The mathematical study of continuous change. Calculus provides the framework for modeling systems in motion, understanding rates of change, and computing areas and volumes with extraordinary precision.
Calculus is one of the greatest intellectual achievements in human history. Developed independently by Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century, calculus provides the mathematical language for describing change and accumulation. While algebra deals with static relationships, calculus gives us the tools to analyze quantities that are constantly in flux.
Calculus is divided into two major branches:
These two branches are intimately connected by the Fundamental Theorem of Calculus, which shows that differentiation and integration are inverse processes — one of the most beautiful results in all of mathematics.
At its heart, calculus tackles two fundamental questions:
The concept of a limit is the foundation upon which all of calculus is built. A limit describes what value a function approaches as the input approaches a certain point — even if the function never actually reaches that value.
We write:
This means: as x gets closer and closer to a (from both sides), the values of f(x) get closer and closer to L. The formal epsilon-delta definition states: for every ε > 0 there exists a δ > 0 such that if 0 < |x - a| < δ, then |f(x) - L| < ε.
Direct substitution gives 0/0 (indeterminate form).
Step 1: Factor the numerator: (x² - 4) = (x + 2)(x - 2)
Step 2: Cancel common factors: (x + 2)(x - 2)/(x - 2) = x + 2
Step 3: Now substitute: lim(x→2) (x + 2) = 2 + 2 = 4
If lim(x→a) f(x) = L and lim(x→a) g(x) = M, then:
Sometimes we need to consider what happens as x approaches a from only one direction:
A two-sided limit exists if and only if both one-sided limits exist and are equal.
We can also ask what happens as x grows without bound:
This means f(x) approaches L as x becomes arbitrarily large. If such a limit exists, the line y = L is called a horizontal asymptote.
Step 1: Divide numerator and denominator by x² (the highest power):
= lim(x→∞) (3 + 2/x)/(5 - 1/x²)
Step 2: As x → ∞, the terms 2/x and 1/x² approach 0:
= (3 + 0)/(5 - 0) = 3/5
A function f is continuous at a point a if three conditions are met:
Intuitively, a function is continuous if you can draw its graph without lifting your pen. Common continuous functions include polynomials, exponentials, sine, and cosine.
When evaluating a limit results in an indeterminate form such as 0/0 or ∞/∞, L'Hôpital's Rule provides a powerful technique:
provided the limit on the right side exists (or is ±∞).
Direct substitution gives 0/0 (indeterminate).
Apply L'Hôpital's Rule: Differentiate numerator and denominator separately:
lim(x→0) sin(x)/x = lim(x→0) cos(x)/1 = cos(0) = 1
This gives ∞/∞ (indeterminate).
Apply L'Hôpital's Rule:
lim(x→∞) ln(x)/x = lim(x→∞) (1/x)/1 = lim(x→∞) 1/x = 0
This tells us that x grows much faster than ln(x).
The derivative of a function measures how the function's output changes as its input changes. Geometrically, the derivative at a point gives the slope of the tangent line to the curve at that point.
The derivative of f(x) is defined as:
This is also called the limit definition of the derivative or the derivative from first principles. The notation f'(x) (Lagrange notation) is equivalent to dy/dx (Leibniz notation) or Df (Euler notation).
f'(x) = lim(h→0) [(x + h)² - x²] / h
= lim(h→0) [x² + 2xh + h² - x²] / h
= lim(h→0) [2xh + h²] / h
= lim(h→0) (2x + h)
= 2x
Rather than using the limit definition every time, we use these established rules:
The derivative of any constant is zero.
This is the most frequently used derivative rule. It works for any real number exponent n.
d/dx [x⁵] = 5x⁴
d/dx [x⁻³] = -3x⁻⁴
d/dx [√x] = d/dx [x^(1/2)] = (1/2)x^(-1/2) = 1/(2√x)
Let f(x) = x² and g(x) = sin(x)
f'(x) = 2x, g'(x) = cos(x)
y' = 2x · sin(x) + x² · cos(x)
f(x) = x² + 1, g(x) = x - 3
f'(x) = 2x, g'(x) = 1
y' = [2x(x - 3) - (x² + 1)(1)] / (x - 3)²
= [2x² - 6x - x² - 1] / (x - 3)²
= (x² - 6x - 1) / (x - 3)²
The chain rule is used to differentiate composite functions — functions within functions:
In Leibniz notation: dy/dx = (dy/du) · (du/dx)
Let u = 3x + 1 so that y = u⁵
dy/du = 5u⁴, du/dx = 3
dy/dx = 5(3x + 1)⁴ · 3 = 15(3x + 1)⁴
Outer function: sin(u), inner function: u = x²
dy/dx = cos(x²) · 2x = 2x cos(x²)
When y is not explicitly defined as a function of x, we differentiate both sides of the equation with respect to x, treating y as a function of x and applying the chain rule:
Differentiate both sides with respect to x:
2x + 2y · (dy/dx) = 0
2y · (dy/dx) = -2x
dy/dx = -x/y
Derivatives have an enormous range of practical applications. They allow us to analyze the behavior of functions, solve optimization problems, and model real-world phenomena.
