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Integrals

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Chapter Notes

Overview
This chapter introduces Integral Calculus as the inverse process of differentiation, motivated by the problem of finding a function when its derivative is given and calculating the area under curves. The one big idea is that integration reverses differentiation, and the indefinite integral represents a family of functions differing by a constant. Students must master the standard integrals, methods of substitution, integration using trigonometric identities, integration by partial fractions, and integration by parts to solve a wide variety of problems.
Key Concepts (14)
Integration
The process of finding an anti-derivative or integral of a function.
e.g. If we know the instantaneous velocity of an object at any instant, integration helps determine its position at any instant.
Anti-derivative (Primitive)
A function F such that F'(x) = f(x) for all x in an interval I.
e.g. sin x is an anti-derivative of cos x because d/dx (sin x) = cos x.
Indefinite Integral
The family of all anti-derivatives of a function f, denoted by ∫ f(x) dx = F(x) + C, where C is the constant of integration.
e.g. ∫ cos x dx = sin x + C.
Constant of Integration
An arbitrary real number C added to an indefinite integral because functions with the same derivative differ by a constant.
e.g. In ∫ 2x dx = x² + C, C is the constant of integration.
Integrand
The function f(x) in the integral ∫ f(x) dx.
e.g. In ∫ x² dx, x² is the integrand.
Variable of Integration
The variable with respect to which integration is performed, denoted by x in ∫ f(x) dx.
e.g. In ∫ t² dt, t is the variable of integration.
Integration by Substitution
A method of integration where the independent variable x is changed to t by substituting x = g(t), transforming the integral into a standard form.
e.g. ∫ 2x sin(x²+1) dx, put x²+1 = t, then 2x dx = dt, giving ∫ sin t dt = -cos t + C = -cos(x²+1) + C.
Integration using Trigonometric Identities
Using known trigonometric identities to rewrite the integrand into a form that can be integrated using standard formulae.
e.g. ∫ cos²x dx = ∫ (1+cos2x)/2 dx = x/2 + sin2x/4 + C.
Integration by Partial Fractions
A technique to integrate rational functions by expressing them as a sum of simpler fractions.
e.g. ∫ 1/(x² - a²) dx = 1/(2a) log |(x-a)/(x+a)| + C.
Proper Rational Function
A rational function P(x)/Q(x) where the degree of P(x) is less than the degree of Q(x).
e.g. 1/(x²+1) is a proper rational function.
Improper Rational Function
A rational function P(x)/Q(x) where the degree of P(x) is greater than or equal to the degree of Q(x).
e.g. (x³+1)/(x²+1) is an improper rational function.
Integration by Parts
A method of integration based on the product rule of differentiation, given by ∫ u v dx = u ∫ v dx - ∫ (u' ∫ v dx) dx.
e.g. ∫ x eˣ dx, take u = x, v = eˣ, gives x eˣ - eˣ + C.
Definite Integral
An integral evaluated over a specific interval [a, b], denoted by ∫ₐᵇ f(x) dx, representing the area under the curve y = f(x) from x = a to x = b.
e.g. ∫₀¹ x² dx = [x³/3]₀¹ = 1/3.
Fundamental Theorem of Calculus
The theorem that connects indefinite and definite integrals, making the definite integral a practical tool for science and engineering.
e.g. If F'(x) = f(x), then ∫ₐᵇ f(x) dx = F(b) - F(a).
Formulas & Equations (19)
∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1
Reads as: Integral of x to the power n dx equals x to the power n plus 1 divided by n plus 1 plus constant C, for n not equal to minus 1.
Variables: n: exponent (real number, n ≠ -1); x: variable of integration; C: constant of integration.
Use when: Integrating any power function except x⁻¹.
⚠ Forgetting the condition n ≠ -1 and applying the formula for ∫ 1/x dx, which gives log|x| + C.
∫ 1/x dx = log|x| + C
Reads as: Integral of 1 by x dx equals natural log of absolute value of x plus constant C.
Variables: x: variable of integration (x ≠ 0); C: constant of integration.
Use when: Integrating the reciprocal function.
