Unit IV: Integral Calculus

Definition of Integration & Anti-Differentiation

Definition of Integration

Integration is a fundamental concept in calculus that deals with finding the accumulation of quantities, such as areas under a curve, volumes, or total change over time. Mathematically, it is the reverse process of differentiation.

Integration as Anti-Differentiation

An anti-derivative of a function f(x) is a differentiable function F(x) whose derivative is equal to the original function, meaning F'(x) = f(x). Because the derivative of a constant is zero, finding an anti-derivative yields a family of functions differing by a constant C.

Definition: Indefinite Integral
∫ f(x) dx = F(x) + C
Where f(x) is the integrand, dx indicates the variable of integration, F(x) is the anti-derivative, and C is the constant of integration.

Integration of Simple Functions

Simple functions follow standard integration formulas derived directly from known differentiation rules.

• Power Rule: ∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, (provided n ≠ -1)
• Reciprocal Rule: ∫ (1 / x) dx = ln |x| + C
• Exponential Rule (Base e): ∫ e^x dx = e^x + C
• Exponential Rule (Base a): ∫ a^x dx = a^x / ln(a) + C
• Trigonometric Functions:
  ∫ sin(x) dx = -cos(x) + C
  ∫ cos(x) dx = sin(x) + C
  ∫ sec²(x) dx = tan(x) + C

Practical Examples

• Find ∫ x³ dx:
  Using the power rule with n = 3:
  ∫ x³ dx = x³⁺¹ / (3 + 1) + C = x⁴ / 4 + C
• Find ∫ (4e^x + 5 / x) dx:
  Using linear properties and standard formulas:
  = 4 ∫ e^x dx + 5 ∫ (1 / x) dx
  = 4e^x + 5 ln |x| + C

Exam-Oriented Notes & Common Mistakes

• Common Mistake: Forgetting to write the constant of integration C at the end of an indefinite integral. Always include + C.
• Important Observation: The power rule fails when n = -1 because division by zero occurs. For n = -1, the integral evaluates to ln |x| + C.

Integration by Substitution

Concept of Integration by Substitution

Integration by substitution is the reverse of the chain rule in differentiation. It is used when an integrand contains a composite function along with its derivative factor, making direct integration difficult. By substituting a new variable, the integral is transformed into a simpler form.

Step-by-Step Procedure

1. Identify a substitution expression within the integrand, typically denoted as u = g(x).
2. Compute the derivative du, which gives du = g'(x) dx.
3. Substitute u and du completely into the integral to eliminate all occurrences of the original variable x.
4. Evaluate the resulting simpler integral with respect to u.
5. Substitute back the original expression in terms of x to obtain the final answer containing C.

Practical Example

• Evaluate ∫ 2x(x² + 1)⁴ dx:
  Step 1: Let u = x² + 1.
  Step 2: Find the differential du = 2x dx.
  Step 3: Substitute into the integral:
  ∫ (x² + 1)⁴ (2x dx) = ∫ u⁴ du
  Step 4: Integrate using the power rule:
  = u⁵ / 5 + C
  Step 5: Substitute back x² + 1 for u:
  = (x² + 1)⁵ / 5 + C

Exam-Oriented Notes

• Look for a function and its derivative multiplied together inside the integral. The part whose derivative is present is the ideal candidate for u.

Integration by Partial Fractions

Concept of Integration by Partial Fractions

Integration by partial fractions is a technique used to integrate rational functions (fractions where both numerator and denominator are polynomials). The method breaks down a complex, difficult-to-integrate rational function into a sum of simpler, easily integrable fractions.

Prerequisite Condition

The degree of the numerator polynomial must be strictly less than the degree of the denominator polynomial. If it is greater or equal, polynomial long division must be performed first.

Standard Partial Fraction Decompositions

Step-by-Step Procedure

1. Factor the denominator completely into linear and/or irreducible quadratic factors.
2. Express the rational function as a sum of partial fractions with unknown constants (A, B, C, etc.).
3. Clear the denominators by multiplying both sides by the original common denominator.
4. Determine the values of the unknown constants by comparing coefficients of like powers of x or by substituting strategic numerical values for x.
5. Integrate each individual partial fraction term separately.

Practical Example

• Evaluate ∫ [1 / (x² - 5x + 6)] dx:
  Step 1: Factor the denominator: x² - 5x + 6 = (x - 2)(x - 3).
  Step 2: Set up partial fractions:
  1 / [(x - 2)(x - 3)] = A / (x - 2) + B / (x - 3)
  Step 3: Multiply by (x - 2)(x - 3):
  1 = A(x - 3) + B(x - 2)
  Step 4: Solve for constants:
  Let x = 2: 1 = A(2 - 3) + B(0) ⇒ A = -1
  Let x = 3: 1 = A(0) + B(3 - 2) ⇒ B = 1
  Step 5: Integrate the decomposed terms:
  ∫ [ -1 / (x - 2) + 1 / (x - 3) ] dx
  = -ln |x - 2| + ln |x - 3| + C
  = ln |(x - 3) / (x - 2)| + C

Exam-Oriented Notes

• Always ensure the denominator is fully factored before setting up the partial fraction structure. Misidentifying factor types leads to incorrect constant setups.

Integration by Parts

Concept of Integration by Parts

Integration by parts is the integral calculus equivalent of the product rule for differentiation. It is primarily used to integrate products of two different types of functions, such as an algebraic function multiplied by a trigonometric or logarithmic function.

Formula: Integration by Parts
∫ u dv = uv - ∫ v du
Where u is the function to be differentiated, and dv is the part to be integrated.

The LIATE Rule for Choosing u

To choose which part of the product should be assigned as u, use the LIATE priority order (choose the function that appears first in the list as u):

• L: Logarithmic functions (e.g., ln(x))
• I: Inverse trigonometric functions (e.g., arctan(x))
• A: Algebraic functions (e.g., x, x²)
• T: Trigonometric functions (e.g., sin(x), cos(x))
• E: Exponential functions (e.g., e^x)

Step-by-Step Procedure

1. Identify the two parts of the product and select u using the LIATE rule, letting the remaining part be dv.
2. Differentiate u to find du = u' dx.
3. Integrate dv to find v = ∫ dv.
4. Substitute u, v, du, and
   into the integration by parts formula: uv - ∫ v du.
5. Evaluate the remaining integral.

Practical Example

• Evaluate ∫ x e^x dx:
  Step 1: Choose u and dv using LIATE (Algebraic comes before Exponential).
  Let u = x and dv = e^x dx.
  Step 2: Differentiate u:
  du = 1 dx = dx.
  Step 3: Integrate dv:
  v = ∫ e^x dx = e^x.
  Step 4: Apply the formula uv - ∫ v du:
  = x e^x - ∫ e^x dx
  Step 5: Evaluate the final integral:
  = x e^x - e^x + C

Exam-Oriented Notes & Common Mistakes

• Common Mistake: Forgetting to include the negative sign or incorrectly assigning dv so that it cannot be easily integrated. Always ensure dv is an expression you know how to integrate directly.
• Multiple Applications: Some integrals (like ∫ x² e^x dx) require applying integration by parts more than once consecutively until the algebraic power is eliminated.