The derivative f'(x) gives the instantaneous rate of change of f at x. This is the foundation of physics, engineering, and economics.
Velocity: v(t) = s'(t) = 3t² - 12t + 9
v(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 (moving backward)
Acceleration: a(t) = v'(t) = 6t - 12
a(2) = 6(2) - 12 = 0 (no acceleration at this instant)
Critical points occur where f'(x) = 0 or f'(x) is undefined. To classify them:
Step 1: Find critical points: f'(x) = 3x² - 3 = 0 → x² = 1 → x = ±1
Step 2: Second derivative test: f''(x) = 6x
At x = -1: f''(-1) = -6 < 0 → local maximum, f(-1) = -1 + 3 + 2 = 4
At x = 1: f''(1) = 6 > 0 → local minimum, f(1) = 1 - 3 + 2 = 0
Optimization uses derivatives to find the maximum or minimum value of a function subject to certain constraints. The typical approach is:
Setup: Let x = width and y = length. Perimeter: 2x + 2y = 200, so y = 100 - x
Objective function: Area A = x · y = x(100 - x) = 100x - x²
Differentiate: A'(x) = 100 - 2x
Set equal to zero: 100 - 2x = 0 → x = 50
Therefore: y = 100 - 50 = 50
The maximum area is 50 × 50 = 2,500 m² (a square!)
Verify: A''(x) = -2 < 0, confirming this is a maximum.
Related rates problems involve finding how fast one quantity changes when a related quantity is changing. The key technique is to differentiate an equation relating the variables with respect to time (t).
Known: dV/dt = 100 cm³/s, r = 5 cm
Relationship: V = (4/3)πr³
Differentiate with respect to t: dV/dt = 4πr² · (dr/dt)
Solve for dr/dt: dr/dt = (dV/dt) / (4πr²) = 100 / (4π · 25) = 100/(100π)
dr/dt = 1/π ≈ 0.318 cm/s
Integration is the reverse process of differentiation. While derivatives measure rates of change, integrals measure accumulation. Geometrically, the definite integral computes the signed area between a function and the x-axis.
An antiderivative of f(x) is a function F(x) such that F'(x) = f(x). We write:
where C is the constant of integration. The constant is necessary because many functions share the same derivative (e.g., x², x² + 5, and x² - 3 all have derivative 2x).
= 3 · x⁵/5 - 2 · x²/2 + 7x + C
= (3/5)x⁵ - x² + 7x + C
A definite integral has specific upper and lower bounds and produces a numerical value:
where F is any antiderivative of f. The expression F(b) - F(a) is often written as [F(x)]ₐᵇ.
Step 1: Find the antiderivative: F(x) = x² + x
Step 2: Evaluate: F(3) - F(1) = (9 + 3) - (1 + 1) = 12 - 2 = 10
Many functions cannot be integrated using basic rules alone. The following techniques extend our ability to evaluate integrals significantly.
Substitution is the integration counterpart of the chain rule. We identify a part of the integrand as u, compute du, and rewrite the integral in terms of u.
Let u = x², then du = 2x dx
∫ 2x · cos(x²) dx = ∫ cos(u) du = sin(u) + C = sin(x²) + C
Let u = x² + 1, then du = 2x dx, so x dx = du/2
∫ x/(x² + 1) dx = (1/2) ∫ (1/u) du = (1/2) ln|u| + C = (1/2) ln(x² + 1) + C
Integration by parts is the integration counterpart of the product rule:
Choose u and dv using the LIATE priority: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.
