Integration as anti differentiation

What is integration?

In plain words: differentiation turns a function into its slope; integration runs it backwards and asks which functions have a given slope. Since sin x, x³/3 and e^x have derivatives cos x, x² and e^x, they are anti derivatives of these. Adding any constant does not change a derivative, so sin x + C is an anti derivative of cos x for every real C.

Functions with same derivatives differ by a constant.

NCERT Mathematics Part II (Class 12), Section 7.2, page 227

So, on an interval, the family F(x) + C contains every anti derivative of f (NCERT’s proof uses an interval I, page 227), and the graphs of the anti derivatives are vertical shifts of one another.

Standard integrals

Each derivative formula gives an integral formula read backwards (NCERT pages 228–229):

For a = 1, ∫a^ˣ dx = x + C. All power formulas apply on intervals where the real powers are defined.

The log |x| needs the absolute value: for x < 0, d/dx log (−x) = (−1)/(−x) = 1/x as well (NCERT Example 1 (iii), page 232). In a particular problem, keep in mind the interval on which each formula holds (NCERT’s Note, page 229).

Properties of the indefinite integral

NCERT Section 7.2.1 (pages 229–231):

• d/dx ∫f(x) dx = f(x), and ∫f′(x) dx = f(x) + C: differentiation and integration undo each other.
• ∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx.
• ∫k f(x) dx = k ∫f(x) dx for any real k, and the same for sums of many terms. (Beyond the NCERT text: strictly k ≠ 0, since ∫0 dx = C while 0 × ∫f dx = 0.)

So a sum of standard terms is integrated term by term, with a single constant at the end. NCERT Example 2 (i) (page 232): ∫(x³ − 1)/x² dx = ∫(x − x⁻²) dx = x²/2 + 1/x + C.

Integration by inspection

To find an anti derivative by inspection, guess a function whose derivative is the integrand, then adjust the constant factor (NCERT page 231). NCERT Example 1 (pages 231–232): since d/dx (sin 2x) = 2 cos 2x, an anti derivative of cos 2x is ½ sin 2x; and x³ + x⁴ is an anti derivative of 3x² + 4x³.

A condition fixes C. NCERT Example 4 (pages 233–234): the anti derivatives of 4x³ − 6 are x⁴ − 6x + C. If F(0) = 3, then C = 3 and F(x) = x⁴ − 6x + 3.

Not every integral is elementary. No function built from polynomials, logarithms, exponentials, trigonometric functions and their inverses has derivative e^−x2, so ∫e^−x2 dx cannot be found by inspection (NCERT’s Remark (ii), page 234).

Solved examples

Common mistakes to avoid

Quick revision summary

• If F′(x) = f(x), then ∫f(x) dx = F(x) + C; on an interval, any two anti derivatives differ by a constant.
• Read the derivative table backwards for the standard integrals; ∫(1/x) dx = log |x| + C and ∫x^ⁿ dx = x^ⁿ⁺¹/(n + 1) + C for n ≠ −1.
• Integrals of sums and constant multiples split term by term; a condition such as F(0) = 3 fixes C.

Questions students ask

What is the difference between an anti derivative and an indefinite integral?

An anti derivative of f is one function F with F′(x) = f(x), while the indefinite integral ∫f(x) dx is the whole family F(x) + C of them on an interval, C being any real constant; NCERT also calls an anti derivative a primitive (page 226). So tan x and tan x + 5 are two anti derivatives of sec² x, and ∫sec² x dx = tan x + C names them all. NCERT’s Table 7.1 (page 227) calls any one of them “an integral of f”, so the two words are often used for each other; the + C is what marks the indefinite integral.

Why is the integral of 1/x equal to log |x| + C and not log x + C?

The integral of 1/x is log |x| + C because 1/x is defined for every x ≠ 0 but log x only for x > 0, and log |x| has derivative 1/x on both sides of 0, as the derivative of log x shows. On x > 0 alone, log x + C is right. The modulus matters whenever the expression inside can be negative: for x < 5, ∫dx/(x − 5) is log (5 − x) + C, which log |x − 5| gives and log (x − 5) cannot. A constant can hide inside too: ∫dx/(3x) is (1/3) log |x| + C or, just as well, (1/3) log |3x| + C, as log |3x| is log 3 + log |x|.

