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Trapping an area between staircases, when the trap closes, and a bounded function it never closes for.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute upper and lower Darboux sums for a monotone function on an equal partition, use the standard sum formulas to evaluate them and take the limit, explain why refining a partition raises the lower sum and lowers the upper one, state the definition of the Riemann integral as the meeting of the two and the criterion that $U - L$ can be made arbitrarily small, name the classes of function that are integrable and give a bounded one that is not, and say why the definition deliberately makes no mention of antiderivatives.
The area under a curve is not defined yet — school area formulas cover rectangles and triangles, and a curved boundary is outside them. The definition is built by trapping.
Take a partition $P$ of $[a,b]$: points $a = x_0 < x_1 < \cdots < x_n = b$ cutting it into pieces. On the $k$th piece let $m_k$ be the infimum of $f$ and $M_k$ the supremum. Then
$$L(f,P) = \sum m_k \Delta x_k, \qquad U(f,P) = \sum M_k \Delta x_k$$
are the lower and upper Darboux sums — rectangles that fit underneath and rectangles that cover over.
Whatever the area is, it is between them: $L(f,P) \le \text{area} \le U(f,P)$, for every $P$.
Refining helps. Adding a point to a partition raises $L$ and lowers $U$, because splitting a piece replaces one infimum by two at least as large. So the lower sums climb, the upper sums fall, and every lower sum is below every upper one — even for unrelated partitions, by comparing both with their common refinement.
The definition. $f$ is Riemann integrable on $[a,b]$ when $$\sup_P L(f,P) = \inf_P U(f,P),$$ and the common value is $\int_a^b f(x)\,dx$.
The criterion. Equivalently: for every $\varepsilon > 0$ there is a partition with $U - L < \varepsilon$. That is the form actually used, and it is the same $\varepsilon$ game as every other definition in this course.
Another way: picture
The region under a curve, with rectangles drawn inside it and rectangles drawn over it. The inner ones under-count, the outer over-count, and the true area is somewhere between. Cut the interval finer and the two staircases close on the curve from both sides; the integral exists when the gap between them can be made as small as you please.
Another way: steps
To compute a sum for $n$ equal pieces:
Three theorems cover everything this course meets, and each is proved by making $U - L$ small.
Continuous on $[a,b]$ $\Rightarrow$ integrable. A continuous function on a closed bounded interval is uniformly continuous, so a fine enough partition makes every oscillation $M_k - m_k$ below $\varepsilon/(b-a)$, and the gap below $\varepsilon$.
Monotone on $[a,b]$ $\Rightarrow$ integrable. With $n$ equal pieces the oscillations telescope: $\sum (M_k - m_k) = f(b) - f(a)$, so $U - L = \dfrac{b-a}{n}\bigl(f(b)-f(a)\bigr) \to 0$. No continuity is needed anywhere — a monotone function may have infinitely many jumps and is integrable regardless.
Bounded with finitely many discontinuities $\Rightarrow$ integrable. Cover the bad points with intervals of total length under $\varepsilon$ and handle the rest by continuity.
And a function that is not. The Dirichlet function, $1$ on the rationals and $0$ on the irrationals, has $M_k = 1$ and $m_k = 0$ on every subinterval, because both kinds of point are everywhere. So $U = b - a$ and $L = 0$ for every partition and the gap never closes. It is bounded, and it is not integrable — which is the limitation that the Lebesgue integral was built to remove.
Taking the left endpoint as the infimum without checking monotonicity. For an increasing function the left end gives the infimum; for a decreasing one it gives the supremum; for a function that turns, neither end need be either.
Confusing a Riemann sum with a Darboux sum. A Riemann sum uses any sample point in each piece; the Darboux sums use the extremes. Every Riemann sum lies between $L$ and $U$, which is why the two definitions agree.
Thinking more discontinuities always means not integrable. A monotone function can have infinitely many jumps and still be integrable. What matters is the total oscillation, not the count.
Forgetting that the integral is signed. Where $f < 0$ the rectangles have negative height, so $\int f$ is area above the axis minus area below. "The integral is the area" is true only for a non-negative integrand.
Assuming a formula is needed. $\int_0^1 e^{-x^2}dx$ has no elementary antiderivative and is a perfectly well-defined number, because this definition never mentions antiderivatives.
"Integration is the opposite of differentiation" is the summary everyone carries, and it is not the definition. This lesson defines $\int_a^b f$ with no derivative anywhere in sight: it is a number trapped between staircases, and its existence is a statement about whether the staircases meet.
That separation is deliberate, and two things depend on it.
