Unit 3: Fundamentals of integral calculus

I. Orientation — Antiderivatives, Accumulation, and the Fundamental Theorem

Integral calculus studies accumulation and reverses differentiation. An indefinite integral represents a family of antiderivatives, while a definite integral represents signed accumulation over an interval. The Fundamental Theorem of Calculus connects these meanings: if (F'(x)=f(x)), then the definite integral of (f) can be evaluated using (F).

• Indefinite integral: If (F'(x)=f(x)), then
  

  TEXT

    ∫ f(x) dx = F(x) + C

  
  where (x) is the variable of integration, (f(x)) is the integrand, (F(x)) is an antiderivative, and (C) is an arbitrary constant.

• Constant of integration: The term (C) is necessary because functions differing by a constant have the same derivative; for example,
  

  TEXT

    d/dx (x² + 3) = d/dx (x² − 5) = 2x.

• Definite integral: For an integrable function (f) on ([a,b]),
  

  TEXT

    ∫[a to b] f(x) dx

  
  gives signed accumulation from the lower limit (a) to the upper limit (b).

• Fundamental Theorem of Calculus: If (f) is continuous on ([a,b]) and (F'(x)=f(x)), then
  

  TEXT

    ∫[a to b] f(x) dx = F(b) − F(a).

  
  Here (F(a)) and (F(b)) are the antiderivative’s values at the integration limits.

• Geometric convention: Area above the (x)-axis contributes positively, whereas area below it contributes negatively. Thus a definite integral is not necessarily the total geometric area.

• Core strategy: Integration requires recognizing the integrand’s structure and selecting an appropriate method: direct rules, substitution, integration by parts, or partial fractions.

II. General Rules — Direct Evaluation of Standard Forms

Direct integration applies known antiderivative formulas together with linearity. These rules are valid wherever the expressions involved are defined, with logarithmic forms interpreted on suitable intervals.

A. general rules of integration

General rules convert standard functions and their linear combinations into antiderivatives without requiring a structural transformation.

• Linearity: Constants may be taken outside an integral, and sums or differences may be integrated term by term:
  

  TEXT

    ∫[αf(x) + βg(x)] dx
      = α∫f(x) dx + β∫g(x) dx.

  
  Here (f) and (g) are integrable functions, while (\alpha) and (\beta) are constants.

• Power rule: For any real (n\neq-1),
  

  TEXT

    ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C.

  
  The restriction (n\neq-1) is essential because division by (n+1) would otherwise be undefined.

• Logarithmic rule: The exceptional power (n=-1) gives
  

  TEXT

    ∫(1/x) dx = ln|x| + C,    x ≠ 0.

  
  The absolute value allows the result to hold on both positive and negative intervals.

• Exponential rules:
  

  TEXT

    ∫eˣ dx = eˣ + C
    ∫aˣ dx = aˣ/ln(a) + C

  
  where (a>0) and (a\neq1).

• Basic trigonometric rules:
  

  TEXT

    ∫sin x dx = −cos x + C
    ∫cos x dx = sin x + C
    ∫sec²x dx = tan x + C
    ∫csc²x dx = −cot x + C
    ∫sec x tan x dx = sec x + C
    ∫csc x cot x dx = −csc x + C

• Inverse-trigonometric forms: For a positive constant (a),
  

  TEXT

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

• Worked example: Linearity and the power rule give
  

  TEXT

    ∫(3x² − 4x + 5) dx
    = 3(x³/3) − 4(x²/2) + 5x + C
    = x³ − 2x² + 5x + C.

  
  Differentiating the result reproduces (3x²-4x+5), which verifies the antiderivative.

III. Substitution — Reversing the Chain Rule

Substitution simplifies a composite integrand by replacing an inner expression with a new variable. It is the integral counterpart of the chain rule and is most effective when the derivative of the inner expression also appears as a factor.

A. integration by substitution

Integration by substitution changes variables so that a complicated integral becomes a standard one.

• Formal rule: If (u=g(x)), then (du=g'(x)\,dx), and
  

  TEXT

    ∫f(g(x))g′(x) dx = ∫f(u) du.

  
  Here (g(x)) is the inner function, (u) is the substituted variable, and (f) is the outer functional form.

• Selection principle: Choose (u) as an expression whose derivative occurs elsewhere in the integrand. Common choices include a denominator, a power’s base, an exponential’s exponent, or a trigonometric argument.

• Constant adjustment: An exact derivative need not appear; a nonzero constant factor may be introduced and compensated for. For example, if (u=3x+1), then (du=3\,dx), so (dx=du/3).

• Definite-integral limits: Definite integrals can be handled in two valid ways:

  1. Change the limits: Convert (x=a,b) into (u=g(a),g(b)), then remain entirely in (u).
  2. Restore the variable: Find the antiderivative in (u), substitute back to (x), and then apply the original limits.

• Worked example: Evaluate an integral containing an inner function and its derivative:
  

  TEXT

    ∫2x cos(x²) dx.

  
  Let (u=x²), so (du=2x\,dx). Therefore,
  

  TEXT

    ∫2x cos(x²) dx
    = ∫cos u du
    = sin u + C
    = sin(x²) + C.

• Limitation: Substitution is ineffective when no useful inner function has a derivative matching the remaining factors; products such as (x e^x) generally require integration by parts instead.

IV. Integration by Parts — Reversing the Product Rule

Integration by parts transforms an integral of a product into another integral. It follows from the derivative identity ((uv)'=u'v+uv') and is useful when differentiating one factor makes it simpler.

A. integration by parts

Integration by parts transfers differentiation from one factor to another in order to simplify the integrand.

• Formula:
  

  TEXT

    ∫u dv = uv − ∫v du.

