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Integration: Class 12 Mathematics

Written by Hassan Raza (Mathematics)

Integration as the reverse of differentiation, standard integrals, substitution, integration by parts, partial fractions, definite integrals with properties, and the area under a curve.

Key points

Integration is the reverse of differentiation. If d/dx F(x) = f(x), then F(x) is an anti-derivative of f(x), and we write ∫ f(x) dx = F(x) + C. This is the indefinite integral. C is the constant of integration, because the derivative of any constant is zero.

Basic rules:

  • ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, for n ≠ −1.
  • ∫ 1/x dx = ln |x| + C.
  • ∫ k f(x) dx = k ∫ f(x) dx, where k is a constant.
  • ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.

Standard integrals (x in radians):

  • ∫ eˣ dx = eˣ + C and ∫ aˣ dx = aˣ/ln a + C.
  • ∫ sin x dx = −cos x + C and ∫ cos x dx = sin x + C.
  • ∫ sec² x dx = tan x + C and ∫ cosec² x dx = −cot x + C.
  • ∫ sec x tan x dx = sec x + C.
  • ∫ 1/(1 + x²) dx = tan⁻¹x + C.

Integration of a linear function inside: if ∫ f(x) dx = F(x) + C, then ∫ f(ax + b) dx = F(ax + b)/a + C. For example, ∫ cos(2x + 1) dx = sin(2x + 1)/2 + C.

Integration by substitution: choose a part of the integrand as u so that du appears too. Replace everything in terms of u, integrate, and then return to x. A very common form is ∫ f′(x)/f(x) dx = ln |f(x)| + C. For example, ∫ 2x/(x² + 1) dx = ln(x² + 1) + C, since the top is the derivative of the bottom.

Useful trigonometric identities for integration are sin² x = (1 − cos 2x)/2 and cos² x = (1 + cos 2x)/2. So ∫ sin² x dx = x/2 − (sin 2x)/4 + C.

Integration by parts is used for a product of two different kinds of function. The formula is ∫ u dv = uv − ∫ v du. Choose u in the order of ILATE: inverse trigonometric, logarithmic, algebraic, trigonometric, exponential. For ∫ x eˣ dx take u = x and dv = eˣ dx, so the answer is x eˣ − eˣ + C.

Partial fractions are used to integrate a rational function whose denominator can be factorised. For example, 1/[(x − 1)(x + 2)] = (1/3)[1/(x − 1) − 1/(x + 2)], so its integral is (1/3) ln |(x − 1)/(x + 2)| + C.

The definite integral from a to b is ∫ₐᵇ f(x) dx = F(b) − F(a), where F is any anti-derivative. There is no constant C because it cancels. This is the fundamental theorem of calculus.

Properties of definite integrals:

  • ∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx.
  • ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx for a point c between a and b.
  • ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx.
  • ∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dx if f is even, and 0 if f is odd.

Area under a curve: the area between y = f(x), the x-axis, and the lines x = a and x = b is ∫ₐᵇ f(x) dx when f(x) ≥ 0 on [a, b]. If the curve lies below the x-axis, the integral is negative, so take its absolute value for the area. The area between two curves is ∫ₐᵇ [upper − lower] dx. When substituting in a definite integral, change the limits to match the new variable.

Study tip: Learn the standard integrals as a table, and always differentiate your answer to check it, because this takes only a few seconds. For substitution, write the choice of u and du clearly, since marks are given for the method. In area questions, sketch the curve first, mark the limits, and give the answer in square units.

Definitions and flashcards

Say the meaning first, then open the card to check.

Anti-derivative

A function whose derivative is the given function.

Indefinite integral

The general anti-derivative ∫ f(x) dx = F(x) + C.

Constant of integration

The arbitrary constant C added to an indefinite integral.

Integrand

The function that is being integrated.

Definite integral

The number ∫ₐᵇ f(x) dx = F(b) − F(a), found between two limits.

Limits of integration

The numbers a and b at the bottom and top of the integral sign.

Substitution

A method that replaces part of the integrand with a new variable u.

Integration by parts

The method based on ∫ u dv = uv − ∫ v du.