⚠ Writing log x without absolute value signs, or forgetting the constant of integration.
∫ eˣ dx = eˣ + C
Reads as: Integral of e to the power x dx equals e to the power x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating the exponential function eˣ.
⚠ Writing eˣ/1 or adding an extra factor.
∫ aˣ dx = aˣ/log a + C, a > 0, a ≠ 1
Reads as: Integral of a to the power x dx equals a to the power x divided by natural log of a plus constant C, for a greater than 0 and a not equal to 1.
Variables: a: base (positive constant, a ≠ 1); x: variable of integration; C: constant of integration.
Use when: Integrating any exponential function with base a.
⚠ Forgetting to divide by log a, or using log₁₀ instead of natural log.
∫ sin x dx = -cos x + C
Reads as: Integral of sin x dx equals minus cos x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating the sine function.
⚠ Writing cos x instead of -cos x.
∫ cos x dx = sin x + C
Reads as: Integral of cos x dx equals sin x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating the cosine function.
⚠ Writing -sin x instead of sin x.
∫ sec² x dx = tan x + C
Reads as: Integral of secant squared x dx equals tan x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating sec² x.
⚠ Writing sec x tan x instead of tan x.
∫ cosec² x dx = -cot x + C
Reads as: Integral of cosecant squared x dx equals minus cot x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating cosec² x.
⚠ Writing cot x instead of -cot x.
∫ sec x tan x dx = sec x + C
Reads as: Integral of secant x times tan x dx equals secant x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating sec x tan x.
⚠ Writing tan x instead of sec x.
∫ cosec x cot x dx = -cosec x + C
Reads as: Integral of cosecant x times cot x dx equals minus cosecant x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating cosec x cot x.
⚠ Writing cosec x instead of -cosec x.
∫ 1/√(1-x²) dx = sin⁻¹ x + C
Reads as: Integral of 1 by square root of 1 minus x squared dx equals inverse sine of x plus constant C.
Variables: x: variable of integration (|x| < 1); C: constant of integration.
Use when: Integrating functions of the form 1/√(a² - x²).
⚠ Forgetting the absolute value or domain restriction.
∫ 1/(1+x²) dx = tan⁻¹ x + C
Reads as: Integral of 1 by 1 plus x squared dx equals inverse tangent of x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating functions of the form 1/(a² + x²).
⚠ Writing cot⁻¹ x instead of tan⁻¹ x.
∫ 1/√(x² ± a²) dx = log|x + √(x² ± a²)| + C
Reads as: Integral of 1 by square root of x squared plus or minus a squared dx equals natural log of absolute value of x plus square root of x squared plus or minus a squared plus constant C.
Variables: a: positive constant; x: variable of integration; C: constant of integration.
Use when: Integrating functions of the form 1/√(x² ± a²).
⚠ Forgetting the absolute value or the plus sign inside the log.
∫ 1/(x² - a²) dx = 1/(2a) log|(x-a)/(x+a)| + C
Reads as: Integral of 1 by x squared minus a squared dx equals 1 by 2a times natural log of absolute value of x minus a over x plus a plus constant C.
Variables: a: positive constant; x: variable of integration (|x| > a); C: constant of integration.
Use when: Integrating rational functions of the form 1/(x² - a²).
⚠ Forgetting the factor 1/(2a) or the absolute value signs.
∫ 1/(a² - x²) dx = 1/(2a) log|(a+x)/(a-x)| + C
Reads as: Integral of 1 by a squared minus x squared dx equals 1 by 2a times natural log of absolute value of a plus x over a minus x plus constant C.
Variables: a: positive constant; x: variable of integration (|x| < a); C: constant of integration.
Use when: Integrating rational functions of the form 1/(a² - x²).
⚠ Confusing with the formula for 1/(x² - a²) and getting the sign wrong.
∫ tan x dx = log|sec x| + C
Reads as: Integral of tan x dx equals natural log of absolute value of secant x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating the tangent function.
⚠ Writing log|sin x| instead of log|sec x|.
∫ cot x dx = log|sin x| + C
Reads as: Integral of cot x dx equals natural log of absolute value of sine x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating the cotangent function.