Choose: u = x (algebraic), dv = eˣ dx
Then: du = dx, v = eˣ
∫ x · eˣ dx = x · eˣ - ∫ eˣ dx = x · eˣ - eˣ + C = eˣ(x - 1) + C
Choose: u = ln(x), dv = dx
Then: du = (1/x) dx, v = x
∫ ln(x) dx = x · ln(x) - ∫ x · (1/x) dx = x · ln(x) - ∫ 1 dx
= x ln(x) - x + C
This technique decomposes a rational function into simpler fractions that are easier to integrate. It applies when the degree of the numerator is less than the degree of the denominator.
Decompose: 1/[(x - 1)(x + 2)] = A/(x - 1) + B/(x + 2)
Multiply both sides by (x - 1)(x + 2): 1 = A(x + 2) + B(x - 1)
Let x = 1: 1 = 3A → A = 1/3
Let x = -2: 1 = -3B → B = -1/3
∫ 1/[(x - 1)(x + 2)] dx = (1/3) ∫ 1/(x - 1) dx - (1/3) ∫ 1/(x + 2) dx
= (1/3) ln|x - 1| - (1/3) ln|x + 2| + C
For integrals involving √(a² - x²), √(a² + x²), or √(x² - a²), use:
The Fundamental Theorem of Calculus (FTC) is arguably the most important theorem in calculus. It establishes the profound connection between differentiation and integration, showing they are inverse processes.
If f is continuous on [a, b] and we define:
then F is differentiable on (a, b) and:
In other words, the derivative of the integral of f is f itself. Integration and differentiation are inverse operations.
By FTC Part 1, the answer is simply: x³ + 2x
(Replace t with x in the integrand.)
The upper limit is x² (not just x), so we need the chain rule:
d/dx [∫₀ˣ² eᵗ dt] = e^(x²) · d/dx[x²] = 2x · e^(x²)
If f is continuous on [a, b] and F is any antiderivative of f, then:
This theorem tells us that to evaluate a definite integral, we only need to find an antiderivative and compute the difference of its values at the endpoints. This is the formula we use in practice when computing definite integrals.
Antiderivative: F(x) = -cos(x)
∫₀^π sin(x) dx = [-cos(x)]₀^π = -cos(π) - (-cos(0))
= -(-1) - (-1) = 1 + 1 = 2
Integration has countless applications in science, engineering, and mathematics. Here are some of the most important geometric and physical applications.
The area between two curves y = f(x) and y = g(x) from x = a to x = b, where f(x) ≥ g(x), is:
On [0, 1], x ≥ x² (the line is above the parabola).
A = ∫₀¹ (x - x²) dx = [x²/2 - x³/3]₀¹
= (1/2 - 1/3) - (0 - 0) = 1/6
When rotating a region around an axis, each cross-section is a disk:
where R(x) is the distance from the curve to the axis of rotation.
V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴
= π (16/2 - 0) = 8π
When there is a hole in the solid (rotating the area between two curves):
where R(x) is the outer radius and r(x) is the inner radius.
An alternative approach using cylindrical shells:
This method is especially useful when rotating about the y-axis.
V = 2π ∫₀² x · x² dx = 2π ∫₀² x³ dx
= 2π [x⁴/4]₀² = 2π (16/4) = 8π
The length of a curve y = f(x) from x = a to x = b is:
f'(x) = (3/2)x^(1/2), so [f'(x)]² = (9/4)x
L = ∫₀⁴ √(1 + 9x/4) dx
Let u = 1 + 9x/4, du = 9/4 dx, dx = 4/9 du
When x = 0: u = 1. When x = 4: u = 10.
L = (4/9) ∫₁¹⁰ √u du = (4/9) · [2u^(3/2)/3]₁¹⁰
= (8/27)[10^(3/2) - 1] = (8/27)(10√10 - 1) ≈ 9.07
The average value of f on [a, b] is:
f_avg = (1/3) ∫₀³ x² dx = (1/3)[x³/3]₀³ = (1/3)(27/3) = (1/3)(9) = 3
A sequence is an ordered list of numbers, and a series is the sum of the terms of a sequence. Infinite series are crucial in calculus because they allow us to represent functions as sums of simpler terms.
A sequence {aₙ} is a function from the positive integers to the real numbers. We say a sequence converges to L if:
If no such limit exists, the sequence diverges.
aₙ = 1/n: {1, 1/2, 1/3, 1/4, ...} → converges to 0
aₙ = (-1)ⁿ: {-1, 1, -1, 1, ...} → diverges (oscillates)
aₙ = (n + 1)/n: {2, 3/2, 4/3, ...} → converges to 1
An infinite series is a sum of infinitely many terms:
The series converges if the sequence of partial sums Sₙ = a₁ + a₂ + ... + aₙ converges to a finite limit.