Formula sheet

NameFormulaWhat it means
Power rule∫ xⁿ dx = xⁿ⁺¹/(n + 1) + CValid for n ≠ −1.
Reciprocal∫ 1/x dx = ln |x| + CThe case n = −1.
Log form∫ f′(x)/f(x) dx = ln |f(x)| + CTop is the derivative of the bottom.
By parts∫ u dv = uv − ∫ v duChoose u by ILATE.
Definite integral∫ₐᵇ f(x) dx = F(b) − F(a)F is an anti-derivative of f.
AreaA = ∫ₐᵇ f(x) dxFor f(x) ≥ 0 on [a, b]; result in square units.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Why do we add a constant C to an indefinite integral? (2 marks)

The derivative of a constant is zero, so many functions that differ only by a constant have the same derivative. The constant C represents this unknown constant.

State the rule for integrating xⁿ and the case where it fails. (2 marks)

∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C. It fails when n = −1, because the denominator would be zero. In that case ∫ 1/x dx = ln |x| + C.

Evaluate ∫ cos(3x) dx. (2 marks)

Since ∫ cos u du = sin u + C and the inside is 3x, divide by 3. So ∫ cos(3x) dx = (sin 3x)/3 + C.

State the formula for integration by parts and use it to find ∫ ln x dx. (5 marks)

The formula is ∫ u dv = uv − ∫ v du. Write ln x as ln x × 1. Take u = ln x and dv = dx, so du = dx/x and v = x. Then ∫ ln x dx = x ln x − ∫ x (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C.

Explain how to find the area under a curve using integration. (3 marks)

To find the area between y = f(x), the x-axis and the lines x = a and x = b, first make sure f(x) is not negative on the interval. Then the area is the definite integral ∫ₐᵇ f(x) dx = F(b) − F(a), where F is an anti-derivative of f. The answer is in square units. If part of the curve is below the x-axis, find that part separately and take its absolute value.

Evaluate ∫ 2x/(x² + 1) dx. (2 marks)

The top is the derivative of the bottom, so the integral is ln(x² + 1) + C. No modulus is needed because x² + 1 is always positive.

Numericals with worked steps

Evaluate ∫ (3x² + 2x − 5) dx.
  1. Integrate each term: ∫ 3x² dx = x³.
  2. ∫ 2x dx = x².
  3. ∫ 5 dx = 5x.
  4. Combine and add the constant.

Answer: x³ + x² − 5x + C

Evaluate ∫₀² (x² + 1) dx.
  1. An anti-derivative is F(x) = x³/3 + x.
  2. F(2) = 8/3 + 2 = 14/3.
  3. F(0) = 0.
  4. The integral is F(2) − F(0) = 14/3.

Answer: 14/3

Evaluate ∫ x eˣ dx.
  1. By ILATE, take u = x (algebraic) and dv = eˣ dx.
  2. Then du = dx and v = eˣ.
  3. ∫ u dv = uv − ∫ v du = x eˣ − ∫ eˣ dx.
  4. = x eˣ − eˣ + C.
  5. Check: d/dx (x eˣ − eˣ) = eˣ + x eˣ − eˣ = x eˣ.

Answer: (x − 1) eˣ + C

Evaluate ∫ x √(x² + 4) dx.
  1. Let u = x² + 4. Then du = 2x dx, so x dx = du/2.
  2. The integral becomes (1/2) ∫ u^(1/2) du.
  3. = (1/2) × u^(3/2)/(3/2) = (1/3) u^(3/2).
  4. Return to x: (1/3)(x² + 4)^(3/2) + C.
  5. Check: derivative = (1/3)(3/2)(x² + 4)^(1/2)(2x) = x √(x² + 4).

Answer: (1/3)(x² + 4)^(3/2) + C

Find the area between the curve y = x², the x-axis, and the lines x = 0 and x = 3.
  1. The curve is above the x-axis for x between 0 and 3.
  2. Area = ∫₀³ x² dx.
  3. An anti-derivative is x³/3.
  4. At x = 3: 27/3 = 9. At x = 0: 0.
  5. Area = 9 − 0 = 9.

Answer: 9 square units

Common mistakes

  • Forgetting the constant C in an indefinite integral, and wrongly adding C in a definite integral.
  • Using the power rule for ∫ 1/x dx, which gives division by zero. The answer is ln |x| + C.
  • Not dividing by the coefficient of x, for example writing ∫ e²ˣ dx = e²ˣ instead of e²ˣ/2.
  • Forgetting to change the limits when substituting in a definite integral, or to return to x in an indefinite one.
  • Taking the area as a negative number. Area is never negative, so use the absolute value for the part below the axis.

Practise this chapter

Practice questions and timed test Printable worksheet

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