Basic MathGuides

Mastering the Basics of Calculus: Limits, Derivatives & Integrals Explained Simply

2 weeks ago
Last Update on April 10, 2025
12 mins
33 Views
0 Comments

Comprehensive Basic Calculus Notes

1. Introduction to Calculus

Calculus is a branch of mathematics that studies continuous change. It has two major branches:

  • Differential Calculus: Concerns rates of change and slopes of curves.
  • Integral Calculus: Concerns accumulation of quantities and the areas under or between curves.

Historical Context

Calculus was developed independently by Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century. Newton developed calculus as a tool to understand motion and physical systems, while Leibniz approached it from a more abstract mathematical perspective.

Why Calculus Matters

Real-World Applications:

  • Physics: Modeling motion, forces, and energy.
  • Engineering: Designing structures, circuits, and systems.
  • Economics: Analyzing marginal costs, profits, and optimization problems.
  • Biology: Population growth models and rates of reactions.
  • Computer Science: Algorithms, machine learning, and computer graphics.
  • Limits: The foundation of calculus - understanding what happens as values approach a specific point.
  • Derivatives: Instantaneous rates of change - the slope of a curve at a specific point.
  • Integrals: Accumulation of quantities - the area under a curve.
  • Series: Sums of infinite sequences of terms.

2. Limits & Continuity

2.1 Definition of a Limit

A limit describes the value that a function approaches as the input approaches a particular value.

limx→a f(x) = L

This means that as x gets closer to a, f(x) gets arbitrarily close to L.

Example 1: Direct Substitution

Find limx→2 (x² + 3x - 1)

  1. Substitute x = 2 into the function: f(2) = 2² + 3(2) - 1
  2. Calculate: f(2) = 4 + 6 - 1 = 9
  3. Therefore, limx→2 (x² + 3x - 1) = 9

Example 2: Limit with Indeterminate Form

Find limx→3 (x² - 9)/(x - 3)

  1. Recognize that direct substitution gives 0/0, which is an indeterminate form.
  2. Factor the numerator: (x² - 9)/(x - 3) = ((x - 3)(x + 3))/(x - 3)
  3. Cancel the common factor: (x² - 9)/(x - 3) = x + 3 for x ≠ 3
  4. As x approaches 3: limx→3 (x + 3) = 3 + 3 = 6
  1. Recognize that direct substitution gives 0/0, an indeterminate form.
  2. Apply L'Hôpital's Rule: differentiate both numerator and denominator.
  3. limx→3 (x² - 9)/(x - 3) = limx→3 2x/1 = 2(3) = 6

2.2 Properties of Limits

For functions f and g, and constants c, the following properties hold (assuming the individual limits exist):

  1. Sum Rule: limx→a [f(x) + g(x)] = limx→a f(x) + limx→a g(x)
  2. Difference Rule: limx→a [f(x) - g(x)] = limx→a f(x) - limx→a g(x)
  3. Product Rule: limx→a [f(x) × g(x)] = limx→a f(x) × limx→a g(x)
  4. Quotient Rule: limx→a [f(x)/g(x)] = limx→a f(x) / limx→a g(x), if limx→a g(x) ≠ 0
  5. Constant Multiple Rule: limx→a [c·f(x)] = c·limx→a f(x)
  6. Power Rule: limx→a [f(x)]n = [limx→a f(x)]n

Example: Using Limit Properties

Find limx→2 (3x² - 4x + 5)

  1. Apply the sum rule: limx→2 (3x² - 4x + 5) = limx→2 3x² - limx→2 4x + limx→2 5
  2. Apply the constant multiple rule: = 3·limx→2 x² - 4·limx→2 x + 5
  3. Evaluate each limit: = 3·(2²) - 4·(2) + 5
  4. Calculate: = 3·4 - 4·2 + 5 = 12 - 8 + 5 = 9
  1. Substitute x = 2 directly: f(2) = 3(2)² - 4(2) + 5
  2. Calculate: f(2) = 3(4) - 4(2) + 5 = 12 - 8 + 5 = 9

2.3 Techniques for Evaluating Limits

Common techniques for evaluating limits include:

  1. Direct Substitution: When the function is continuous at the point.
  2. Factoring: Useful for rational functions with common factors.
  3. Rationalization: For limits involving square roots.
  4. Using Special Limits: Such as limx→0 sin(x)/x = 1.
  5. L'Hôpital's Rule: For indeterminate forms (0/0, ∞/∞).

