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Mathematics Grade XII

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Chapter 7: Integrals

Class 12 • Mathematics • Chapter 7

Integrals

Undoing differentiation, and adding up infinitely many slivers to measure an area.

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Chapter Roadmap

Integration as Anti-Differentiation • Standard Integrals • Integration by Substitution • Integration by Parts • Partial Fractions • Definite Integrals • Fundamental Theorem of Calculus • Properties of Definite Integrals

Reversing the Derivative

Differentiation breaks a quantity down into its rate of change. Integration runs the process backwards: given a rate, it rebuilds the original quantity. Because many derivatives can lead to the same shape once a constant is dropped, every indefinite integral carries an arbitrary constant C. Integration is therefore the reverse of differentiation, which is why it is also called anti-differentiation.

Integration has a second, equally powerful meaning. The definite integral adds up infinitely many thin slices under a curve to give the exact area of a region. The Fundamental Theorem of Calculus ties these two ideas together, turning area calculations into simple antiderivative evaluations. This chapter builds the standard integrals and the main techniques, substitution, parts and partial fractions, then applies them to definite integrals.

Foundation

If d/dx[F(x)] = f(x), then the indefinite integral ∫ f(x) dx = F(x) + C. The definite integral from a to b is F(b) − F(a), and it measures the signed area between the curve and the x-axis.

Key Terms You Must Know

Term

Meaning

Example

Antiderivative

A function whose derivative is the given function.

F(x) with F′ = f

Indefinite integral

The family F(x) + C of all antiderivatives.

∫ x dx = x²/2 + C

Definite integral

A number giving signed area between limits a and b.

∫ from 0 to 1 of x² dx = 1/3

Substitution

Swapping in a new variable to simplify an integral.

let u = inner function

Integration by parts

A rule from the product rule for products of functions.

∫ u dv = uv − ∫ v du

Partial fractions

Splitting a rational function into simpler pieces.

1/[x(x+1)] = 1/x − 1/(x+1)

Core Concepts, Step by Step

1. Integration as Anti-Differentiation

To integrate is to ask: which function has this as its derivative? Since adding a constant does not change a derivative, the answer is a whole family F(x) + C. Reversing the power rule gives the most-used result: the integral of xⁿ is xⁿ⁺¹/(n + 1) + C, valid for every n except −1, where the integral is ln|x| + C instead.

2. Standard Integrals

A short table of standard integrals covers most needs: powers, 1/x giving ln|x|, eₓ giving eₓ, sin and cos, sec² giving tan, and the inverse-trig forms such as 1/(1 + x²) giving tan⁻¹x. Integration is linear, so constants pull out and sums integrate term by term. Knowing these by heart turns most integrals into quick recognition.

3. Integration by Substitution

When an integral contains a function and (a multiple of) its derivative, we substitute a new variable u for the inner function. This converts the integral into a simpler one in u, which we integrate and then change back. Substitution is the integral version of the chain rule and is the first technique to try on a composite expression.

4. Integration by Parts

For a product of two functions we use ∫ u dv = uv − ∫ v du. The art is choosing u so that differentiating it simplifies the problem; the common guide is the order logarithmic, inverse-trig, algebraic, trig, exponential. Integration by parts is the integral form of the product rule and handles cases such as x eₓ and ln x.

5. Integration by Partial Fractions

A proper rational function with a factorable denominator can be split into simpler fractions, each easy to integrate. For example 1/[x(x + 1)] becomes 1/x − 1/(x + 1), and integrating gives a difference of logarithms. This technique reduces complicated rational integrals to standard log and power forms.

6. Definite Integrals and Their Properties

A definite integral is evaluated as F(b) − F(a) by the Fundamental Theorem of Calculus, and it equals the signed area under the curve. Helpful properties let us reverse limits (which flips the sign), split the interval, and exploit symmetry: for example ∫ from 0 to a of f(x) dx equals ∫ from 0 to a of f(a − x) dx, a trick that cracks many otherwise hard integrals.

A definite integral measures the exact area under the curve between two limits

Shaded area under a rising curve between x equals a and x equals b representing a definite integral

Key Results with Proofs

Result 1: Fundamental Theorem of Calculus

Statement. If F is an antiderivative of a continuous f, then ∫ from a to b of f(x) dx = F(b) − F(a).

Proof

The area built up to x grows at the rate f(x), which makes the area function an antiderivative.

Let A(x) be the area function, the area under f from a up to x. – define the area so far

A thin extra strip of width h has area about f(x) h, so A′(x) = f(x). – rate of area is the height

Thus A is an antiderivative of f, and any other antiderivative F differs from A by a constant. – A and F differ by a constant

Since A(a) = 0, the total area A(b) = F(b) − F(a). – evaluate at the limits

This is why evaluating a definite integral reduces to plugging the limits into any antiderivative.

Result 2: Integration by Parts

Statement. ∫ u dv = uv − ∫ v du.

Proof

The formula is just the product rule, integrated and rearranged.

