Integrals

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Integrals #

CalculusDifficulty: ★★☆☆☆Depth: 3Unlocks: 7

Area under a curve. Antiderivatives and Riemann sums.

Interactive Visualization #

⏮◀◀▶▶STEP0.25x1xZOOM

t=0s

Core Concepts #

Key Symbols & Notation #

Definite integral notation: "∫_a^b f(x) dx"Differential symbol: "dx" (the integration variable/infinitesimal width)

Essential Relationships #

Prerequisites (1) #

Derivatives6 atoms

Unlocks (5) #

Consumer Surpluslvl 3Fundamental Theorem of Calculuslvl 2Multiple Integralslvl 3Measure Theorylvl 5Stochastic Processeslvl 4

Advanced Learning Details

Graph Position #

34

Depth Cost

7

Fan-Out (ROI)

5

Bottleneck Score

3

Chain Length

Cognitive Load #

6

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33

Total Elements

L1

Percentile Level

L4

Atomic Level

All Concepts (15) #

Teaching Strategy #

Deep-dive lesson - accessible entry point but dense material. Use worked examples and spaced repetition.

Derivatives tell you how something changes at an instant. Integrals tell you what those tiny changes add up to over time or over a range. If derivatives are “speed right now,” integrals are “distance traveled.”

TL;DR:

A definite integral ∫ₐᵇ f(x) dx is the limit of Riemann sums and measures net accumulation (often area). An indefinite integral ∫ f(x) dx is an antiderivative F(x) + C. The Fundamental Theorem of Calculus links them: if F′(x) = f(x), then ∫ₐᵇ f(x) dx = F(b) − F(a).

What Is an Integral? #

Why integrals exist (motivation) #

Derivatives answer questions like:

But many real questions are about accumulation:

These questions all share the same pattern:

  1. 1)Break the interval into tiny pieces.
  2. 2)Approximate the contribution on each piece.
  3. 3)Add them up.
  4. 4)Take a limit as the pieces become infinitely fine.

That “add up tiny contributions and take a limit” is the central idea of integration.

There are two closely related notions:

  1. Definite integral (a number):
  1. Indefinite integral (a family of functions):

It’s common to say “the integral is the area under the curve,” but that’s only fully accurate when f(x) ≥ 0. In general, a definite integral measures signed area (area above the x-axis counts positive, below counts negative).

What does dx mean? #

In ∫ₐᵇ f(x) dx:

In Riemann-sum language, dx corresponds to Δx in the limit as Δx → 0.

A key habit: treat f(x) as the “height” and dx as the “infinitesimal width,” so f(x) dx is an infinitesimal “slice of area/accumulation.”

A first geometric picture #

Imagine approximating the region under y = f(x) from x = a to x = b by rectangles.

As Δx gets smaller, this approximation improves. The limit (if it exists) is the definite integral.

Core Mechanic 1: Definite Integrals as Limits of Riemann Sums #

Why a limit definition matters #

If we only said “area under a curve,” we’d be stuck with vague pictures. The power of calculus is that it turns pictures into precise computations.

The definite integral is defined as a limit of sums, which makes it:

Partitioning the interval #

Take an interval [a, b] and split it into n subintervals.

Now, on each subinterval [xᵢ₋₁, xᵢ], pick a sample point xᵢ*.

Common choices:

The Riemann sum #

The approximate accumulated value is:

∑ᵢ₌₁ⁿ f(xᵢ*) Δx

Think “height × width” for each rectangle, then add them.

The limit definition of the definite integral #

If the limit exists as n → ∞ (and thus Δx → 0), we define:

∫ₐᵇ f(x) dx = limₙ→∞ ∑ᵢ₌₁ⁿ f(xᵢ*) Δx

This is the definite integral.

Signed area and interpretation #

So a better mental model is:

Examples of “net accumulation”:

Properties you can trust (and why they’re reasonable) #

These follow naturally from sums and limits:

  1. Linearity

∫ₐᵇ (αf(x) + βg(x)) dx = α∫ₐᵇ f(x) dx + β∫ₐᵇ g(x) dx

Why: sums distribute over addition and scalar multiplication.

  1. Additivity over intervals

If a < c < b, then:

∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx

Why: splitting an interval splits the corresponding sum.

  1. Reversing bounds flips sign

∫ᵇₐ f(x) dx = −∫ₐᵇ f(x) dx

Why: you’re “accumulating backward,” so net accumulation reverses.

