Area under a curve. Antiderivatives and Riemann sums.
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.”
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).
Derivatives answer questions like:
But many real questions are about accumulation:
These questions all share the same pattern:
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):
2) 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).
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.”
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.
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:
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 approximate accumulated value is:
∑ᵢ₌₁ⁿ f(xᵢ*) Δx
Think “height × width” for each rectangle, then add them.
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.
So a better mental model is:
Examples of “net accumulation”:
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.
2) 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.
3) Reversing bounds flips sign
∫ᵇₐ f(x) dx = −∫ₐᵇ f(x) dx
Why: you’re “accumulating backward,” so net accumulation reverses.
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.
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.
We write:
∫ f(x) dx = F(x) + C
where:
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.
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/x | ln | x | d/dx [ln | x | ] = 1/x | |
| eˣ | eˣ | derivative stays eˣ | ||||
| cos(x) | sin(x) | (sin x)′ = cos x | ||||
| sin(x) | −cos(x) | (−cos x)′ = sin x |
In ∫ f(x) dx, the dx tells you:
Example:
Same pattern, different variable.
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.).
So far we have two stories:
The Fundamental Theorem of Calculus says these are not separate topics—they are two sides of one idea.
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.
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).
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.
1) Find an antiderivative F(x) of f(x).
2) Evaluate F(b) − F(a).
This idea generalizes:
But the core intuition stays the same: add up infinitesimal contributions.
Compute ∫₀¹ x dx from the limit definition using a right-endpoint Riemann sum.
Partition [0, 1] into n equal pieces.
Δx = (1 − 0)/n = 1/n
xᵢ = i/n (right endpoints)
Form the Riemann sum:
∑ᵢ₌₁ⁿ f(xᵢ) Δx = ∑ᵢ₌₁ⁿ (xᵢ)(1/n)
= ∑ᵢ₌₁ⁿ (i/n)(1/n)
= ∑ᵢ₌₁ⁿ i / n²
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)
Take the limit:
limₙ→∞ (n+1)/(2n)
= limₙ→∞ (1/2)(1 + 1/n)
= 1/2
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.
Find ∫ (6x² − 4x + 5) dx and verify by differentiating.
Integrate term-by-term using linearity:
∫ (6x² − 4x + 5) dx
= ∫ 6x² dx − ∫ 4x dx + ∫ 5 dx
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
Combine and add the constant:
∫ (6x² − 4x + 5) dx = 2x³ − 2x² + 5x + C
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.
Compute ∫₂⁵ (3x² − 1) dx using an antiderivative.
Find an antiderivative:
∫ (3x² − 1) dx
= ∫ 3x² dx − ∫ 1 dx
= x³ − x + C
Apply FTC:
∫₂⁵ (3x² − 1) dx = [x³ − x]₂⁵
= (5³ − 5) − (2³ − 2)
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.”
A definite integral ∫ₐᵇ f(x) dx is defined as a limit of Riemann sums ∑ f(xᵢ*) Δx (net accumulation).
The symbol dx corresponds to an infinitesimal width in the integration variable; in sums it’s the Δx that goes to 0.
Definite integrals measure signed area: parts below the x-axis contribute negatively.
An indefinite integral ∫ f(x) dx is the set of all antiderivatives: F(x) + C where F′(x) = f(x).
The constant C is required because derivatives erase constants, so antiderivatives are never unique.
Linearity holds: integrals distribute over addition and pull out constants.
Fundamental Theorem of Calculus: if F′ = f, then ∫ₐᵇ f(x) dx = F(b) − F(a), connecting accumulation with net change.
Forgetting the +C in an indefinite integral (losing the whole family of antiderivatives).
Treating ∫ₐᵇ f(x) dx as “area” even when f(x) is negative (it’s signed area / net accumulation).
Mixing up variables and differentials (e.g., writing ∫ f(x) dt without changing the function to f(t)).
Using FTC without evaluating at both endpoints (computing F(b) but forgetting −F(a)).
Compute ∫₀² (x + 1) dx and interpret it as net area.
Hint: Find an antiderivative F(x) and use F(2) − F(0).
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.
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).
∫ (1/x + 2cos(x)) dx = ∫ 1/x dx + 2∫ cos(x) dx
= ln|x| + 2sin(x) + C.
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.
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.