Unit 3: Integral Calculus

Table of Contents

1. Review of Integration and Definite Integral

Indefinite Integral

Integration is the reverse process of differentiation. If d/dx(F(x)) = f(x), then the indefinite integral of f(x) is:

∫ f(x) dx = F(x) + C

Where 'C' is the arbitrary constant of integration.

Definite Integral (Fundamental Theorem of Calculus)

The definite integral represents the signed area under the curve y = f(x) from x = a to x = b.

ab f(x) dx = [ F(x) ]ab = F(b) - F(a)

Key Properties:


2. Differentiation Under Integral Sign (Leibniz Rule)

This rule (also called Leibniz Integral Rule) tells us how to find the derivative of a definite integral whose limits and/or integrand depend on the variable we are differentiating with respect to.

Leibniz Rule for Differentiation Under the Integral Sign:
Let I(x) = ∫a(x)b(x) f(x, t) dt
Then, its derivative dI/dx is:
dI/dx = f(x, b(x)) · b'(x) - f(x, a(x)) · a'(x) + ∫a(x)b(x) (∂f/∂x) dt

In words:
(Derivative) = (Integrand at upper limit) · (Derivative of upper limit)
... - (Integrand at lower limit) · (Derivative of lower limit)
... + (Integral of the partial derivative of the integrand)

Simplified Case: If the limits 'a' and 'b' are constants, then a'(x)=0 and b'(x)=0. The formula simplifies to:
d/dx [ ∫ab f(x, t) dt ] = ∫ab (∂f/∂x) dt

3. Double Integral

A double integral is used to integrate a function of two variables, f(x, y), over a 2D region R in the xy-plane. It is often used to calculate Volume under a surface or Area of a region.

R f(x, y) dA

We evaluate double integrals as iterated integrals. The key is to set up the limits of integration correctly.

Type 1: Integrating with `dy dx` (Vertical Strips)

Region R is defined by a ≤ x ≤ b and g1(x) ≤ y ≤ g2(x).

R f(x, y) dA = ∫ab [ ∫g1(x)g2(x) f(x, y) dy ] dx

How-to: Integrate the *inner* integral w.r.t 'y' first (treating x as constant), then integrate the resulting function of 'x' w.r.t 'x'.

Type 2: Integrating with `dx dy` (Horizontal Strips)

Region R is defined by c ≤ y ≤ d and h1(y) ≤ x ≤ h2(y).

R f(x, y) dA = ∫cd [ ∫h1(y)h2(y) f(x, y) dx ] dy

How-to: Integrate the *inner* integral w.r.t 'x' first (treating y as constant), then integrate the resulting function of 'y' w.r.t 'y'.

Calculating Area: The area of a region R is found by setting f(x, y) = 1.
Area(R) = ∬R 1 dA

4. Change of Order of Integration

Sometimes, an iterated integral is very difficult or impossible to solve in the given order (e.g., ∫ e(x²) dx). By changing the order of integration (from `dy dx` to `dx dy`, or vice-versa), the integral often becomes much simpler.

Step-by-Step Procedure:

  1. Identify the Region: From the given limits, sketch the 2D region R of integration.
    Example:01x1 f(x,y) dy dx
    Limits are: x goes from 0 to 1, and y goes from y=x (bottom) to y=1 (top).
    This is a triangle bounded by y=x, y=1, and x=0.
  2. Describe the Region in the Other Order: Look at the sketch and describe the same region using horizontal strips (`dx dy`) instead of vertical ones.
    Example: For the same triangle, y now goes from 0 to 1. For any 'y', x goes from x=0 (left) to x=y (right).
  3. Write the New Integral:
    Example:010y f(x,y) dx dy
Exam Tip: This is a very common exam question. The most important step is accurately sketching the region R. Get the sketch wrong, and the new limits will be wrong.

5. Transformation of Variables (Jacobian)

This is the "u-substitution" for double integrals. We change variables from (x, y) to a new coordinate system (u, v) to simplify the region R or the integrand f(x, y).

Example: Converting from Cartesian (x, y) to Polar (r, θ).

Transformation: x = g(u, v) and y = h(u, v)

The Jacobian (J)

The "scaling factor" for the area element is the absolute value of the Jacobian determinant.

J = ∂(x, y) / ∂(u, v) = det | [∂x/∂u ∂x/∂v] | | [∂y/∂u ∂y/∂v] | J = (∂x/∂u)(∂y/∂v) - (∂x/∂v)(∂y/∂u)

The Transformation Formula:

R f(x, y) dx dy = ∬S f(g(u,v), h(u,v)) · |J| du dv

Where S is the new, simpler region in the uv-plane.

Example: Polar Coordinates


6. Beta and Gamma Functions

These are "special functions" defined by improper integrals, which are extremely useful in engineering and statistics.

The Gamma Function (Γ(n))

A generalization of the factorial function to non-integers.

Definition:
Γ(n) = ∫0 e-x · xn-1 dx (for n > 0)

Key Properties:

The Beta Function (B(m, n))

Definition (Type 1):
B(m, n) = ∫01 xm-1 · (1 - x)n-1 dx (for m > 0, n > 0)

Key Properties:

Relationship between Beta and Gamma

This is the key formula that connects them and is used to solve many integrals.

B(m, n) = [ Γ(m) · Γ(n) ] / [ Γ(m + n) ]
Exam Tip: You'll be given an integral that *looks like* the Beta or Gamma function. Your job is to use substitution to make it match the exact definition, identify 'm' and 'n', and then use the properties (especially the relationship formula) to get the answer in terms of Γ or π.