Calculus

Limits

The concept of limits is fundamental in calculus. It describes the value that a function approaches as the input approaches some point.

Definition: For a function \( f(x) \), the limit as \( x \) approaches \( a \) is \( L \) if for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that whenever \( 0 < |x - a| < \delta \), it follows that \( |f(x) - L| < \varepsilon \).

\(\displaystyle \lim_{x \to a} f(x) = L \)

Example 1: Find \( \lim_{x \to 3} (2x + 1) \).

  1. Substitute \( x = 3 \) into the function: \( 2(3) + 1 = 7 \).
  2. Since the function is continuous at 3, the limit is simply the function value.

The limit is \( 7 \) because \( 2x + 1 \) is a polynomial and continuous everywhere.

Example 2: Evaluate \( \lim_{x \to 0} \frac{\sin x}{x} \).

  1. Direct substitution gives \( \frac{0}{0} \), which is indeterminate.
  2. Use the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

This is a classic trigonometric limit, which evaluates to 1.

Continuity

A function \( f(x) \) is continuous at a point \( a \) if the following three conditions hold:

  • \( f(a) \) is defined.
  • \( \lim_{x \to a} f(x) \) exists.
  • \( \lim_{x \to a} f(x) = f(a) \).
Continuity means no breaks, jumps, or holes at the point \( a \).

Example: Show that \( f(x) = x^2 \) is continuous at \( x = 2 \).

  1. Calculate \( f(2) = 4 \).
  2. Calculate \( \lim_{x \to 2} x^2 = 4 \).
  3. Since the limit equals the function value, \( f \) is continuous at 2.

The function \( x^2 \) is continuous everywhere because it's a polynomial.

Derivatives

The derivative of a function measures the rate at which the function value changes as its input changes. It is defined as the limit:

\(\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

The derivative represents the slope of the tangent line to the graph of \( f \) at point \( x \).

Example: Find the derivative of \( f(x) = x^3 \).

  1. Use the definition: \( f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} \).
  2. Expand numerator: \( (x^3 + 3x^2h + 3xh^2 + h^3) - x^3 = 3x^2h + 3xh^2 + h^3 \).
  3. Divide by \( h \): \( 3x^2 + 3xh + h^2 \).
  4. Take the limit as \( h \to 0 \): \( 3x^2 \).

The derivative is \( f'(x) = 3x^2 \).

Basic Derivative Rules

  • Power Rule: \( \frac{d}{dx} x^n = nx^{n-1} \)
  • Sum Rule: \( \frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x) \)
  • Product Rule: \( \frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \)
  • Quotient Rule: \( \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \)
  • Chain Rule: \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \)

Applications of Derivatives

Derivatives are used to find maxima, minima, and points of inflection of functions.

Critical points occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined.

Example: Find critical points and determine maxima or minima for \( f(x) = x^3 - 3x^2 + 4 \).

  1. Find derivative: \( f'(x) = 3x^2 - 6x \).
  2. Set \( f'(x) = 0 \): \( 3x^2 - 6x = 0 \Rightarrow 3x(x - 2) = 0 \Rightarrow x = 0, 2 \).
  3. Use second derivative test: \( f''(x) = 6x - 6 \).
  4. At \( x=0 \), \( f''(0) = -6 < 0 \) → local maximum.
  5. At \( x=2 \), \( f''(2) = 6 > 0 \) → local minimum.

Critical points at \( x=0 \) (max) and \( x=2 \) (min).

Integrals

Integration is the inverse process of differentiation. It is used to calculate areas, volumes, and accumulations.

\(\displaystyle \int f(x) \, dx \)

Indefinite Integral: The collection of all antiderivatives of \( f(x) \), plus a constant \( C \).

Example: Find \( \int 3x^2 \, dx \).

  1. Use the power rule for integration: \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \).
  2. Here, \( n = 2 \), so \( \int 3x^2 dx = 3 \times \frac{x^{3}}{3} + C = x^3 + C \).

The integral is \( x^3 + C \), where \( C \) is the constant of integration.