Example: Rationalization Technique

Find limx→4 (√x - 2)/(x - 4)

  1. Direct substitution gives 0/0, an indeterminate form.
  2. Multiply by the conjugate of the numerator:
    limx→4 (√x - 2)/(x - 4) · (√x + 2)/(√x + 2)
  3. Simplify:
    limx→4 (x - 4)/((x - 4)(√x + 2)) = limx→4 1/(√x + 2)
  4. Now direct substitution works: 1/(√4 + 2) = 1/(2 + 2) = 1/4
  1. Direct substitution gives 0/0, an indeterminate form.
  2. Apply L'Hôpital's Rule by taking the derivative of numerator and denominator:
    limx→4 (√x - 2)/(x - 4) = limx→4 (1/(2√x))/1
  3. Evaluate: 1/(2√4) = 1/(2·2) = 1/4

Example: Special Limit

Find limx→0 sin(3x)/x

  1. Recall the special limit: limx→0 sin(x)/x = 1
  2. Rewrite the expression:
    limx→0 sin(3x)/x = limx→0 3 · sin(3x)/(3x)
  3. Apply the special limit with the substitution u = 3x:
    = 3 · limu→0 sin(u)/u = 3 · 1 = 3
  1. Rewrite the expression:
    limx→0 sin(3x)/x = limx→0 3 · sin(3x)/(3x)
  2. Let u = 3x, so as x → 0, u → 0:
    = 3 · limu→0 sin(u)/u
  3. Use the fact that limu→0 sin(u)/u = 1:
    = 3 · 1 = 3

2.4 Continuity

A function f is continuous at a point x = a if:

  1. f(a) is defined (the function exists at x = a)
  2. limx→a f(x) exists (the limit exists as x approaches a)
  3. limx→a f(x) = f(a) (the limit equals the function value)

Types of Discontinuities:

  • Removable Discontinuity: The limit exists but doesn't equal the function value (or the function is undefined at that point).
  • Jump Discontinuity: The left and right limits exist but are not equal.
  • Infinite Discontinuity: The limit approaches infinity or negative infinity.

Example: Analyzing Continuity

Determine if the function f(x) = (x² - 4)/(x - 2) is continuous at x = 2.

  1. Check if f(2) is defined:

    f(2) = (2² - 4)/(2 - 2) = 0/0, which is undefined.

    So, f(2) is not defined, which means the function is not continuous at x = 2.

  2. However, we can check if the limit exists:

    limx→2 (x² - 4)/(x - 2) = limx→2 ((x - 2)(x + 2))/(x - 2) = limx→2 (x + 2) = 4

  3. The limit exists (equals 4), but f(2) is undefined. This is a removable discontinuity.

Quick Quiz: Limits & Continuity

1. Evaluate limx→1 (x³ - 1)/(x - 1)
  • a) 0
  • b) 1
  • c) 3
  • d) Undefined
2. Which technique is most appropriate for evaluating limx→0 (1 - cos(x))/x²?
  • a) Direct substitution
  • b) Factoring
  • c) L'Hôpital's Rule
  • d) Taylor series expansion
3. A function f is continuous at x = a if:
  • a) f(a) is defined
  • b) limx→a f(x) exists
  • c) limx→a f(x) = f(a)
  • d) All of the above

3. Derivatives

3.1 Definition of a Derivative

The derivative of a function f at a point x = a represents the instantaneous rate of change or the slope of the tangent line to the curve at that point.

f'(a) = limh→0 [f(a+h) - f(a)]/h

Alternatively, using the limit definition:

f'(x) = limΔx→0 [f(x+Δx) - f(x)]/Δx

Example: Finding a Derivative Using the Definition

Find the derivative of f(x) = x² using the limit definition.

