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Definite Integrals & Riemann Sums

Partition · Sample · Sum · Take the Limit

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From Straight Edges to a Curved One

A straight-sided region has a simple area formula
But here the top edge is a curve
Elementary geometry formulas no longer apply
We need a new idea — the definite integral

Goal: starting from geometry and physics, see why the definite integral is needed — and the one idea behind it.

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Setting Up the Problem

Assume $f$ is continuous on $[a,b]$
and $f(x)\ge 0$, so the graph sits above the $x$-axis
The curve $y=f(x)$, the lines $x=a,b$, and the $x$-axis bound a region
We want the area of this curved region

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Trade the Curve for Straight Pieces

The whole is hard — so chop the interval into pieces
Hug the curve with many thin rectangles
Each thin curved strip ≈ one straight rectangle
Straight-for-curved, constant-for-varying

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Partition the Interval, Sample Each Piece

Partition: cut $[a,b]$ into $n$ subintervals
Each has width $\Delta x_i = x_i - x_{i-1}$
Sample: pick any $\xi_i \in [x_{i-1},x_i]$ in each
Let $f(\xi_i)$ be that rectangle’s height

$$ a = x_0 < x_1 < \cdots < x_n = b $$

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General Partitions and the Mesh

Partition points need not be evenly spaced
The $i$-th width is $\Delta x_i = x_i - x_{i-1}$
The mesh is $\lambda = \max_i \Delta x_i$
$\lambda \to 0$ means the partition is refined without bound

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Building a General Riemann Sum

Pick any point $\xi_i$ in each subinterval
Approximate $f$ on that piece by $f(\xi_i)$
Rectangle area: $f(\xi_i)\,\Delta x_i$
Add them: $\sum_{i=1}^{n} f(\xi_i)\,\Delta x_i$

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One Rectangle: Height × Width

Width = Δxᵢ (the subinterval length)
Height = f(ξᵢ) (the value at the sample point)
Area ≈ f(ξᵢ)·Δxᵢ

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The Riemann Sum: Area as a Sum

Add up every rectangle’s area
This approximates the curved region’s area
This sum is called a Riemann sum
More rectangles → a sharper approximation

$$ S_n = \sum_{i=1}^{n} f(\xi_i)\,\Delta x_i $$

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More Rectangles, Better Fit (n = 4)

n = 4: coarse — a clear over- or under-estimate

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Refining the Partition (n = 8)

n = 8: the staircase hugs the curve more closely

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Keep Refining (n = 20)

n = 20: nearly exact — the area approaches 1/3

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Let n → ∞: the Sum Becomes the Integral

As the rectangles keep refining, the fit sharpens
The Riemann sum tends to $\tfrac13$
The limit is the area under the curve

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From Approximate to Exact: Take the Limit

The partition is arbitrary; each $\xi_i$ is arbitrary
Key: as $\lambda \to 0$, $S_n$ has a limit
and that limit ignores how we partitioned or chose $\xi_i$
That limit is the definite integral

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The Riemann Definition of the Integral

Prerequisite: $f$ is bounded on $[a,b]$
If every Riemann sum tends to the same number $I$
then $f$ is integrable on $[a,b]$
and that limit $I$ is the definite integral

When the limit exists and is independent of the partition and the sample points, $f$ is integrable on $[a,b]$.

$$ \int_a^b f(x)\,\mathrm dx=\lim_{\lambda\to0}\sum_{i=1}^n f(\xi_i)\,\Delta x_i $$

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Distance as Area Under a Velocity Curve

Horizontal axis: time $t$; vertical axis: velocity $v$
Distance $s$ is the area under the velocity curve
Approximate it with rectangles, just as before
The same construction, in a physical guise

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Worked Example: $\int_0^1 x^2\,\mathrm{d}x$

Take $f(x)=x^2$ on $[0,1]$
$f$ is continuous, hence integrable
Any sample point works — we use right endpoints
Split $[0,1]$ into $n$ equal parts

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Rewrite the Riemann Sum as a Sum of Squares

$\Delta x = \tfrac1n$, right endpoint $\xi_i = \tfrac in$
Substitute, then factor out $\tfrac{1}{n^3}$

$$ S_n = \sum_{i=1}^{n}\left(\tfrac in\right)^2 \tfrac1n = \frac{1}{n^3}\sum_{i=1}^{n} i^2 $$

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Apply the Sum-of-Squares Formula

$\sum_{i=1}^{n} i^2 = \tfrac{n(n+1)(2n+1)}{6}$
Regroup into a product that is easy to limit

$$ S_n = \frac{(n+1)(2n+1)}{6n^2} = \frac16\left(1+\tfrac1n\right)\left(2+\tfrac1n\right) $$

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Take the Limit to Get the Value

As $n \to \infty$, $\tfrac1n \to 0$
$S_n \to \tfrac16\cdot 1\cdot 2 = \tfrac13$

$$ \int_0^1 x^2\,\mathrm{d}x = \frac13 $$

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Geometric Check: the Sample Choice Doesn’t Matter

Left endpoints: underestimate
Right endpoints: overestimate
Midpoints: most accurate
All three converge to the same limit, 1/3

Left sum (under)

Right sum (over)

Midpoint sum (best)

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One Thread: Partition — Sample — Sum — Limit

Partition: cut [a,b] into pieces
Sample: pick a point ξᵢ in each
Sum: Σ f(ξᵢ)Δxᵢ (the Riemann sum)
Limit: λ→0 gives the definite integral

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Next Time: You Won’t Always Need the Limit

Summing and limiting by definition is tedious
The Fundamental Theorem (Newton–Leibniz) is the shortcut
If F′(x)=f(x), then ∫ₐᵇ f = F(b)−F(a)
e.g. F(x)=x³/3 ⇒ ∫₀¹ x² = 1/3