⚠ Writing log|cos x| instead of log|sin x|.
∫ sec x dx = log|sec x + tan x| + C
Reads as: Integral of secant x dx equals natural log of absolute value of secant x plus tan x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating the secant function.
⚠ Forgetting the absolute value or writing log|sec x| alone.
∫ cosec x dx = log|cosec x - cot x| + C
Reads as: Integral of cosecant x dx equals natural log of absolute value of cosecant x minus cot x plus constant C.
Variables: x: variable of integration; C: constant of integration.
Use when: Integrating the cosecant function.
⚠ Writing log|cosec x + cot x| instead of minus.
How It Works
Integration as Inverse Process of Differentiation
  1. Step 1: Identify the given function f(x) as the derivative of some known function F(x).
  2. Step 2: Write the integral ∫ f(x) dx = F(x) + C, where C is an arbitrary constant.
  3. Step 3: Verify by differentiating F(x) + C to get back f(x).
💡 Functions with the same derivative differ only by a constant, so always add +C for indefinite integrals.
Integration by Substitution
  1. Step 1: Choose a substitution x = g(t) or t = h(x) such that the integrand simplifies.
  2. Step 2: Compute dx = g'(t) dt or dt = h'(x) dx.
  3. Step 3: Rewrite the entire integral in terms of t.
  4. Step 4: Integrate using standard formulae.
  5. Step 5: Substitute back t = h(x) to express the result in terms of x.
💡 Always include the derivative of the substitution; a common mistake is forgetting to change dx correctly.
Integration using Trigonometric Identities
  1. Step 1: Identify the trigonometric function in the integrand.
  2. Step 2: Use appropriate identities (e.g., cos²x = (1+cos2x)/2, sin³x = (3sinx - sin3x)/4) to rewrite the integrand.
  3. Step 3: Integrate term by term using standard integrals.
  4. Step 4: Add the constant of integration.
💡 Double-angle and triple-angle identities are frequently used; practice rewriting powers of sine and cosine.
Integration by Partial Fractions
  1. Step 1: Ensure the rational function is proper; if improper, perform long division first.
  2. Step 2: Factor the denominator Q(x) into linear and irreducible quadratic factors.
  3. Step 3: Write the partial fraction decomposition based on the factors (e.g., for distinct linear factors: A/(x-a) + B/(x-b)).
  4. Step 4: Solve for the constants A, B, etc., by equating coefficients or substituting suitable x values.
  5. Step 5: Integrate each term separately using standard integrals (e.g., ∫ 1/(x-a) dx = log|x-a| + C).
💡 For repeated linear factors, include terms for each power; for irreducible quadratics, use linear numerators.
Integration by Parts
  1. Step 1: Identify the integrand as a product of two functions u and v.
  2. Step 2: Choose u as the function that simplifies when differentiated (LIATE rule: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential).
  3. Step 3: Apply the formula ∫ u v dx = u ∫ v dx - ∫ (u' ∫ v dx) dx.
  4. Step 4: Compute the integrals step by step.
  5. Step 5: Add the constant of integration.
💡 The choice of u and v is critical; a wrong choice can make the integral more complicated.
Evaluating Integrals of the Form ∫ dx/(ax²+bx+c)
  1. Step 1: Complete the square in the denominator: ax²+bx+c = a[(x + b/(2a))² + (c/a - b²/(4a²))].
  2. Step 2: Substitute t = x + b/(2a) to get the form ∫ dt/(a(t² ± k²)).
  3. Step 3: Use standard formulae for ∫ dt/(t² ± a²) or ∫ dt/(a² - t²).
  4. Step 4: Substitute back to x.
💡 Always factor out the coefficient a before completing the square.
Evaluating Integrals of the Form ∫ (px+q)/√(ax²+bx+c) dx
  1. Step 1: Write px + q = A d/dx (ax²+bx+c) + B = A(2ax+b) + B.
  2. Step 2: Equate coefficients to find A and B.
  3. Step 3: Split the integral into A∫ (2ax+b)/√(ax²+bx+c) dx + B∫ dx/√(ax²+bx+c).
  4. Step 4: The first integral is solved by substitution (let t = ax²+bx+c).
  5. Step 5: The second integral is solved by completing the square and using standard forms.