  1. Apply the definition:
    f'(x) = limh→0 [f(x+h) - f(x)]/h
  2. Substitute f(x) = x²:
    f'(x) = limh→0 [(x+h)² - x²]/h
  3. Expand (x+h)²:
    f'(x) = limh→0 [x² + 2xh + h² - x²]/h
  4. Simplify:
    f'(x) = limh→0 [2xh + h²]/h = limh→0 [2x + h]
  5. Apply the limit:
    f'(x) = 2x

Notation for Derivatives:

  • Leibniz notation: dy/dx or df/dx
  • Lagrange notation: f'(x) or y'
  • Newton's notation: ẏ or ḟ(x)
  • For higher-order derivatives: f''(x), f'''(x), f⁽⁴⁾(x), etc., or d²y/dx², d³y/dx³, etc.

3.2 Basic Differentiation Rules

Basic Differentiation Rules:

Rule Formula Example
Constant Rule d/dx [c] = 0 d/dx [5] = 0
Power Rule d/dx [x^n] = n·x^(n-1) d/dx [x³] = 3x²
Constant Multiple Rule d/dx [c·f(x)] = c·f'(x) d/dx [4x²] = 4·2x = 8x
Sum Rule d/dx [f(x) + g(x)] = f'(x) + g'(x) d/dx [x² + 3x] = 2x + 3
Difference Rule d/dx [f(x) - g(x)] = f'(x) - g'(x) d/dx [x² - 3x] = 2x - 3

Example: Using Basic Differentiation Rules

Find the derivative of f(x) = 3x⁴ - 2x² + 5x - 7

  1. Use the power rule for each term:
    • d/dx [3x⁴] = 3 · 4x³ = 12x³
    • d/dx [-2x²] = -2 · 2x = -4x
    • d/dx [5x] = 5 · 1 = 5
    • d/dx [-7] = 0
  2. Apply the sum rule to combine the results:
    f'(x) = 12x³ - 4x + 5
  1. Apply the sum, difference, and constant multiple rules together with the power rule:
    f'(x) = 3 · 4x³ - 2 · 2x + 5 · 1 - 0
  2. Simplify:
    f'(x) = 12x³ - 4x + 5

3.3 Advanced Differentiation Rules

Advanced Differentiation Rules:

Rule Formula
Product Rule d/dx [f(x)·g(x)] = f'(x)·g(x) + f(x)·g'(x)
Quotient Rule d/dx [f(x)/g(x)] = [f'(x)·g(x) - f(x)·g'(x)]/[g(x)]²
Chain Rule d/dx [f(g(x))] = f'(g(x))·g'(x)

Example: Product Rule

Find the derivative of f(x) = (x² + 1)(3x - 4)

  1. Identify f(x) = x² + 1 and g(x) = 3x - 4
  2. Find their derivatives:
    • f'(x) = 2x
    • g'(x) = 3
  3. Apply the product rule: f'(x)·g(x) + f(x)·g'(x)
    = 2x·(3x - 4) + (x² + 1)·3
  4. Expand and simplify:
    = 6x² - 8x + 3x² + 3 = 9x² - 8x + 3
  1. Expand the product first:
    f(x) = (x² + 1)(3x - 4) = 3x³ - 4x² + 3x - 4
  2. Now differentiate term by term:
    f'(x) = 9x² - 8x + 3

Example: Quotient Rule

Find the derivative of f(x) = (x² - 1)/(x + 2)

  1. Identify f(x) = x² - 1 and g(x) = x + 2
  2. Find their derivatives:
    • f'(x) = 2x
    • g'(x) = 1
  3. Apply the quotient rule: [f'(x)·g(x) - f(x)·g'(x)]/[g(x)]²
    = [2x·(x + 2) - (x² - 1)·1]/[x + 2]²
  4. Simplify the numerator:
    = [2x² + 4x - x² + 1]/[x + 2]² = [x² + 4x + 1]/[x + 2]²

Example: Chain Rule

Find the derivative of f(x) = √(2x² + 3)

