Integrals

CBSE Class 12 Maths · 11 questions · 26 marks

Integration reverses differentiation: given a function, this chapter finds every function whose derivative it is. Substitution, integration by parts and partial fractions extend the small set of standard integrals to a huge range of expressions, while the Fundamental Theorem of Calculus turns definite integrals into a simple subtraction, F(b) − F(a). Clever use of symmetry and the property ∫f(x)dx = ∫f(a+b−x)dx often shortcuts what would otherwise be a long computation.

Quick revision

∫xⁿdx = xⁿ⁺¹/(n+1) + C for n≠−1; ∫(1/x)dx = ln|x| + C; ∫eˣdx = eˣ + C; ∫cos x dx = sin x + C; ∫sin x dx = −cos x + C; ∫sec²x dx = tan x + C.
∫1/(x²+a²)dx = (1/a)tan⁻¹(x/a) + C; ∫1/√(a²−x²)dx = sin⁻¹(x/a) + C.
Integration by substitution: if u=g(x), then ∫f(g(x))g'(x)dx = ∫f(u)du, converting the integral into a simpler variable.
Integration by parts: ∫u dv = uv − ∫v du; the ILATE order (Inverse, Log, Algebraic, Trig, Exponential) helps choose which factor is u.
Rational functions are integrated by first splitting them into partial fractions, then integrating each simple fraction separately.
The Fundamental Theorem of Calculus: if F'(x)=f(x), then ∫[a,b]f(x)dx = F(b) − F(a).
Property: ∫[a,b]f(x)dx = ∫[a,b]f(a+b−x)dx; this symmetry often simplifies definite integrals involving trig functions.
If f is even, ∫[−a,a]f(x)dx = 2∫[0,a]f(x)dx; if f is odd, ∫[−a,a]f(x)dx = 0.
Every indefinite integral carries an arbitrary constant C, since functions differing by a constant have the same derivative.

Important questions

1
MCQ1 mark
The indefinite integral ∫x³ dx equals:
1. A.x⁴/4 + C
2. B.x²/2 + C
3. C.4x³ + C
4. D.3x² + C
Answer
∫x³ dx = x⁴/4 + C. Using the power rule ∫xⁿdx = xⁿ⁺¹/(n+1) + C with n=3, we get x⁴/4 + C.
2
MCQ1 mark
The indefinite integral ∫sec²x dx equals:
1. A.tan x + C
2. B.sec x + C
3. C.-cot x + C
4. D.cot x + C
Answer
∫sec²x dx = tan x + C. This is a standard integral, the reverse of the known derivative d/dx(tan x) = sec²x.
3
MCQ1 mark
The value of ∫[0,π/2] sin x dx is:
1. A.1
2. B.0
3. C.-1
4. D.π/2
Answer
The value is 1. ∫sin x dx = -cos x + C, so ∫[0,π/2] sin x dx = [-cos x] from 0 to π/2 = -cos(π/2) - (-cos(0)) = 0 + 1 = 1.
4
MCQ1 mark
The indefinite integral ∫1/(1+x²) dx equals:
1. A.tan⁻¹x + C
2. B.sin⁻¹x + C
3. C.cot⁻¹x + C
4. D.log(1+x²) + C
Answer
∫1/(1+x²) dx = tan⁻¹x + C. This is a standard integral, the reverse of the derivative d/dx(tan⁻¹x) = 1/(1+x²).
5
Assertion & Reason1 mark
Assertion (A): ∫[-2,2] x³ dx = 0. Reason (R): If f is an odd function, then ∫[-a,a] f(x) dx = 0.
1. A.Both A and R are true and R is the correct explanation of A
2. B.Both A and R are true but R is not the correct explanation of A
3. C.A is true but R is false
4. D.A is false but R is true
Answer
Both A and R are true and R is the correct explanation of A. Reason R is the standard symmetry property for odd functions integrated over a symmetric interval. For Assertion A, f(x) = x³ satisfies f(-x) = -x³ = -f(x), so f is odd, and by the property in R, ∫[-2,2] x³ dx = 0. This confirms A using exactly the reasoning in R.
6
Short answer2 marks
Evaluate ∫ x eˣ dx using integration by parts.
7
Short answer2 marks
Evaluate ∫ 2x cos(x² + 1) dx.
8
Short answer3 marks
Evaluate ∫ 1/(x²-1) dx using partial fractions.
9
Long answer5 marks
Evaluate ∫ x² eˣ dx using integration by parts.
10
Long answer5 marks
Evaluate ∫[0,π] x sin x/(1+cos²x) dx.
11
Case study4 marks
Read the following and answer the questions that follow: The velocity of a particle moving along a straight line is v(t) = 3t² + 2t metres per second, for t ≥ 0, and its initial displacement is s(0) = 0. (a) Find the displacement function s(t). (b) Write the definite integral that gives the displacement of the particle from t=1 to t=3. (c) Evaluate this definite integral using the fundamental theorem of calculus. (d) Verify that s(3) - s(1) matches your answer in (c).