💡 The derivative of the quadratic appears in the numerator; this method is standard for such forms.
Exam Traps — Don't Make These Mistakes
⚠ Forgetting the Constant of Integration
✗ ∫ 2x dx = x²
→ The indefinite integral represents a family of functions; without +C, the answer is incomplete.
✓ ∫ 2x dx = x² + C
⚠ Misapplying the Power Rule for n = -1
✗ ∫ 1/x dx = x⁰/0 + C (or some incorrect expression)
→ The power rule ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C is valid only for n ≠ -1; for n = -1, the integral is log|x| + C.
✓ ∫ 1/x dx = log|x| + C
⚠ Incorrect Sign in Trigonometric Integrals
✗ ∫ sin x dx = cos x + C
→ The derivative of cos x is -sin x, so the anti-derivative of sin x is -cos x.
✓ ∫ sin x dx = -cos x + C
⚠ Forgetting Absolute Values in Logarithmic Integrals
✗ ∫ 1/x dx = log x + C
→ The domain of 1/x includes negative x, so the anti-derivative must be log|x| to be valid for all x ≠ 0.
✓ ∫ 1/x dx = log|x| + C
⚠ Wrong Substitution in Integration by Substitution
✗ For ∫ 2x sin(x²+1) dx, they write: let t = x², then dt = 2x dx, giving ∫ sin(t+1) dt = -cos(t+1) + C = -cos(x²+1) + C (this is actually correct, but a common trap is choosing t = x²+1 and forgetting to adjust dx)
→ If a student writes let t = x²+1 but then writes dt = 2x (instead of 2x dx), they will get the wrong answer.
✓ Let t = x²+1, then dt = 2x dx, so the integral becomes ∫ sin t dt = -cos t + C = -cos(x²+1) + C.
⚠ Incorrect Partial Fraction Decomposition for Repeated Roots
✗ For ∫ 1/(x-1)² dx, they write 1/(x-1)² = A/(x-1) + B/(x-1) (same denominator)
→ For repeated linear factors, the decomposition must include terms with increasing powers: A/(x-1) + B/(x-1)².
✓ 1/(x-1)² = A/(x-1) + B/(x-1)²
Compare & Contrast
Presence of limits
Indefinite Integral
Definite Integral
Indefinite integral has no limits and includes an arbitrary constant C; definite integral is evaluated over a specific interval [a, b] and gives a numerical value (area).
Method of transformation
Integration by Substitution
Integration by Parts
Substitution changes the variable to simplify the integrand, often used when the derivative of a function appears; parts splits the product of two functions using the formula ∫ u dv = uv - ∫ v du, used when the integrand is a product of two different types of functions.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 Very Short Answer (2 marks), 1 Short Answer (3 marks), 1 Long Answer (5 marks) — total approximately 12-15 marks.
Most asked: Integration by substitution · Integration using trigonometric identities · Integration by partial fractions · Integration by parts · Definite integrals and their properties · Fundamental Theorem of Calculus
Don't skip: Integration by parts with logarithmic or inverse trigonometric functions; Integration by partial fractions with quadratic denominators; Definite integral evaluation using substitution
One-Line Revision (read aloud before the exam)
  1. Integration is the inverse process of differentiation.
  2. The indefinite integral ∫ f(x) dx = F(x) + C, where F'(x) = f(x).
  3. Functions with the same derivative differ by a constant.
  4. ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ -1.
  5. ∫ 1/x dx = log|x| + C.
  6. ∫ eˣ dx = eˣ + C.
  7. ∫ sin x dx = -cos x + C.
  8. ∫ cos x dx = sin x + C.
  9. ∫ sec² x dx = tan x + C.
  10. ∫ cosec² x dx = -cot x + C.
  11. ∫ sec x tan x dx = sec x + C.
  12. ∫ cosec x cot x dx = -cosec x + C.
  13. ∫ 1/√(1-x²) dx = sin⁻¹ x + C.
  14. ∫ 1/(1+x²) dx = tan⁻¹ x + C.
  15. ∫ tan x dx = log|sec x| + C.
  16. ∫ cot x dx = log|sin x| + C.
  17. ∫ sec x dx = log|sec x + tan x| + C.
  18. ∫ cosec x dx = log|cosec x - cot x| + C.
  19. Integration by substitution: let x = g(t), then dx = g'(t) dt.
  20. Integration by parts: ∫ u v dx = u ∫ v dx - ∫ (u' ∫ v dx) dx.
  21. Partial fractions decompose proper rational functions into simpler fractions.
  22. The Fundamental Theorem of Calculus connects indefinite and definite integrals.
  23. Always add the constant of integration C for indefinite integrals.
  24. For definite integrals, ∫ₐᵇ f(x) dx = F(b) - F(a), where F'(x) = f(x).