Some integrals exist with no antiderivative to find. $\int_0^1 e^{-x^2}dx$ is a definite number — the staircases close — and no elementary function differentiates to $e^{-x^2}$. If the integral were defined as an antiderivative, this quantity would not exist.
The fundamental theorem has content. It says two separately defined things coincide: a limit of sums, and a difference of antiderivative values. If the second were the definition of the first, there would be nothing to prove and no theorem to be surprised by — and it is genuinely surprising that area and slope are inverse to one another.
$\int_0^1 x^2dx$ with $n$ equal pieces of width $1/n$. Increasing, so the upper sum takes right endpoints $k/n$.
Set up the partition.
$U = \sum_{k=1}^{n}\left(\dfrac{k}{n}\right)^2\dfrac1n = \dfrac{1}{n^3}\sum k^2 = \dfrac{1}{n^3}\cdot\dfrac{n(n+1)(2n+1)}{6}$.
The sum formula does the work.
$= \dfrac{(n+1)(2n+1)}{6n^2} \to \dfrac{2n^2}{6n^2} = \dfrac13$. The lower sum gives $\dfrac13$ too, so the integral is $\dfrac13$ — obtained with no antiderivative anywhere.
Both sums, same limit.
The Dirichlet function on $[0,1]$. Take any partition, and any subinterval of it, however short.
The argument works at every scale.
That subinterval contains a rational (where $f = 1$) and an irrational (where $f = 0$), because both are dense. So $M_k = 1$ and $m_k = 0$ on every piece.
Density is what does it.
$U = \sum 1 \cdot \Delta x_k = 1$ and $L = 0$, for every partition. The gap is $1$ always, so the function is not integrable — bounded, defined everywhere, and outside the theory.
Width $\tfrac14$; increasing, so the supremum on each piece is at its right end: $\tfrac14, \tfrac12, \tfrac34, 1$.
Right endpoints.
$U = \tfrac14\left(\tfrac14 + \tfrac12 + \tfrac34 + 1\right) = \tfrac14 \cdot \tfrac52 = \tfrac58$.
The exact integral is $\tfrac12$, and $\tfrac58 > \tfrac12$ as an upper sum must be. The excess $\tfrac18$ is $\tfrac{1}{2n}$, closing as $n$ grows.
Match each Riemann-sum observation to its consequence.
| Underestimates the signed integral | Overestimates the signed integral | Usually improves the approximation toward the integral | Rectangle contributions are negative | |
|---|---|---|---|---|
| Left-endpoint sum for an increasing function | ||||
| Right-endpoint sum for an increasing function | ||||
| Use more, narrower subintervals | ||||
| The function is below the axis |
Match each Riemann-sum fact to the conclusion it supports.
| A lower bound on the integral | An upper bound on the integral | Raises $L$ and lowers $U$ or leaves them unchanged | The function is Riemann integrable | |
|---|---|---|---|---|
| Use infimum heights on each subinterval | ||||
| Use supremum heights on each subinterval | ||||
| Refine an existing partition | ||||
| $U(f,P)-L(f,P)$ can be made arbitrarily small |
Find the upper Darboux sum $U(f, P)$ for $f(x) = 6x$ on $[\,0, 4\,]$, with $P$ the partition into $4$ equal subintervals.
Answer:
For $f(x) = 5x$ on $[\,0, 2\,]$ and $P$ the partition into $7$ equal pieces, find $U(f, P) - L(f, P)$.
Answer:
For $f(x) = x^2$ on $[\,0, 3\,]$ cut into $6$ equal pieces of width $1/2$, what is the lower sum $L(f, P)$?
Answer:
For $f(x) = 6x$ on $[\,0, 1\,]$, what single number satisfies $L(f, P) \le N \le U(f, P)$ for **every** partition $P$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $f(x) = x^2$ on $[\,0, 3\,]$, a partition gives $L = 55/8$ and $U = 91/8$. What does that tell you about $\int_{0}^{3} f$?
| Does not exceed the integral | Is not below the integral | Places the integral in $[L,U]$ | Can narrow the upper-lower gap | |
|---|---|---|---|---|
| Lower Darboux sum $L$ | ||||
| Upper Darboux sum $U$ | ||||
| One partition gives $L\le U$ | ||||
| Refining a partition |
You can compute Darboux sums, say when the gap closes, and name a bounded function that is not integrable. Say in your own words why the integral is defined without antiderivatives. Next: the definite integral and its properties.
8. Your turn: the upper sum for $f(x) = x$ on $[0, 1]$ with $4$ equal pieces, step 3