Start from the product rule: d/dx(uv) = u (dv/dx) + v (du/dx). – product rule

Integrate both sides with respect to x: uv = ∫ u dv + ∫ v du. – integrate the identity

Rearrange to isolate the wanted integral. – make ∫ u dv the subject

∫ u dv = uv − ∫ v du. – the formula

Choosing u by the logarithmic, inverse-trig, algebraic, trig, exponential order usually simplifies the new integral.

Result 3: A Symmetry Property

Statement. ∫ from 0 to a of f(x) dx = ∫ from 0 to a of f(a − x) dx.

Proof

A single substitution t = a − x maps the interval onto itself.

In the right-hand integral substitute t = a − x, so dt = −dx. – substitute

The limits swap: x = 0 gives t = a and x = a gives t = 0. – change the limits

The two sign changes (from dx and from reversing limits) cancel. – signs cancel

This returns ∫ from 0 to a of f(t) dt, the same as the left side. – equality follows

Adding the integral to its mirror often collapses tough definite integrals, as in ∫ sin x/(sin x + cos x).

Worked Examples

Example 1

Question: Find ∫ x³ dx.

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Reverse the power rule.

Add one to the power and divide: x⁴/4. – power rule for integrals

Add the constant of integration. – never forget + C

Answer: x⁴/4 + C.

Example 2

Question: Find ∫ (3x² + 2x) dx.

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Integrate term by term.

∫ 3x² dx = x³ and ∫ 2x dx = x². – each term

Combine and add C. – sum

Answer: x³ + x² + C.

Example 3

Question: Find ∫ (1/x) dx.

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This is the exceptional power.

The power rule fails for n = −1; instead the integral of 1/x is ln|x|. – special case

Answer: ln|x| + C.

Example 4

Question: Find ∫ eₓ dx.

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Standard integral.

The exponential is its own integral. – recall

Answer: eₓ + C.

Example 5

Question: Find ∫ cos x dx.

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Standard integral.

The integral of cos x is sin x. – recall

Answer: sin x + C.

Example 6

Question: Find ∫ x e^(x²) dx.

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Use substitution.

Let u = x², so du = 2x dx and x dx = du/2. – substitute the inner function

The integral becomes (1/2) ∫ eₕ du = (1/2) eₕ. – integrate in u

Change back: (1/2) e^(x²). – return to x

Answer: (1/2) e^(x²) + C.

Example 7

Question: Find ∫ x cos x dx.

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Use integration by parts.

Take u = x (du = dx) and dv = cos x dx (v = sin x). – choose u and dv

∫ u dv = uv − ∫ v du = x sin x − ∫ sin x dx. – apply the formula

= x sin x + cos x. – finish the easy integral

Answer: x sin x + cos x + C.

Example 8

Question: Find ∫ ln x dx.

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Use parts with the whole thing as u.

Take u = ln x (du = dx/x) and dv = dx (v = x). – clever choice

= x ln x − ∫ x · (1/x) dx = x ln x − ∫ 1 dx. – apply parts

= x ln x − x. – finish

Answer: x ln x − x + C.

Example 9

Question: Find ∫ 1/[x(x + 1)] dx.

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Use partial fractions.

Split: 1/[x(x + 1)] = 1/x − 1/(x + 1). – partial fractions

Integrate each: ln|x| − ln|x + 1|. – logs

= ln|x/(x + 1)|. – combine logs

Answer: ln|x/(x + 1)| + C.

Example 10

Question: Evaluate ∫ from 0 to 1 of x² dx.

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Use the Fundamental Theorem of Calculus.

Antiderivative is x³/3. – integrate

Evaluate: 1³/3 − 0 = 1/3. – apply limits

Answer: 1/3.

Example 11

Question: Evaluate ∫ from 0 to π/2 of sin x dx.

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Antiderivative then limits.

Antiderivative of sin x is −cos x. – integrate

= −cos(π/2) + cos 0 = 0 + 1 = 1. – apply limits

Answer: 1.

Example 12

Question: Evaluate I = ∫ from 0 to π/2 of sin x/(sin x + cos x) dx.

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Use the symmetry property.

Replacing x by π/2 − x swaps sin and cos, giving J = ∫ cos x/(sin x + cos x). – mirror integral

Then I + J = ∫ from 0 to π/2 of 1 dx = π/2, and by symmetry I = J. – add the two

So 2I = π/2, giving I = π/4. – solve

Answer: π/4.

Where You Meet This in Real Life

Physics

Distance is the integral of speed and work is the integral of force, so integration recovers totals from rates.

Engineering

Areas, volumes, centres of mass and the load on a beam are all computed with definite integrals.

Economics

Total cost and consumer surplus are found by integrating marginal quantities over a range.

Probability and statistics

For continuous data, probabilities are areas under a density curve, which are definite integrals.

Medicine

The total exposure of the body to a drug over time is the area under its concentration curve.

Practice Sets A–D

Practice Set A – Basics

A1. Find ∫ x⁴ dx.

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Power rule.

x⁵/5 + C.