A note on existence (at this level) #

Most functions you meet early on (polynomials, exponentials, trig functions, continuous functions) are integrable on closed intervals. You don’t need the advanced theory yet—just know the Riemann-sum definition is what makes everything precise.

Core Mechanic 2: Indefinite Integrals as Antiderivatives (and the +C) #

Why antiderivatives matter #

Riemann sums are conceptually foundational, but not a practical way to compute integrals by hand.

So we look for a function whose derivative gives the integrand.

If F′(x) = f(x), then F is an antiderivative of f.

Indefinite integral definition #

We write:

∫ f(x) dx = F(x) + C

where:

Why the constant C must be there #

If F′(x) = f(x), then (F(x) + C)′ = F′(x) + 0 = f(x).

So there isn’t just one antiderivative—there’s a whole family of them.

Geometrically, adding C shifts a graph up or down without changing slopes.

Common antiderivative patterns (starter kit) #

You already know derivative rules; integration reverses them.

f(x)One antiderivative F(x)Check by differentiating
xⁿ (n ≠ −1)xⁿ⁺¹/(n+1)d/dx [xⁿ⁺¹/(n+1)] = xⁿ
1/xlnx
eˣeˣderivative stays eˣ
cos(x)sin(x)(sin x)′ = cos x
sin(x)−cos(x)(−cos x)′ = sin x

The “dx” in indefinite integrals #

In ∫ f(x) dx, the dx tells you:

Example:

Same pattern, different variable.

Quick sanity check habit #

After you integrate, differentiate your answer.

If you don’t get back the integrand, something went wrong.

This is especially useful because integration has more “gotchas” than differentiation (constants, absolute values for ln|x|, etc.).

Application/Connection: The Fundamental Theorem of Calculus (FTC) #

Why the FTC is the bridge #

So far we have two stories:

The Fundamental Theorem of Calculus says these are not separate topics—they are two sides of one idea.

FTC (evaluation form) #

If f is continuous on [a, b] and F is any antiderivative of f (so F′(x) = f(x)), then:

∫ₐᵇ f(x) dx = F(b) − F(a)

This is the main computational tool for definite integrals.

Why this makes sense (intuition before algebra) #

Think of accumulation as “total change.”

If F′(x) = f(x), then f describes how F changes.

So accumulating f from a to b should give the net change in F from a to b.

That net change is F(b) − F(a).

Mini-derivation sketch (connecting to Riemann sums) #

Start with a partition a = x₀ < x₁ < … < xₙ = b.

Because F′ = f, changes in F over each subinterval are approximately:

F(xᵢ) − F(xᵢ₋₁) ≈ f(xᵢ*) (xᵢ − xᵢ₋₁)

Now add over i = 1…n:

∑ᵢ₌₁ⁿ (F(xᵢ) − F(xᵢ₋₁)) ≈ ∑ᵢ₌₁ⁿ f(xᵢ*) Δxᵢ

The left side telescopes:

(F(x₁) − F(x₀)) + (F(x₂) − F(x₁)) + … + (F(xₙ) − F(xₙ₋₁))

= F(xₙ) − F(x₀)

= F(b) − F(a)

As the partition gets finer (max Δxᵢ → 0), the approximation becomes exact, and the right side becomes the integral:

F(b) − F(a) = ∫ₐᵇ f(x) dx

That’s the FTC in action.

Practical workflow for computing a definite integral #

  1. Find an antiderivative F(x) of f(x).

  2. Evaluate F(b) − F(a).

Interpreting results #

Looking ahead #

This idea generalizes:

But the core intuition stays the same: add up infinitesimal contributions.

Worked Examples (3) #

Riemann sum → definite integral for f(x) = x on [0, 1] #

Compute ∫₀¹ x dx from the limit definition using a right-endpoint Riemann sum.

