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MathematicsClass 12Integrals

Integrals | Class 12 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 3 min read

Integrals – this guide gives you a concise, exam-ready overview of Integrals from Class 12 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

Introduction

The chapter on Integrals introduces integration as the inverse process of differentiation. While differentiation is concerned with finding the rate of change or slope of a function at any point, integration focuses on finding the original function when its derivative is known. This fundamental relationship between differentiation and integration forms the basis of integral calculus. Integration is used to determine quantities like area under curves, displacement from velocity, and accumulated change from rates. The chapter begins by explaining the concept of an antiderivative or indefinite integral, which represents a family of functions differing by a constant. The notation ∫f(x) dx is introduced to denote the integral of a function f(x) with respect to x. The constant of integration, denoted by C, is emphasized as essential because differentiation of a constant is zero, so the original function can only be determined up to an additive constant. The chapter sets the stage for learning various techniques of integration, including substitution, integration by parts, and integration of special functions. It also introduces definite integrals, which assign numerical values to the area under a curve between two points, connecting integration to geometric interpretation and applications.

📊 Diagram: The chapter typically includes a diagram showing a curve y = f(x) and the area under the curve between two points, illustrating the geometric interpretation of integration as area.

🧪 Activity: No specific activity in this introductory section.

🔗 Connection: Leads to the next section 'Integral as an Antiderivative' which formally defines the concept of antiderivative and indefinite integral.

Frequently asked questions

Find an anti derivative (or integral) of the following functions by the method of inspection. 1. sin 2x 2. cos 3x 3. e2x 4. (ax + b)2 5. sin 2x – 4 e3x

1. ∫ sin 2x dx = -1/2 cos 2x + C

2. ∫ cos 3x dx = 1/3 sin 3x + C

3. ∫ e2x dx = 1/2 e2x + C

4. ∫ (ax + b)^2 dx = ∫ (a^2 x^2 + 2abx + b^2) dx = a^2 ∫ x^2 dx + 2ab ∫ x dx + b^2 ∫ dx = a^2 (x^3/3) + 2ab (x^2/2) + b^2 x + C = (a^2 x^3)/3 + ab x^2 + b^2 x + C

5. ∫ (sin 2x – 4 e3x) dx = ∫ sin 2x dx – 4 ∫ e3x dx = (-1/2) cos 2x – 4 (1/3) e3x + C = (-1/2) cos 2x – (4/3) e3x + C

6. ∫(4 e^{3x} + 1) dx

To evaluate ∫(4 e^{3x} + 1) dx, split the integral:

∫4 e^{3x} dx + ∫1 dx

First integral: ∫4 e^{3x} dx = 4 ∫ e^{3x} dx Using substitution u = 3x, du = 3 dx => dx = du/3 So, ∫ e^{3x} dx = (1/3) e^{3x} + C Therefore, ∫4 e^{3x} dx = 4 * (1/3) e^{3x} = (4/3) e^{3x}

Second integral: ∫1 dx = x + C

Combining: ∫(4 e^{3x} + 1) dx = (4/3) e^{3x} + x + C

7. ∫ x^2 (1 – 1/x^2) dx

Rewrite the integrand:

x^2 (1 – 1/x^2) = x^2 – 1

So, ∫ x^2 (1 – 1/x^2) dx = ∫ (x^2 – 1) dx = ∫ x^2 dx – ∫ 1 dx

Integrate each term: ∫ x^2 dx = (x^3)/3 + C ∫ 1 dx = x + C

Therefore, ∫ x^2 (1 – 1/x^2) dx = (x^3)/3 – x + C

8. ∫ (a x^2 + b x + c) dx

Integrate each term separately:

∫ a x^2 dx = a ∫ x^2 dx = a (x^3 / 3) + C ∫ b x dx = b ∫ x dx = b (x^2 / 2) + C ∫ c dx = c x + C

Therefore, ∫ (a x^2 + b x + c) dx = (a x^3)/3 + (b x^2)/2 + c x + C

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