  1. Rewrite as f(x) = (2x² + 3)^(1/2)
  2. Identify the outer function f(u) = u^(1/2) and inner function u = g(x) = 2x² + 3
  3. Find their derivatives:
    • f'(u) = (1/2)·u^(-1/2)
    • g'(x) = 4x
  4. Apply the chain rule: f'(g(x))·g'(x)
    = (1/2)·(2x² + 3)^(-1/2)·4x
  5. Simplify:
    = 2x/√(2x² + 3)

Derivatives of Common Functions:

Function Derivative
sin(x) cos(x)
cos(x) -sin(x)
tan(x) sec²(x)
e^x e^x
ln(x) 1/x
a^x a^x·ln(a)
log_a(x) 1/(x·ln(a))

3.4 Applications of Derivatives

Key Applications of Derivatives:

  1. Rate of Change: How quickly a quantity is changing with respect to another.
  2. Slope of Tangent Line: The slope of the curve at a specific point.
  3. Optimization: Finding maximum or minimum values of functions.
  4. Related Rates: Relationships between rates of change of related quantities.
  5. Linear Approximation: Approximating functions near a point.
  6. Curve Sketching: Analyzing function behavior.

Example: Optimization Problem

Find the dimensions of a rectangle with perimeter 100 units that has the maximum possible area.

  1. Let x = width and y = length of the rectangle.
  2. The perimeter equation gives us: 2x + 2y = 100, so y = 50 - x.
  3. The area function is: A(x) = x·y = x·(50 - x) = 50x - x².
  4. To find the maximum area, we differentiate and set equal to zero:
    A'(x) = 50 - 2x = 0
  5. Solve for x: 2x = 50, so x = 25.
  6. Calculate y: y = 50 - 25 = 25.
  7. Verify it's a maximum using the second derivative test: A''(x) = -2 < 0, which confirms a maximum.
  8. Therefore, the rectangle with maximum area has width = 25 and length = 25 (a square), with area = 625 square units.

Example: Related Rates

Water is flowing into a conical tank at a rate of 5 cubic feet per minute. The tank has height 12 feet and radius at the top 6 feet. At what rate is the water level rising when the water is 8 feet deep?

  1. Set up the variables: r = radius at water surface, h = water depth, V = water volume.
  2. We know dV/dt = 5 ft³/min and we want to find dh/dt when h = 8 ft.
  3. Using similar triangles: r/h = 6/12 = 1/2, so r = h/2.
  4. The volume of a cone is V = (1/3)πr²h, so V = (1/3)π(h/2)²h = (π/12)h³.
  5. Differentiate with respect to time:
    dV/dt = (π/12) · 3h² · dh/dt = (π/4)h² · dh/dt
  6. Substitute what we know: 5 = (π/4)(8)² · dh/dt
  7. Solve for dh/dt: dh/dt = 5/((π/4)(64)) = 5/(16π) ≈ 0.1 ft/min

Quick Quiz: Derivatives

1. Find the derivative of f(x) = x³ - 5x² + 2x - 7.
  • a) 3x² - 10x + 2
  • b) x² - 5x + 2
  • c) 3x² - 5x + 2
  • d) 3x² - 10x - 7
2. Using the product rule, find the derivative of f(x) = x·sin(x).
  • a) sin(x) + x·cos(x)
  • b) cos(x)
  • c) x·cos(x)
  • d) sin(x) - x·cos(x)
3. At which value of x does the function f(x) = x³ - 3x² - 9x + 7 have a critical point?
  • a) x = -1
  • b) x = 0
  • c) x = 1
  • d) x = 3

4. Integrals

4.1 Definition of an Integral

Two Types of Integrals:

  • Indefinite Integrals: The collection of all antiderivatives of a function.
  • Definite Integrals: The net signed area between a function and the x-axis over a given interval.