  
  Here (u) is the factor selected for differentiation, (du) is its differential, (dv) is the remaining factor including (dx), and (v=\int dv).

• Definite form:
  

  TEXT

    ∫[a to b] u dv = [uv]ₐᵇ − ∫[a to b] v du,

  
  where ([uv]ₐᵇ=u(b)v(b)-u(a)v(a)).

• Choosing (u): A useful preference order is logarithmic, inverse-trigonometric, algebraic, trigonometric, and exponential functions. This is a heuristic rather than a theorem; the resulting integral must actually be simpler.

• Typical uses:

  • Products such as (x e^x), (x\sin x), or (x^2\ln x).
  • Single functions such as (\ln x), treated as ((\ln x)\cdot1).
  • Repeated application for polynomial factors of degree greater than one.
  • Cyclic integrals, such as products of exponentials and trigonometric functions, where the original integral reappears algebraically.

• Worked example: For
  

  TEXT

    ∫x eˣ dx,

  
  choose (u=x) and (dv=e^x dx). Then (du=dx) and (v=e^x), giving
  

  TEXT

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

• Limitation: A poor choice of (u) can produce a more difficult integral. For instance, differentiating (e^x) does not simplify it, whereas differentiating the algebraic factor (x) reduces its degree.

V. Partial Fractions — Integrating Rational Functions

Partial-fraction decomposition rewrites a proper rational function as a sum of simpler fractions with standard antiderivatives. It applies after the denominator has been factored over the real numbers.

A. integration by partial fraction

Integration by partial fraction reduces a rational integrand to logarithmic, power, and inverse-trigonometric forms.

• Precondition: For (P(x)/Q(x)), the fraction must be proper:
  

  TEXT

    degree P < degree Q.

  
  If (\deg P\geq\deg Q), polynomial division must be performed first. Here (P) and (Q) are polynomials.

• Distinct linear factors: If
  

  TEXT

    Q(x) = (x − a)(x − b),

  
  then use
  

  TEXT

    P(x)/Q(x) = A/(x − a) + B/(x − b),

  
  where (A) and (B) are constants determined by equating coefficients or substituting convenient values of (x).

• Repeated linear factors: A factor ((x-a)^m) requires every power:
  

  TEXT

    A₁/(x−a) + A₂/(x−a)² + ··· + Aₘ/(x−a)ᵐ,

  
  where (m) is a positive integer and (A_1,\ldots,A_m) are constants.

• Irreducible quadratic factors: For a real quadratic (x²+px+q) that cannot be factored into real linear factors, use
  

  TEXT

    (Ax + B)/(x² + px + q).

  
  Repeated quadratic factors require one linear numerator for each denominator power.

• Worked example:
  

  TEXT

    ∫(3x + 5)/[(x + 1)(x + 2)] dx.

  
  Write
  

  TEXT

    (3x + 5)/[(x + 1)(x + 2)]
    = A/(x + 1) + B/(x + 2).

  
  Thus (3x+5=A(x+2)+B(x+1)). Setting (x=-1) gives (A=2), and setting (x=-2) gives (B=1). Therefore,
  

  TEXT

    ∫(3x + 5)/[(x + 1)(x + 2)] dx
    = 2ln|x + 1| + ln|x + 2| + C.

• Limitation: The method directly targets rational functions. Expressions involving radicals, exponentials, or trigonometric functions require preliminary transformations or another integration technique.

VI. Definite Integrals — Structural Properties of Accumulation

A definite integral is a number determined by an integrand and an oriented interval. Its algebraic, interval, comparison, and symmetry properties can simplify calculations before an antiderivative is found.

A. properties of definite integrals

The properties of definite integrals express how accumulated quantities respond to changed limits, decomposed intervals, linear combinations, bounds, and symmetry.

• Equal and reversed limits:
  

  TEXT

    ∫[a to a] f(x) dx = 0
    ∫[a to b] f(x) dx = −∫[b to a] f(x) dx.

  
  Reversing orientation changes the sign but not the magnitude.

• Additivity over intervals: For any intermediate point (c),
  

  TEXT

    ∫[a to b] f(x) dx
    = ∫[a to c] f(x) dx + ∫[c to b] f(x) dx.

  
  This remains valid regardless of the relative ordering of (a), (b), and (c).

• Linearity:
  

  TEXT

    ∫[a to b] [αf(x) + βg(x)] dx
    = α∫[a to b] f(x) dx + β∫[a to b] g(x) dx,

  
  where (\alpha) and (\beta) are constants.

• Positivity and comparison: If (f(x)\geq0) on ([a,b]), then its integral is nonnegative. If (f(x)\leq g(x)) throughout the interval, then
  

  TEXT

    ∫[a to b] f(x) dx ≤ ∫[a to b] g(x) dx.

• Bounds: If (m\leq f(x)\leq M) on ([a,b]), with (a<b), then
  

  TEXT

    m(b − a) ≤ ∫[a to b] f(x) dx ≤ M(b − a),

  
  where (m) and (M) are lower and upper bounds of (f).

• Absolute-value inequality:
  

  TEXT

    |∫[a to b] f(x) dx| ≤ ∫[a to b] |f(x)| dx.

  
  Signed cancellation can reduce the magnitude of the first integral.

• Symmetry on ([-a,a]):

  1. Even function: If (f(-x)=f(x)), then
     
     TEXT

          ∫[−a to a] f(x) dx = 2∫[0 to a] f(x) dx.

  2. Odd function: If (f(-x)=-f(x)), then
     
     TEXT

          ∫[−a to a] f(x) dx = 0.

• Worked example: Since (x^3) is odd and (x^2) is even,
  

  TEXT

    ∫[−2 to 2] (x³ + x²) dx
    = 0 + 2∫[0 to 2] x² dx
    = 2[x³/3]₀²
    = 16/3.