MCQs — 10 Questions with Answers

Q1. If the derivative of F(x) is f(x), which statement defines the indefinite integral ∫f(x)dx?

  • A. ∫f(x)dx = F(x) + C, where C is an arbitrary constant ✓
  • B. ∫f(x)dx = F(x), where C must be zero
  • C. ∫f(x)dx = F'(x) + C
  • D. ∫f(x)dx represents only one specific antiderivative of f

Answer: A — By definition, the indefinite integral gives the family of all antiderivatives, differing by arbitrary constant C.

Q2. Evaluate ∫(5x⁴ + 3x² - 2)dx.

  • A. x⁵ + x³ - 2x + C ✓
  • B. 5x⁵ + 3x³ - 2x + C
  • C. x⁵ + x³ - 2 + C
  • D. 5x⁵/5 + 3x³/3 - 2x + C = x⁵ + x³ - 2x + C

Answer: A — Apply power rule term-by-term: ∫5x⁴dx = 5·x⁵/5 = x⁵; ∫3x²dx = 3·x³/3 = x³; ∫-2dx = -2x; add C.

Q3. Which of the following is correct regarding antiderivatives of the same function?

  • A. All antiderivatives of f(x) are identical
  • B. Any two antiderivatives of f(x) differ by a constant ✓
  • C. Antiderivatives differ by trigonometric functions
  • D. There is only one antiderivative for each function

Answer: B — If F and G both satisfy F'(x) = G'(x) = f(x), then (F-G)' = 0, so F(x) - G(x) = constant.

Q4. What is ∫(1/x)dx for x > 0?

  • A. -1/x² + C
  • B. log(x) + C
  • C. log|x| + C ✓
  • D. 1/x + C

Answer: C — The antiderivative of 1/x is log|x| + C; absolute value is used because log is defined for negative x too.

Q5. If F(x) = x³ + 2x + 5, what is ∫f(x)dx where f(x) = F'(x)?

  • A. x³ + 2x + 5
  • B. x³ + 2x + 5 + C ✓
  • C. 3x² + 2 + C
  • D. 3x² + 2

Answer: B — F'(x) = 3x² + 2, so ∫(3x² + 2)dx = x³ + 2x + C; the original function F plus constant C.

Q6. Which property correctly states the relationship between differentiation and integration?

  • A. d/dx[∫f(x)dx] = ∫f(x) (without C)
  • B. d/dx[∫f(x)dx] = f(x) ✓
  • C. ∫[d/dx(f(x))]dx = f(x) (without C)
  • D. d/dx[∫f(x)dx] = f'(x)

Answer: B — Differentiation of an indefinite integral recovers the original integrand f(x); the constant C differentiates to zero.

Q7. When evaluating ∫x⁻¹dx, why cannot we use the power rule ∫xⁿdx = xⁿ⁺¹/(n+1) + C?

  • A. Because x⁻¹ is negative
  • B. Because n = -1 makes the denominator (n+1) = 0, causing division by zero ✓
  • C. Because negative powers always need special handling
  • D. Because ∫x⁻¹dx has no antiderivative

Answer: B — The power rule fails when n = -1 since xⁿ⁺¹/(n+1) = x⁰/0, which is undefined; instead, ∫x⁻¹dx = log|x| + C.

Q8. Consider two statements: (I) Every continuous function has an antiderivative. (II) The constant of integration C represents infinitely many antiderivatives. Which is correct?

  • A. Both (I) and (II) are correct ✓
  • B. Only (I) is correct
  • C. Only (II) is correct
  • D. Neither (I) nor (II) is correct

Answer: A — Both statements are true: (I) is a theorem in calculus; (II) follows because differentiating constant gives zero, allowing C to vary.

Q9. If ∫f(x)dx = x⁴/4 - x²/2 + 7, what is f(x)?

