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Mathematics — Class 12 — L30: Integration

NIOS Code 311 · Module 8 · Calculus

Notes extracted from NIOS Mathematics Course (311), Lesson 30 — Integration (ilovepdf_merged (6).pdf). Content covers sections 30.1–30.x.
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Overview — Integration (L30)

Integration is the reverse of differentiation. An indefinite integral is a family of antiderivatives; the constant of integration C must appear.

∫ f(x) dx = F(x) + C where F′ = f
Check by differentiating back
area ≈ ∫ y dx
Definite integral as net signed area

30.1 Standard integrals

∫ xⁿ dx = xⁿ⁺¹/(n+1)+C (n≠−1) · ∫ eˣ dx = eˣ+C · ∫ 1/x dx = ln|x|+C
Power, exp, log
  • ∫ sin x dx = −cos x + C; ∫ cos x dx = sin x + C.
  • ∫ sec² x dx = tan x + C.

30.2 Methods

Substitution: set u=g(x), du=g′(x)dx. By parts: ∫u dv = uv − ∫v du. Choose u via LIATE heuristic when helpful.

P tangent, slope f′(x)
Derivative as slope of the tangent
Check: Always differentiate your answer to verify it recovers the integrand.

MCQ Quiz — L30 Integration

0 / 10 correct

Flashcards — L30

1 / 16

Golden Rules — L30 Integration

Most exam-important points from this chapter:

Know the definitions of L30

Start every answer with the key definition or standard form from Integration.

Write the formula first

Copy the formula, then substitute values; never jump to the number alone.

Watch conditions

Check domain, quadrant, non-zero denominators, and applicability of the theorem.

Master result 1

Memorise and apply: a dx,

Master result 2

Memorise and apply: x dx,

∫ xⁿ dx = x^{n+1}/(n+1)
∫ eˣ dx = eˣ
∫ 1/x dx = ln|x|
∫ uv' = uv−∫u'v
Substitution
Standard forms

1. Formulas & Definitions

Full Ch 30 — Integration study guide: definitions, derivations, why each formula works, history, deep understanding, pencil diagrams, step-by-step use, and 4-level solved examples.

∫ xⁿ dx = x^{n+1}/(n+1) + C (n≠−1)

Definition: Power rule for integration.

Derivation

Antiderivative reverse of power rule differentiation.

Variables

n ≠ −1 · +C constant

Why it works

Reverse engineering the derivative.

Historical context

Fundamental theorem links area and antiderivative.

Deep understanding

n=−1 case is ln|x|+C.

2. Diagrams & Visuals

∫ xⁿ dx = x^{n+1}/(n+1) + C (n≠−1) Power +1 Divide +C

Pencil sketch · labelled · step-by-step breakdown below

  1. Check n≠−1
  2. Add 1 to power
  3. Divide by new power
  4. Add +C

3. Solved Examples

Basic

Q: ∫ x³ dx

Solution: x⁴/4+C

Answer: x⁴/4 + C

Intermediate

Q: ∫ √x dx

Solution: (2/3)x^{3/2}+C

Answer: (2/3) x^{3/2} + C

Advanced

Q: ∫ 1/x² dx

Solution: −1/x+C

Answer: −1/x + C

Exam

Q: ∫ x⁻¹ dx

Solution: ln|x|+C

Answer: ln|x| + C

∫ u dv = uv − ∫ v du

Definition: Integration by parts (product reverse).

Derivation

From (uv)' = u'v + uv'.

Variables

Choose u to simplify when differentiated

Why it works

Moves derivative from one factor to another.

Historical context

Attributed to integration-by-parts formula of calculus texts (from product rule).

Deep understanding

LIATE heuristic for choosing u.

2. Diagrams & Visuals

∫ u dv = uv − ∫ v du uv − ∫v du Choose u wisely LIATE

Pencil sketch · labelled · step-by-step breakdown below

  1. Pick u and dv
  2. Compute du, v
  3. uv − ∫v du
  4. Simplify +C

3. Solved Examples

Basic

Q: ∫ x eˣ dx

Solution: x eˣ − ∫ eˣ = eˣ(x−1)+C

Answer: eˣ(x−1) + C

Intermediate

Q: ∫ x sin x dx

Solution: −x cos x + ∫ cos x = −x cos x + sin x + C

Answer: −x cos x + sin x + C

Advanced

Q: ∫ ln x dx

Solution: x ln x − x + C

Answer: x ln x − x + C

Exam

Q: Formula for parts.