What is the difference between integration and differentiation?

Differentiation takes a function to its derivative, one answer, while integration goes back from a derivative to the functions that have it, so an indefinite integral is a family F(x) + C: d/dx (x⁶ + 4) = 6x⁵, but ∫6x⁵ dx = x⁶ + C, as the 4 is lost in differentiating. The rules differentiate any elementary function where it is differentiable, but integration has only methods that may fail, and some elementary functions, such as e^−x2, have no elementary anti derivative (NCERT page 234). Geometrically, a derivative is a slope and a definite integral a signed area.

What is the integral of 0, of 1 and of a constant k?

The integral of 0 is an arbitrary constant, ∫0 dx = C, not 0, because on an interval the functions with derivative 0 are exactly the constants, such as 5 or −2. The integral of 1 is ∫dx = x + C (NCERT page 228), since dx alone means 1 dx and the derivative of x is 1. For any constant k, ∫k dx = kx + C, because d/dx (kx) = k: for example, ∫7 dx = 7x + C.

Why can ∫dx/√(1 − x²) be written both as sin^⁻¹ x + C and as −cos^⁻¹ x + C?

Both forms of ∫dx/√(1 − x²) are right, and NCERT’s table of standard integrals lists both (page 228), because sin^⁻¹ x and −cos^⁻¹ x differ only by a constant: sin^⁻¹ x + cos^⁻¹ x = π/2 for −1 ≤ x ≤ 1 (a JEE identity beyond the NCERT text), so −cos^⁻¹ x is sin^⁻¹ x − π/2. At x = ½, for example, sin^⁻¹ x = π/6 and −cos^⁻¹ x = −π/3, which is π/6 − π/2. In the same way, ∫dx/(1 + x²) is tan^⁻¹ x + C or −cot^⁻¹ x + C: the C of one form absorbs the π/2.

Is integration in the JEE Main 2027 syllabus?

Yes, integration is in the JEE Main 2027 syllabus: Unit 8 (Integral Calculus) opens with “Integral as an anti-derivative, Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions”, then names “Integration by substitution, by parts and by partial fractions” and “Integration using trigonometric identities”. It goes on to the fundamental theorem of calculus, properties of definite integrals and areas bounded by simple curves. NCERT covers all of this in Class 12 Chapter 7 (indefinite integrals in Sections 7.2–7.6, definite ones in 7.7–7.10) and Chapter 8, Application of Integrals.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find an anti derivative of (i) sin 2x; (ii) e^2x; (iii) (ax + b)² by inspection (NCERT Exercise 7.1, Q1, Q3 and Q4, page 234).

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   (i) −½ cos 2x; (ii) ½e^2x; (iii) for a ≠ 0, (ax + b)³/(3a); for a = 0, an anti derivative is b^²x. In the first case, the derivative of (ax + b)³ is 3a(ax + b)².

2. Q2. Find ∫(4e^3x + 1) dx and ∫x²(1 − 1/x²) dx (NCERT Exercise 7.1, Q6 and Q7, page 234).

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   (4/3)e^3x + x + C, and ∫(x² − 1) dx = x³/3 − x + C.

3. Q3. Find ∫sec x (sec x + tan x) dx (NCERT Exercise 7.1, Q18, page 235).

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   ∫sec² x dx + ∫sec x tan x dx = tan x + sec x + C.

4. Q4. The anti derivative of √x + 1/√x is (A) (1/3)x^1/3 + 2x^1/2 + C; (B) (2/3)x^2/3 + ½x² + C; (C) (2/3)x^3/2 + 2x^1/2 + C; (D) (3/2)x^3/2 + ½x^1/2 + C (NCERT Exercise 7.1, Q21, page 235).

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   ∫x^1/2 dx = (2/3)x^3/2 and ∫x^−1/2 dx = 2x^1/2: (C).