Answer: x⁵/5 + C.

A2. Find ∫ sin x dx.

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Standard.

−cos x + C.

Answer: −cos x + C.

A3. Find ∫ 5 dx.

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Constant.

5x + C.

Answer: 5x + C.

A4. Find ∫ sec² x dx.

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Standard.

tan x + C.

Answer: tan x + C.

Practice Set B – Conceptual

B1. Why does an indefinite integral always include + C?

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Think about derivatives of constants.

Any constant differentiates to 0, so infinitely many functions share the same derivative. – same derivative

The + C captures that whole family. – family of answers

Answer: Because constants vanish on differentiation, giving a family of antiderivatives.

B2. Which standard power has a logarithmic integral instead of a power?

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Recall the exception.

The power n = −1, that is 1/x, integrates to ln|x|. – exception

Answer: 1/x integrates to ln|x|.

B3. Which technique reverses the chain rule?

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Name it.

Integration by substitution.

Answer: Substitution.

B4. Which technique reverses the product rule?

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Name it.

Integration by parts.

Answer: Integration by parts.

Practice Set C – Application / Numerical

C1. Find ∫ (2x + 1)⁴ dx.

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Substitute u = 2x + 1.

u = 2x + 1, du = 2 dx, so dx = du/2. – substitute

(1/2) ∫ u⁴ du = (1/2) u⁵/5 = (2x + 1)⁵/10. – integrate and return

Answer: (2x + 1)⁵/10 + C.

C2. Evaluate ∫ from 1 to 2 of (1/x) dx.

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Antiderivative then limits.

Antiderivative ln|x|; evaluate ln 2 − ln 1 = ln 2. – apply limits

Answer: ln 2.

C3. Find ∫ x eₓ dx.

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Parts.

u = x, dv = eₓ dx, so v = eₓ. – choose

= x eₓ − ∫ eₓ dx = x eₓ − eₓ = eₓ(x − 1). – apply parts

Answer: eₓ(x − 1) + C.

C4. Evaluate ∫ from 0 to 1 of eₓ dx.

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FTC.

Antiderivative eₓ; e¹ − e⁰ = e − 1. – apply limits

Answer: e − 1.

Practice Set D – HOTS / Multi-step

D1. Evaluate ∫ from 0 to 1 of x eₓ dx.

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Parts, then limits.

Antiderivative is eₓ(x − 1). – from parts

Evaluate: at 1 it is e¹(0) = 0; at 0 it is e⁰(−1) = −1. – limits

0 − (−1) = 1. – subtract

Answer: 1.

D2. Find ∫ (2x)/(x² + 1) dx.

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Spot the derivative on top.

The numerator 2x is exactly the derivative of the denominator x² + 1. – recognise

So the integral is ln|x² + 1|. – log form

Answer: ln(x² + 1) + C.

D3. Use partial fractions to find ∫ 1/(x² − 1) dx.

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Factor first.

x² − 1 = (x − 1)(x + 1); split as ½[1/(x − 1) − 1/(x + 1)]. – partial fractions

Integrate: ½[ln|x − 1| − ln|x + 1|] = ½ ln|(x − 1)/(x + 1)|. – logs

Answer: ½ ln|(x − 1)/(x + 1)| + C.

D4. Evaluate I = ∫ from 0 to π/2 of cos x/(sin x + cos x) dx using symmetry.

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Pair it with the sine version.

Its mirror under x → π/2 − x is the sine integral, and the two sum to ∫ 1 dx = π/2. – symmetry pairing

By symmetry both equal half of π/2. – equal halves

Answer: π/4.

Chapter Summary

Everything in One Glance

Anti-Differentiation

∫ f dx = F + C where F′ = f; the integral of xⁿ is xⁿ⁺¹/(n+1), except 1/x → ln|x|.

Standard Integrals

Know powers, 1/x, eₓ, sin, cos, sec² and the inverse-trig forms.

Substitution

Reverses the chain rule; set u to the inner function.

By Parts

∫ u dv = uv − ∫ v du; choose u by the LIATE order.

Partial Fractions

Split a rational function into simple fractions, then integrate each.

Definite Integrals

F(b) − F(a) gives signed area; use symmetry and splitting properties.

Are You Exam-Ready?

8-Point Exam Quick-Check

Find ∫ x⁵ dx.

Find ∫ (1/x) dx.

Find ∫ cos x dx.

Find ∫ x e^(x²) dx by substitution.

Find ∫ x sin x dx by parts.

Evaluate ∫ from 0 to 2 of x dx.

Evaluate ∫ from 0 to π of sin x dx.

Split 1/[x(x + 2)] into partial fractions.

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Class 12 Mathematics Chapter 7: Integrals, Complete Notes and Practice

These free Class 12 Maths Integrals notes follow the NCERT 2026 to 27 syllabus and cover indefinite and definite integrals, standard integrals, substitution, integration by parts, partial fractions, the Fundamental Theorem of Calculus and properties of definite integrals, with twelve worked examples and sixteen graded practice questions. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.