The integral is built from scratch — upper and lower sums squeezing the area under a curve. Getting the definition right is what makes the Fundamental Theorem (next unit) meaningful rather than magical.
"Area under a curve" sounds obvious until you try to define it for a wiggly function. Riemann's upper and lower sums make both "area" and "accumulated total" precise.
Any accumulated quantity is an integral: distance from speed, total cost from a rate, work in physics, the center of mass — and, crucially for statistics, probabilities for continuous distributions are areas under a density curve.
Archimedes was integrating in ~250 BC, finding areas by his "method of exhaustion." Riemann gave the modern definition in 1854. Half a century later Lebesgue rebuilt integration on measure theory — the foundation that makes modern probability and statistics possible.
Upper and lower sums measure the worst possible disagreement inside each subinterval. If the total disagreement can be forced below every $\varepsilon>0$, then the function has one well-defined area. This is why the definition works even before antiderivatives exist.
This is also the right way to think about why a function like the Dirichlet function fails: every interval contains rationals and irrationals, so the oscillation never shrinks.
For a bounded function $f$ and a partition $P=\{a=x_0<\dots<x_n=b\}$ of $[a,b]$, with $m_i=\inf_{[x_{i-1},x_i]}f$ and $M_i=\sup_{[x_{i-1},x_i]}f$, $$L(f,P)=\sum_i m_i\Delta x_i,\qquad U(f,P)=\sum_i M_i\Delta x_i.$$
Every lower sum is $\le$ every upper sum. Refining a partition raises lower sums and lowers upper sums, squeezing toward the area.
A bounded $f$ is integrable on $[a,b]$ if $\sup_P L(f,P)=\inf_P U(f,P)$. That common value is $\displaystyle\int_a^b f$.
A bounded $f$ is integrable iff for every $\varepsilon>0$ there is a partition $P$ with $U(f,P)-L(f,P)<\varepsilon$.
If $f$ is continuous on $[a,b]$, it is integrable there. (Monotone functions are integrable too.)
For integrable functions on $[a,b]$, linearity $\int_a^b(\alpha f+\beta g)=\alpha\!\int_a^b f+\beta\!\int_a^b g$; additivity $\int_a^b f=\int_a^c f+\int_c^b f$; monotonicity $f\le g\Rightarrow \int_a^b f\le\int_a^b g$; and $\left|\int_a^b f\right|\le\int_a^b|f|$.
• $m_i,M_i$ are the inf/sup over each subinterval, not the endpoint values.
• Integrability is about the gap $U-L$ closing, not about finding an
antiderivative.
• Not every bounded function is integrable; the definition really is needed.
Problems adapted from the MAT137 question bank by Alfonso Gracia-Saz, used under CC BY-SA 4.0. Write full solutions before opening the answer.
Using equal subintervals, show $\int_0^1 x^2\,dx=\tfrac13$.
Explain why $f(x)=c$ (constant) is integrable and $\int_a^b c=c(b-a)$.
If $f$ is integrable and $0\le f(x)\le 3$ on $[1,4]$, bound $\int_1^4 f$.
Which of these are partitions of $[0,2]$? $\{0,2\}$, $\{1,2\}$, $\{0.5,1,1.5\}$, $\{0,1,2\}$.
Given $\int_0^2 f=3$, $\int_0^4 f=9$, $\int_0^4 g=2$, compute $\int_2^4 f$, $\int_2^0 f$, and $\int_0^4(f-2g)$.
Express $\displaystyle\lim_{n\to\infty}\sum_{i=1}^n\frac1n\sin\frac in$ as a definite integral and evaluate it.
State the integrability criterion: a bounded $f$ is integrable on $[a,b]$ if and only if what holds?
Prove: if $a<b$ and $f$ is continuous on $[a,b]$, there is $c\in[a,b]$ with $f(c)=\dfrac{1}{b-a}\int_a^b f$.
Course coverage. This unit is part of the complete MAT137 topic map. It should cover the following items before you move on:
Use the Alfonso Gracia-Saz MAT137 materials as the primary local spine, with the current UofT single-variable calculus topic split as the completeness check. The added items are original study prompts, not copied textbook exercises.
The definition chain is: partitions approximate the area, upper and lower sums trap all tagged sums, and integrability means the trap can be made arbitrarily tight.
Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.
State the upper-lower sum criterion for Riemann integrability.
Compute $\int_a^b c\,dx$ from the definition.
Is the function $f=1$ at $0$ and $0$ elsewhere on $[0,1]$ Riemann integrable?
Explain why $\int_{-1}^1 x\,dx=0$ without antiderivatives.