  • A. f(x) = x³ - x ✓
  • B. f(x) = x⁴/4 - x²/2 + 7
  • C. f(x) = x⁵/5 - x³/3 + 7x
  • D. f(x) = x³ - x + 7

Answer: A — Differentiate: d/dx[x⁴/4 - x²/2 + 7] = 4x³/4 - 2x/2 + 0 = x³ - x; constant 7 differentiates to zero.

Q10. A common mistake: Student A writes ∫sin(x)dx = cos(x) + C. Is this correct? If not, what is the error?

  • A. Correct; no error
  • B. Incorrect; missing negative sign; should be -cos(x) + C ✓
  • C. Incorrect; should be sin(x) because integration undoes differentiation
  • D. Incorrect; should include d/dx notation

Answer: B — Since d/dx[-cos(x)] = sin(x), the antiderivative of sin(x) is -cos(x) + C, not cos(x) + C; neglecting the negative is a critical exam error.

Flashcards

What is an antiderivative (primitive) of a function?

A function F(x) such that F'(x) = f(x), i.e., when differentiated it gives the original function f(x).

Why does the constant of integration C appear in indefinite integrals?

Because the derivative of any constant is zero, so infinitely many functions differing only by a constant have the same derivative.

State the Power Rule for Integration: ∫xⁿdx = ?

∫xⁿdx = xⁿ⁺¹/(n+1) + C for n ≠ -1; when n = -1, use ∫(1/x)dx = log|x| + C.

What is ∫cos(x)dx and ∫sin(x)dx?

∫cos(x)dx = sin(x) + C and ∫sin(x)dx = -cos(x) + C.

State the fundamental property: d/dx[∫f(x)dx] = ?

d/dx[∫f(x)dx] = f(x), showing differentiation and integration are inverse processes.

Find ∫eˣdx.

∫eˣdx = eˣ + C.

What is ∫(1/x)dx for x ≠ 0?

∫(1/x)dx = log|x| + C; absolute value is essential because log is undefined for negative arguments.

Evaluate ∫(3x² + 2x + 5)dx.

∫(3x² + 2x + 5)dx = 3·x³/3 + 2·x²/2 + 5x + C = x³ + x² + 5x + C.

If F and G both have the same derivative on interval I, how do they relate?

F(x) - G(x) = C (constant), meaning any two antiderivatives of the same function differ by only a constant.

Why do we write ∫f(x)dx = F(x) + C instead of just F(x)?

The '+ C' represents the entire family of antiderivatives; omitting it means missing infinitely many valid solutions.

Important Board Questions

Define antiderivative and indefinite integral. If F'(x) = f(x), write the relationship between F and the indefinite integral ∫f(x)dx. [2 marks]

State that antiderivative is function whose derivative gives f(x); write ∫f(x)dx = F(x) + C showing why arbitrary constant C appears (derivative of constant is zero).

Prove that if two functions F(x) and G(x) have the same derivative on an interval I, then F(x) - G(x) = C, where C is a constant. Use this result to justify why the indefinite integral includes an arbitrary constant. [5 marks]

Define h(x) = F(x) - G(x); show h'(x) = F'(x) - G'(x) = 0 on I, implying h is constant; conclude that any two antiderivatives differ by constant only, so indefinite integral is entire family F(x) + C.

Using standard integration formulas and the concept of antiderivatives, evaluate ∫(6x⁵ + 4cos(x) - 3/x + eˣ)dx. Show all working steps and verify your answer by differentiation. [6 marks]

Apply power rule to 6x⁵, use ∫cos(x)dx = sin(x) + C, use ∫(1/x)dx = log|x| + C, and ∫eˣdx = eˣ + C; combine to get result, then differentiate to confirm it equals the original integrand.

Assertion-Reason Questions for Integrals

Q1. Assertion (A):

The integral ∫f(x) dx represents the family of all antiderivatives of f(x).

Reason (R):

Two functions with the same derivative differ by a constant.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The reason correctly explains why ∫f(x) dx = F(x) + C: since antiderivatives differ only by a constant, the integral represents an entire family of curves parameterized by C.

Q2. Assertion (A):

If d/dx[F(x)] = f(x), then ∫f(x) dx = F(x) + C.

Reason (R):

Integration is the inverse process of differentiation, so the constant C arises because differentiation of any constant is zero.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. The reason correctly justifies why C appears: d/dx[C] = 0 for any constant, so F(x) and F(x) + C have the same derivative, making them both valid antiderivatives of f(x).