Indefinite Integral: ∫f(x) dx = F(x) + C, where F'(x) = f(x)

Definite Integral: ∫abf(x) dx = F(b) - F(a), where F'(x) = f(x)

Example: Finding an Indefinite Integral

Find ∫(3x² + 2x - 5) dx

  1. Apply the linearity property of integration:
    ∫(3x² + 2x - 5) dx = 3∫x² dx + 2∫x dx - 5∫dx
  2. Use the power rule for integration:
    = 3(x³/3) + 2(x²/2) - 5x + C
  3. Simplify:
    = x³ + x² - 5x + C

Example: Evaluating a Definite Integral

Evaluate ∫02(x² + 3x) dx

  1. First find the indefinite integral:
    ∫(x² + 3x) dx = x³/3 + 3x²/2 + C
  2. Apply the fundamental theorem of calculus:
    ∫02(x² + 3x) dx = [x³/3 + 3x²/2]02
  3. Evaluate at the upper and lower limits:
    = (2³/3 + 3·2²/2) - (0³/3 + 3·0²/2) = (8/3 + 6) - 0 = 8/3 + 6 = 26/3

4.2 Integration Techniques

Basic Integration Rules:

Rule Formula
Power Rule ∫xn dx = xn+1/(n+1) + C, n ≠ -1
Constant Multiple Rule ∫c·f(x) dx = c·∫f(x) dx
Sum Rule ∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx
Difference Rule ∫[f(x) - g(x)] dx = ∫f(x) dx - ∫g(x) dx

Common Integration Formulas:

Function Integral
∫sin(x) dx -cos(x) + C
∫cos(x) dx sin(x) + C
∫tan(x) dx -ln|cos(x)| + C
∫sec²(x) dx tan(x) + C
∫ex dx ex + C
∫(1/x) dx ln|x| + C
∫ax dx ax/ln(a) + C

Advanced Integration Techniques:

  1. Substitution Method (u-substitution): Used when the integrand contains a function and its derivative.
  2. Integration by Parts: Based on the product rule for differentiation.
  3. Partial Fractions: Used for integrating rational functions.
  4. Trigonometric Substitution: Useful for integrals involving √(a² - x²), √(a² + x²), or √(x² - a²).
  5. Improper Integrals: Integrals with infinite limits or integrand with a vertical asymptote.

Example: Integration by Substitution

Evaluate ∫cos(3x) dx

  1. Let u = 3x, so du = 3 dx or dx = du/3
  2. Substitute:
    ∫cos(3x) dx = ∫cos(u) · (du/3) = (1/3)∫cos(u) du
  3. Integrate:
    = (1/3)sin(u) + C
  4. Substitute back:
    = (1/3)sin(3x) + C

Example: Integration by Parts

Evaluate ∫x·sin(x) dx

  1. Use the formula: ∫u·dv = u·v - ∫v·du
  2. Choose u = x and dv = sin(x) dx
  3. Then du = dx and v = -cos(x)
  4. Apply the formula:
    ∫x·sin(x) dx = x·(-cos(x)) - ∫(-cos(x)) dx = -x·cos(x) + ∫cos(x) dx = -x·cos(x) + sin(x) + C

4.3 Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC):

The FTC establishes the connection between differentiation and integration.

First Part (FTC Part I):

If f is continuous on [a, b], then the function F defined by:

F(x) = ∫axf(t) dt

is differentiable on (a, b), and F'(x) = f(x).

Second Part (FTC Part II):

If f is continuous on [a, b] and F is any antiderivative of f, then:

∫abf(x) dx = F(b) - F(a)

Example: Using the Fundamental Theorem

If F(x) = ∫0xt² dt, find F'(x).

  1. By the First Fundamental Theorem of Calculus:
    F'(x) = x²

Alternatively, we can solve this by evaluating the integral:

  1. Evaluate the indefinite integral: ∫t² dt = t³/3 + C
  2. Apply the definite integral: F(x) = ∫0xt² dt = [t³/3]0x = x³/3 - 0 = x³/3
  3. Now differentiate: F'(x) = d/dx(x³/3) = x²

4.4 Applications of Integrals

Key Applications of Integrals:

  1. Area Between Curves: Finding the area enclosed by two or more functions.
  2. Volume of Solids: Calculating the volume of three-dimensional objects.
  3. Arc Length: Determining the length of a curve.
  4. Surface Area: Finding the surface area of a solid of revolution.
  5. Work and Energy: Calculating work done by a variable force.
  6. Probability Distributions: Computing probabilities and expected values.

Example: Area Between Curves

Find the area between the curves y = x² and y = x + 2 over the interval where they intersect.