Definite Integral: Represents the signed area under \( f(x) \) between limits \( a \) and \( b \): \[ \int_a^b f(x) \, dx = F(b) - F(a) \] where \( F(x) \) is any antiderivative of \( f(x) \).

Example: Calculate \( \int_1^4 (2x) \, dx \).

  1. Find the antiderivative of \( 2x \): \( F(x) = x^2 \).
  2. Evaluate \( F(4) - F(1) = 16 - 1 = 15 \).

The definite integral equals 15, which is the area under the curve from 1 to 4.

Applications of Integrals

Integrals are used to find areas, volumes, work done, and many physical quantities.

Common applications include:
  • Area between curves
  • Volume of solids of revolution
  • Average value of a function
  • Work done by a variable force

Example: Find the area between \( y = x^2 \) and \( y = x + 2 \) from \( x=0 \) to \( x=2 \).

  1. Find points of intersection: solve \( x^2 = x + 2 \).
  2. At \( x=0 \), \( y = 0 \) and \( y=2 \), so from 0 to 2, \( x + 2 \) is above \( x^2 \).
  3. Area = \( \int_0^2 [(x + 2) - x^2] dx \).
  4. Calculate integral: \( \int_0^2 (x + 2 - x^2) dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_0^2 = \left(2 + 4 - \frac{8}{3}\right) - 0 = \frac{10}{3} \).

The area between the curves is \( \frac{10}{3} \) square units.

Fundamental Theorem of Calculus

This theorem connects differentiation and integration, showing they are inverse processes.

  1. If \( F \) is an antiderivative of \( f \), then \[ \int_a^b f(x) dx = F(b) - F(a) \]
  2. If \[ g(x) = \int_a^x f(t) dt, \] then \( g'(x) = f(x) \).

Example: If \( F'(x) = f(x) = 3x^2 \), evaluate \( \int_1^3 3x^2 dx \).

  1. Find antiderivative \( F(x) = x^3 + C \).
  2. Evaluate \( F(3) - F(1) = 27 - 1 = 26 \).

The definite integral equals 26 by the Fundamental Theorem of Calculus.

Techniques of Integration

Various methods are used to solve integrals that are not straightforward:

  • Substitution: Change of variables to simplify integral.
  • Integration by Parts: Based on the product rule of differentiation.
  • Partial Fractions: Decompose rational functions into simpler fractions.
  • Trigonometric Integrals: Use identities to simplify.

Example (Substitution): Evaluate \( \int 2x \cos(x^2) dx \).

  1. Let \( u = x^2 \), so \( du = 2x dx \).
  2. Integral becomes \( \int \cos(u) du = \sin(u) + C \).
  3. Substitute back: \( \sin(x^2) + C \).

The integral evaluates to \( \sin(x^2) + C \).

Example (Integration by Parts): Evaluate \( \int x e^x dx \).

  1. Let \( u = x \), so \( du = dx \).
  2. Let \( dv = e^x dx \), so \( v = e^x \).
  3. Use formula: \( \int u dv = uv - \int v du = x e^x - \int e^x dx = x e^x - e^x + C \).

The integral equals \( e^x (x - 1) + C \).

Sequences & Series

A sequence is an ordered list of numbers. A series is the sum of the terms of a sequence.

Arithmetic Sequence: Terms increase by a constant difference \( d \). \[ a_n = a_1 + (n-1)d \]
Geometric Sequence: Terms multiply by a constant ratio \( r \). \[ a_n = a_1 r^{n-1} \]

Example (Arithmetic): Find the 10th term of the sequence where \( a_1=3 \), \( d=5 \).

  1. Use formula: \( a_{10} = 3 + (10-1) \times 5 = 3 + 45 = 48 \).

The 10th term is 48.

Example (Geometric): Find the sum of the first 5 terms with \( a_1=2 \), \( r=3 \).

  1. Sum of first \( n \) terms: \( S_n = a_1 \frac{r^n - 1}{r - 1} \).
  2. \( S_5 = 2 \times \frac{3^5 - 1}{3 - 1} = 2 \times \frac{243 - 1}{2} = 2 \times 121 = 242 \).

The sum of the first 5 terms is 242.