Q3. Assertion (A):

The indefinite integral ∫sin x dx = -cos x + C is correct.

Reason (R):

The derivative of -cos x with respect to x is sin x.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason is the correct explanation. We verify by differentiation: d/dx[-cos x] = sin x, confirming that -cos x is an antiderivative of sin x.

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Case-Based Questions (CBQ) for Integrals

Case Study:

An environmental scientist studies the rate of forest deforestation in a region. She models the forest area A(t) (in square kilometers) at time t (in years) using the deforestation rate dA/dt = -50 + 8t, where the initial forest area is A(0) = 2000 km². The negative rate indicates loss, while the positive term 8t suggests environmental recovery efforts reducing deforestation over time. The scientist wants to predict the forest area after 5 years and determine when the forest area reaches a minimum.

Q1. What is the forest area A(t) as a function of time?

  • A. A(t) = -50t + 8t² + 2000
  • B. A(t) = -50t + 4t² + 2000 ✓
  • C. A(t) = -50 + 8t + 2000
  • D. A(t) = 2000 - 50t + 8t²

Answer: B

Q2. What is the forest area after 5 years?

  • A. 1650 km²
  • B. 1700 km²
  • C. 1750 km² ✓
  • D. 1550 km²

Answer: C

Q3. At what time does the forest area reach its minimum value?

  • A. t = 2.5 years
  • B. t = 3.125 years ✓
  • C. t = 6.25 years
  • D. t = 4 years

Answer: B

Q4. What is the minimum forest area?

  • A. 1896.875 km² ✓
  • B. 1890.625 km²
  • C. 1875 km²
  • D. 1903.125 km²

Answer: A

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Previous Year Questions (PYQ) for Integrals

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 1 mark

If ∫ (3ax)/(b² + c²x²) dx = A log |b² + c²x²| + K, then the value of A is

  • 3a
  • 3a/(2b²)
  • 3a/(b²c²)
  • 3a/(2c²)

CBSE 2026 · Set 65/1/1 · 1 mark

The value of ∫₋₁¹ x³/(x² + 2|x| + 1) dx is

  • 0
  • log 2
  • 2 log 2
  • (1/2) log 2

CBSE 2026 · Set 65/1/1 · 3 marks

Evaluate: ∫₀¹ x tan⁻¹x dx.

CBSE 2026 · Set 65/1/1 · 3 marks

Find ∫ (x + 2)/(x – 2) dx.

CBSE 2026 · Set 65/1/1 · 3 marks

If I₁ = ∫₍₋π/₄₎^(π/4) 1/(1 + cos 2x) dx and I₂ = ∫₍₋₁/₂₎^(1/2) |x| dx, then show that I₁ – 4I₂ = 0.

CBSE 2026 · Set 65/1/2 · 1 mark

If ∫(3ax)/(b² + c²x²) dx = A log |b² + c²x²| + K, then the value of A is

  • 3a
  • 3a/(2b²)
  • 3a/(b²c²)
  • 3a/(2c²)

CBSE 2026 · Set 65/1/2 · 1 mark

The value of ∫₋₁¹ (x³)/(x² + 2|x| + 1) dx is

  • 0
  • log 2
  • 2 log 2
  • (1/2) log 2

CBSE 2026 · Set 65/1/2 · 3 marks

If I₁ = ∫₋π/₄^(π/4) dx/(1 + cos 2x) and I₂ = ∫₋₁/₂^(1/2) |x| dx, then show that I₁ – 4I₂ = 0.

CBSE 2026 · Set 65/1/2 · 3 marks

Evaluate: ∫₀¹ x sin⁻¹x dx

CBSE 2026 · Set 65/1/2 · 3 marks

Find: ∫(x + 2)/(x - 2) dx

CBSE 2026 · Set 65/1/3 · 3 marks

Evaluate: ∫₀¹ log(1 + x²) dx

CBSE 2026 · Set 65/1/3 · 3 marks

Find ∫(x + 2)/(x - 2) dx

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Class 12 Maths Formula Sheet

Every formula from all 13 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Mathematics Previous Year Questions, Chapter-wise

All 1048 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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