Solution: uv−∫v du

Answer: uv − ∫ v du

Substitution: ∫ f(g(x)) g'(x) dx = ∫ f(u) du

Definition: Reverse chain rule.

Derivation

Set u=g(x), du=g'(x)dx.

Variables

u = inner function

Why it works

Compress composite derivative structure.

Historical context

Core technique with power and chain reverses.

Deep understanding

Always substitute back to x unless definite with limits change.

2. Diagrams & Visuals

Substitution: ∫ f(g(x)) g'(x) dx = ∫ f(u u = inner du = g' dx Integrate f(u)

Pencil sketch · labelled · step-by-step breakdown below

  1. Spot inner g(x)
  2. du = g' dx
  3. Integrate in u
  4. Back-substitute

3. Solved Examples

Basic

Q: ∫ 2x cos(x²) dx

Solution: sin(x²)+C

Answer: sin(x²) + C

Intermediate

Q: ∫ e^{3x} dx

Solution: (1/3)e^{3x}+C

Answer: (1/3) e^{3x} + C

Advanced

Q: ∫ tan x dx

Solution: −ln|cos x|+C

Answer: −ln|cos x| + C

Exam

Q: Idea of substitution?

Solution: Reverse chain rule

Answer: u = inner

5. Special Features & Extras

Complete study guide — exam tips, common mistakes, memory aids, and quick reference.

Exam Tips & Tricks

  • Always +C for indefinite integrals.
  • Differentiate answer to check.
  • Parts: let u be log/inverse/poly often.

Common Student Mistakes

  • Forgetting +C
  • n=−1 power rule wrongly
  • Not changing limits in definite sub

Memory Aids & Mnemonics

Parts: “u v minus integral v du”.
Power: up one, divide.

Which Formula When?

  • xⁿ → power
  • product different types → parts
  • f(g)g' → sub

Quick reference box

• x^{n+1}/(n+1) · uv−∫vdu · u-sub

PYQ — Previous Year Questions

English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.

L30 — Integration

4 question(s) · Sources: Board-style, Board-style parts, Board-style sub, Sample QP 2024

PYQ1. ∫ sec²(mx) dx (m≠0) equals:

1 mark(s) · SA · Sample QP 2024 · Q13(ii) OR

Model Answer

1)  (1/m) tan(mx) + C

Explanation

1)  Because d/dx tan(mx)=m sec²(mx).

PYQ2. Integrate ∫ (3x² − 2x + 1) dx.

1 mark(s) · SA · Board-style

Model Answer

1)  x³ − x² + x + C

Explanation

1)  Power rule term by term.

PYQ3. Evaluate ∫ 2x e^(x²) dx.

2 mark(s) · SA · Board-style sub

Model Answer

1)  e^(x²) + C

Explanation

1)  u=x², du=2x dx ⇒ ∫e^u du = e^(x²)+C.

PYQ4. Find ∫ x eˣ dx.

2 mark(s) · SA · Board-style parts

Model Answer

1)  eˣ(x − 1) + C

Explanation

1)  Parts: u=x, dv=eˣdx → xeˣ − ∫eˣdx = eˣ(x−1)+C.

Problem Solving — L30 Integration

10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.

Question 1 of 10Power

Integrate ∫ (3x² − 2x + 1) dx.

∫xⁿ dx = xⁿ⁺¹/(n+1)+C

1)  = x³ − x² + x + C.

Answer:  x³ − x² + x + C

Formula used

∫xⁿ dx = xⁿ⁺¹/(n+1)+C

Textbook formal language

Termwise antidifferentiation with +C.

Easy language (same calculation)

Raise powers and divide by new power; 3/3=1 for x³.

Why this formula

Always include C for indefinite integrals.

Exam tip

3x² → x³ not 3x³.

Common mistakes

  • x³−x²+x
  • 3x³−x²+x+C
Question 2 of 10trig

Find ∫ cos x dx.

∫cos x dx = sin x + C

1)  sin x + C.

Answer:  sin x + C

Formula used

∫cos x dx = sin x + C

Textbook formal language

Standard integral of cosine.

Easy language (same calculation)

Cos becomes sin.

Why this formula

Check by differentiating.

Exam tip

Not −sin x.

Common mistakes

  • −sin x + C
  • cos x + C
Question 3 of 10sub

Evaluate ∫ 2x e^(x²) dx.

u-sub

1)  Let u=x², du=2x dx.