  1. Partition [0, 1] into n equal pieces.

    Δx = (1 − 0)/n = 1/n

    xᵢ = i/n (right endpoints)

  2. Form the Riemann sum:

    ∑ᵢ₌₁ⁿ f(xᵢ) Δx = ∑ᵢ₌₁ⁿ (xᵢ)(1/n)

    = ∑ᵢ₌₁ⁿ (i/n)(1/n)

    = ∑ᵢ₌₁ⁿ i / n²

  3. Use the formula ∑ᵢ₌₁ⁿ i = n(n+1)/2:

    ∑ᵢ₌₁ⁿ i / n² = (1/n²) · n(n+1)/2

    = (n(n+1)) / (2n²)

    = (n+1)/(2n)

  4. Take the limit:

    limₙ→∞ (n+1)/(2n)

    = limₙ→∞ (1/2)(1 + 1/n)

    = 1/2

  5. Therefore:

    ∫₀¹ x dx = 1/2

Insight: The “area under y = x from 0 to 1” is a triangle with area 1/2, and the Riemann-sum limit matches the geometry. The limit definition turns that geometric fact into a general computation method.

Indefinite integral as an antiderivative (and checking by differentiation) #

Find ∫ (6x² − 4x + 5) dx and verify by differentiating.

  1. Integrate term-by-term using linearity:

    ∫ (6x² − 4x + 5) dx

    = ∫ 6x² dx − ∫ 4x dx + ∫ 5 dx

  2. Apply the power rule for integrals (reverse of derivative power rule):

    ∫ 6x² dx = 6 · x³/3 = 2x³

    ∫ 4x dx = 4 · x²/2 = 2x²

    ∫ 5 dx = 5x

  3. Combine and add the constant:

    ∫ (6x² − 4x + 5) dx = 2x³ − 2x² + 5x + C

  4. Check by differentiating:

    d/dx [2x³ − 2x² + 5x + C]

    = 6x² − 4x + 5 + 0

    = 6x² − 4x + 5

    Matches the integrand.

Insight: Indefinite integration is “find a function whose slope is the given function.” The +C is not optional; it represents the whole family of functions with the same derivative.

Using the FTC to evaluate a definite integral quickly #

Compute ∫₂⁵ (3x² − 1) dx using an antiderivative.

  1. Find an antiderivative:

    ∫ (3x² − 1) dx

    = ∫ 3x² dx − ∫ 1 dx

    = x³ − x + C

  2. Apply FTC:

    ∫₂⁵ (3x² − 1) dx = [x³ − x]₂⁵

    = (5³ − 5) − (2³ − 2)

  3. Compute:

    (125 − 5) − (8 − 2)

    = 120 − 6

    = 114

Insight: Riemann sums define the definite integral, but the FTC is what makes it practical: convert a “limit of sums” into “evaluate an antiderivative at endpoints.”

Key Takeaways #

Common Mistakes #

Practice #

easy

Compute ∫₀² (x + 1) dx and interpret it as net area.

Hint: Find an antiderivative F(x) and use F(2) − F(0).

Show solution

Antiderivative: ∫ (x + 1) dx = x²/2 + x + C.

Evaluate:

∫₀² (x + 1) dx = [x²/2 + x]₀² = (4/2 + 2) − 0 = 2 + 2 = 4.

Since x + 1 ≥ 0 on [0, 2], this equals the ordinary area under the curve.

medium

Find the most general antiderivative: ∫ (1/x + 2cos(x)) dx.

Hint: Use ln|x| for ∫ 1/x dx and remember ∫ cos(x) dx = sin(x).

Show solution

∫ (1/x + 2cos(x)) dx = ∫ 1/x dx + 2∫ cos(x) dx

= ln|x| + 2sin(x) + C.

hard

Use a right-endpoint Riemann sum with n subintervals to compute ∫₀¹ x² dx (set it up and take the limit).

Hint: With Δx = 1/n and xᵢ = i/n, the sum becomes (1/n)∑ (i/n)². Use ∑ i² = n(n+1)(2n+1)/6.

Show solution

Partition [0, 1] into n equal parts: Δx = 1/n, right endpoints xᵢ = i/n.

Riemann sum:

∑ᵢ₌₁ⁿ f(xᵢ)Δx = ∑ᵢ₌₁ⁿ (i/n)² (1/n)

= (1/n³) ∑ᵢ₌₁ⁿ i².

Use ∑ᵢ₌₁ⁿ i² = n(n+1)(2n+1)/6:

(1/n³) · n(n+1)(2n+1)/6

= (n+1)(2n+1) / (6n²).

Take the limit:

limₙ→∞ (n+1)(2n+1)/(6n²)

= limₙ→∞ (2n² + 3n + 1)/(6n²)

= 2/6

= 1/3.

Therefore ∫₀¹ x² dx = 1/3.

Connections #

Quality: B (4.1/5)

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