A geometric series converges if and only if |r| < 1.
Here a = 1 and r = 1/2. Since |1/2| < 1, the series converges:
Sum = 1/(1 - 1/2) = 1/(1/2) = 2
The special case p = 1 gives the harmonic series ∑ 1/n, which diverges.
To determine whether a series converges, use these tests:
If lim(n→∞) aₙ ≠ 0, then the series ∑ aₙ diverges.
If f is positive, continuous, and decreasing on [1, ∞) and aₙ = f(n), then ∑ aₙ and ∫₁^∞ f(x) dx either both converge or both diverge.
If 0 ≤ aₙ ≤ bₙ for all n:
Apply the Ratio Test:
|aₙ₊₁/aₙ| = [(n + 1)/2ⁿ⁺¹] / [n/2ⁿ] = (n + 1)/(2n)
lim(n→∞) (n + 1)/(2n) = 1/2 < 1
The series converges by the Ratio Test.
An alternating series ∑ (-1)ⁿ bₙ converges if:
A power series centered at a is:
Every power series has a radius of convergence R. The series converges for |x - a| < R and diverges for |x - a| > R.
The Taylor series of a function f(x) centered at x = a is:
A Maclaurin series is a Taylor series centered at a = 0.
f(x) = eˣ, so f⁽ⁿ⁾(x) = eˣ for all n, and f⁽ⁿ⁾(0) = 1 for all n.
eˣ = 1 + x + x²/2! + x³/3! + ...
= 1 + x + x²/2 + x³/6 + ...
This gives us the polynomial approximation e ≈ 1 + 1 + 1/2 + 1/6 = 2.667 (actual value ≈ 2.718).
So far, we have studied functions of a single variable. Multivariable calculus extends these ideas to functions of two or more variables, such as f(x, y) or f(x, y, z). This is essential for modeling real-world phenomena that depend on multiple factors — temperature across a surface, fluid flow in three dimensions, or profit as a function of multiple inputs.
A function of two variables, z = f(x, y), maps each ordered pair (x, y) in its domain to a real number z. The graph of such a function is a surface in three-dimensional space.
f(x, y) = x² + y² describes a paraboloid — a bowl-shaped surface.
f(1, 2) = 1 + 4 = 5 (the point (1, 2, 5) lies on the surface)
f(0, 0) = 0 (the vertex of the paraboloid is at the origin)
A partial derivative measures how a multivariable function changes when we vary one variable while holding the others constant.
The symbol ∂ (partial) is used instead of d to indicate that other variables are being held constant.
∂f/∂x (treat y as a constant): = 3x²y + 2y²
∂f/∂y (treat x as a constant): = x³ + 4xy - 1
We can take partial derivatives of partial derivatives:
The gradient of a function f(x, y) is a vector of its partial derivatives:
The gradient points in the direction of the steepest increase of the function, and its magnitude gives the rate of increase in that direction. This concept is fundamental in optimization, machine learning (gradient descent), and physics.
∂f/∂x = 2x + 3y, ∂f/∂y = 3x
∇f = (2x + 3y, 3x)
At (1, 2): ∇f(1, 2) = (2 + 6, 3) = (8, 3)
The function increases most steeply in the direction of the vector (8, 3).
Just as single integrals compute area, double integrals compute volume and triple integrals compute quantities in three dimensions.
This computes the volume under the surface z = f(x, y) over the region R.
Inner integral (with respect to y, treating x as constant):
∫₀¹ (x + 2y) dy = [xy + y²]₀¹ = (x · 1 + 1²) - 0 = x + 1
Outer integral (with respect to x):
∫₀² (x + 1) dx = [x²/2 + x]₀² = (4/2 + 2) - 0 = 4
Innermost (z): ∫₀¹ xyz dz = xy[z²/2]₀¹ = xy/2
Middle (y): ∫₀¹ xy/2 dy = (x/2)[y²/2]₀¹ = x/4
Outermost (x): ∫₀¹ x/4 dx = (1/4)[x²/2]₀¹ = 1/8
Just as substitution simplifies single integrals, we can change coordinate systems for multiple integrals: