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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.

Integral as an Antiderivative

This section elaborates on the fundamental idea that integration is the reverse process of differentiation. Given a function f(x) continuous on an interval, if there exists a function F(x) such that the derivative of F(x) is f(x), i.e., F'(x) = f(x), then F(x) is called an antiderivative of f(x). Since the derivative of a constant is zero, any two antiderivatives of f(x) differ by a constant. Therefore, the most general antiderivative is expressed as F(x) + C, where C is an arbitrary constant. The indefinite integral of f(x) with respect to x is denoted by ∫f(x) dx and represents the family of all antiderivatives of f(x). This section also explains that the process of finding an antiderivative is called integration. It emphasizes the importance of the constant of integration and provides examples to illustrate the concept. The section also discusses the linearity of integration, meaning the integral of a sum is the sum of the integrals, and constants can be factored out of the integral.

📊 Diagram: No specific diagram, but conceptual illustrations showing functions and their antiderivatives may be included.

🧪 Activity: No specific activity in this section.

🔗 Connection: Prepares for the next section 'Basic Integration Formulas' by establishing the foundation of integration.

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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