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Integrals

CBSE Class 12 Maths · 11 questions · 26 marks

Quick revision

∫xⁿdx = xⁿ⁺¹/(n+1) + C for n≠−1; ∫(1/x)dx = ln|x| + C; ∫eˣdx = eˣ + C; ∫cos x dx = sin x + C; ∫sin x dx = −cos x + C; ∫sec²x dx = tan x + C.

∫1/(x²+a²)dx = (1/a)tan⁻¹(x/a) + C; ∫1/√(a²−x²)dx = sin⁻¹(x/a) + C.

Integration by substitution: if u=g(x), then ∫f(g(x))g'(x)dx = ∫f(u)du, converting the integral into a simpler variable.

Integration by parts: ∫u dv = uv − ∫v du; the ILATE order (Inverse, Log, Algebraic, Trig, Exponential) helps choose which factor is u.

Rational functions are integrated by first splitting them into partial fractions, then integrating each simple fraction separately.

The Fundamental Theorem of Calculus: if F'(x)=f(x), then ∫[a,b]f(x)dx = F(b) − F(a).

Property: ∫[a,b]f(x)dx = ∫[a,b]f(a+b−x)dx; this symmetry often simplifies definite integrals involving trig functions.

If f is even, ∫[−a,a]f(x)dx = 2∫[0,a]f(x)dx; if f is odd, ∫[−a,a]f(x)dx = 0.

Every indefinite integral carries an arbitrary constant C, since functions differing by a constant have the same derivative.

Important questions

MCQ1 mark

The indefinite integral ∫x³ dx equals:

A.x⁴/4 + C
B.x²/2 + C
C.4x³ + C
D.3x² + C

MCQ1 mark

The indefinite integral ∫sec²x dx equals:

A.tan x + C
B.sec x + C
C.-cot x + C
D.cot x + C

MCQ1 mark

The value of ∫[0,π/2] sin x dx is:

A.1
B.0
C.-1
D.π/2

MCQ1 mark

The indefinite integral ∫1/(1+x²) dx equals:

A.tan⁻¹x + C
B.sin⁻¹x + C
C.cot⁻¹x + C
D.log(1+x²) + C

Assertion & Reason1 mark

Assertion (A): ∫[-2,2] x³ dx = 0. Reason (R): If f is an odd function, then ∫[-a,a] f(x) dx = 0.

A.Both A and R are true and R is the correct explanation of A
B.Both A and R are true but R is not the correct explanation of A
C.A is true but R is false
D.A is false but R is true

Short answer2 marks

Evaluate ∫ x eˣ dx using integration by parts.

Short answer2 marks

Evaluate ∫ 2x cos(x² + 1) dx.

Short answer3 marks

Evaluate ∫ 1/(x²-1) dx using partial fractions.

Long answer5 marks

Evaluate ∫ x² eˣ dx using integration by parts.

Long answer5 marks

Evaluate ∫[0,π] x sin x/(1+cos²x) dx.

Case study4 marks

Read the following and answer the questions that follow: The velocity of a particle moving along a straight line is v(t) = 3t² + 2t metres per second, for t ≥ 0, and its initial displacement is s(0) = 0. (a) Find the displacement function s(t). (b) Write the definite integral that gives the displacement of the particle from t=1 to t=3. (c) Evaluate this definite integral using the fundamental theorem of calculus. (d) Verify that s(3) - s(1) matches your answer in (c).