function.

What are the key properties of definite integrals?

Integral from a to a = 0, integral from a to b = -integral from b to a, and the integral splits at intermediate points. Non-negative functions give non-negative integrals.

What is the definite integral as a limit of sum?

The definite integral equals the limit as n approaches infinity of the sum of f(a + ih) * h where h = (b-a)/n. This connects integration to area approximation.

What are properties for symmetric limits?

For even functions f(-x) = f(x): integral from -a to a = 2 * integral from 0 to a. For odd functions f(-x) = -f(x): integral from -a to a = 0.

How to evaluate definite integrals using properties?

Check for simplifying properties (even/odd, periodicity), find the antiderivative, then apply the Fundamental Theorem: evaluate at upper limit minus lower limit.

Frequently Asked Questions — Integrals

What is Definite Integrals and Properties in NCERT Class 12 Mathematics?

Definite Integrals and Properties is a key concept covered in NCERT Class 12 Mathematics, Chapter 7: Integrals. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Definite Integrals and Properties step by step?

To solve problems on Definite Integrals and Properties, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 7: Integrals?

The essential formulas of Chapter 7 (Integrals) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Definite Integrals and Properties important for the Class 12 board exam?

Definite Integrals and Properties is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Definite Integrals and Properties?

Common mistakes in Definite Integrals and Properties include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Definite Integrals and Properties?

End-of-chapter NCERT exercises for Definite Integrals and Properties cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 7, and solve at least one previous-year board paper to consolidate your understanding.

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