  1. Find the points of intersection by solving x² = x + 2:
    x² - x - 2 = 0 (x - 2)(x + 1) = 0 x = 2 or x = -1
  2. For x in [-1, 2], the curve y = x + 2 is above y = x², so the area is:
    Area = ∫-12[(x + 2) - x²] dx
  3. Expand and integrate:
    = ∫-12[x + 2 - x²] dx = [x²/2 + 2x - x³/3]-12
  4. Evaluate:
    = [(2²/2 + 2·2 - 2³/3) - ((-1)²/2 + 2·(-1) - (-1)³/3)] = [(2 + 4 - 8/3) - (1/2 - 2 + 1/3)] = [6 - 8/3 - 1/2 + 2 - 1/3] = [6 - 8/3 - 3/6 + 12/6 - 2/6] = [36/6 - 16/6 - 3/6 + 12/6 - 2/6] = [27/6] = 9/2

Example: Volume of a Solid of Revolution

Find the volume of the solid formed by rotating the region bounded by y = √x, y = 0, and x = 4 about the x-axis.

  1. Use the disk method: V = π∫ab[R(x)]² dx, where R(x) is the radius function
  2. In this case, R(x) = √x, so:
    V = π∫04(√x)² dx = π∫04x dx
  3. Integrate:
    V = π[x²/2]04 = π(4²/2 - 0²/2) = π(16/2) = 8π

Quick Quiz: Integrals

1. Evaluate ∫(2x³ - 5x + 3) dx
  • a) (1/2)x⁴ - (5/2)x² + 3x + C
  • b) (1/2)x⁴ - (5/2)x² + 3 + C
  • c) (1/2)x⁴ - (5/2)x² - 3x + C
  • d) (1/2)x⁴ - (5/2)x + 3x + C
2. Using substitution, evaluate ∫cos(2x) dx
  • a) (1/2)sin(2x) + C
  • b) sin(2x) + C
  • c) (1/2)cos(2x) + C
  • d) 2sin(2x) + C
3. Evaluate ∫01x·ex dx
  • a) e - 1
  • b) 1
  • c) e - 2
  • d) 0

5. Comprehensive Quiz

Test Your Calculus Knowledge

1. For the function f(x) = 3x⁴ - 4x³ + 2x - 5, which of the following statements is true?
  • a) f'(x) = 12x³ - 12x² + 2
  • b) The function has a critical point at x = 0
  • c) limx→∞ f(x)/x⁴ = 3
  • d) All of the above
2. The integral ∫-11|x| dx equals:
  • a) 0
  • b) 1
  • c) -1
  • d) 2
3. Using the limit definition, the derivative of f(x) = 1/x at x = 2 is:
  • a) -1/4
  • b) 1/4
  • c) -1
  • d) 1
4. The area of the region bounded by y = x², y = 0, x = 0, and x = 2 is:
  • a) 8/3
  • b) 4
  • c) 2/3
  • d) 4/3
5. If f(x) = x² and g(x) = x + 4, then the derivative of f(g(x)) at x = 1 is:
  • a) 5
  • b) 10
  • c) 2
  • d) 2x
6. For which value of x is the function f(x) = x³ - 6x² + 9x + 2 increasing?
  • a) x = 0
  • b) x = 1
  • c) x = 3
  • d) x = 4
7. Evaluate limx→0 (e^x - 1)/x
  • a) 0
  • b) 1
  • c) e
  • d) Does not exist
8. The volume of the solid obtained by rotating the region bounded by y = √x, y = 0, and x = 1 about the y-axis is:
  • a) π/2
  • b) π
  • c) 2π/3
  • d) π/3
9. Find the value of c in the Mean Value Theorem for the function f(x) = x³ over the interval [0, 2].
  • a) 2/√3
  • b) 2/3
  • c) √(2/3)
  • d) 4/3
10. The derivative of f(x) = ln(sin(x)) is:
  • a) cos(x)/sin(x)
  • b) 1/sin(x)
  • c) cot(x)
  • d) tan(x)
Tags:
Shares:

Leave a Reply

Your email address will not be published. Required fields are marked *