2)  ∫ e^u du = e^u + C = e^(x²) + C.

Answer:  e^(x²) + C

Formula used

u-sub

Textbook formal language

Substitution matches the chain rule reverse.

Easy language (same calculation)

2x is exactly du.

Why this formula

Differentiate to verify: 2x e^(x²).

Exam tip

Do not answer 2x e^(x²).

Common mistakes

  • e^(x²)/2 + C
  • 2e^(x²)+C
Question 4 of 101/x

Integrate ∫ dx/(x+3).

∫ dx/x = ln|x|+C

1)  ln|x+3| + C.

Answer:  ln|x+3| + C

Formula used

∫ dx/x = ln|x|+C

Textbook formal language

Standard log form with linear argument.

Easy language (same calculation)

Same as 1/u du with u=x+3.

Why this formula

Absolute value for real log.

Exam tip

Not ln(x+3) without abs in many mark schemes, but both often accepted if domain clear.

Common mistakes

  • 1/(x+3)+C
  • ln|x|+3
Question 5 of 10by parts

Find ∫ x eˣ dx.

∫u dv=uv−∫v du

1)  u=x, dv=eˣ dx ⇒ du=dx, v=eˣ.

2)  xeˣ − ∫ eˣ dx = eˣ(x−1) + C.

Answer:  eˣ(x − 1) + C

Formula used

∫u dv=uv−∫v du

Textbook formal language

Integration by parts once.

Easy language (same calculation)

Parts: let u be algebraic, dv exponential.

Why this formula

Factor eˣ.

Exam tip

Sign: minus the remaining integral.

Common mistakes

  • xeˣ + C
  • eˣ(x+1)+C
Question 6 of 10sec²

∫ sec² x dx = ?

∫sec² x dx = tan x + C

1)  tan x + C.

Answer:  tan x + C

Formula used

∫sec² x dx = tan x + C

Textbook formal language

Standard integral inverse to derivative of tan.

Easy language (same calculation)

sec² becomes tan.

Why this formula

Differentiate tan to check.

Exam tip

Not sec x tan x (that is derivative of sec).

Common mistakes

  • sec x + C
  • sec x tan x + C
Question 7 of 10Power integrate

Integrate ∫ (4x³ − 1) dx.

∫xⁿ dx = x^{n+1}/(n+1)+C

1)  ∫ = x⁴ − x + C.

Answer:  x⁴ − x + C

Formula used

∫xⁿ dx = x^{n+1}/(n+1)+C

Textbook formal language

Antiderivative term by term.

Easy language (same calculation)

Raise powers; constant −x.

Why this formula

Always +C for indefinite integrals.

Exam tip

Check each algebraic step carefully.

Common mistakes

  • 4x⁴
  • Missing +C
Question 8 of 10eˣ

Find ∫ e^{2x} dx.

∫eˣ dx = eˣ + C

1)  Let u=2x, du=2 dx ⇒ (1/2)∫ e^u du = (1/2)e^{2x} + C.

Answer:  (1/2) e^{2x} + C

Formula used

∫eˣ dx = eˣ + C

Textbook formal language

Substitution / reverse chain rule.

Easy language (same calculation)

Divide by the inner derivative 2.

Why this formula

Check by differentiating.

Exam tip

Check each algebraic step carefully.

Common mistakes

  • e^{2x}
  • 2e^{2x}
Question 9 of 101/x

Integrate ∫ dx/(x+3).

∫ dx/x = ln|x|+C

1)  ln|x+3| + C.

Answer:  ln|x + 3| + C

Formula used

∫ dx/x = ln|x|+C

Textbook formal language

Standard log form with linear argument.

Easy language (same calculation)

Shift of 1/x.

Why this formula

Absolute value in the log.

Exam tip

Check each algebraic step carefully.

Common mistakes

  • 1/(x+3)
  • ln(x+3) without abs as only form
Question 10 of 10Parts

Evaluate ∫ x eˣ dx.

∫u dv = uv − ∫v du

1)  u=x, dv=eˣ dx ⇒ du=dx, v=eˣ.

2)  ∫ = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x−1)+C.

Answer:  eˣ(x − 1) + C

Formula used

∫u dv = uv − ∫v du

Textbook formal language

Integration by parts.

Easy language (same calculation)

Let u be the polynomial factor.

Why this formula

Factor eˣ at the end.

Exam tip

Check each algebraic step carefully.

Common mistakes

  • x eˣ only
  • eˣ