Two symbols, one object AB & BC

The expression

$$\sum_{i=1}^{4} (3i - 1)$$

says: run $i$ through $1, 2, 3, 4$, evaluate $3i - 1$ each time, and add the four results. That is $2 + 5 + 8 + 11 = 26$. Four parts, four jobs:

1. The index $i$ is a counter. It takes whole-number values only, and it does not appear in the answer.
2. The lower bound says where the count starts. Starting at $0$ instead of $1$ adds a whole extra term.
3. The upper bound says where it stops, and the stop is inclusive.
4. The term is a formula in $i$, evaluated fresh at each value.

Two counting facts follow, and both get lost. A sum from $i = 1$ to $i = n$ has $n$ terms. A sum from $i = 2$ to $i = 5$ has $5 - 2 + 1 = 4$ terms, not $3$: subtracting the bounds counts gaps, and terms are one more than gaps.

Anything not carrying the index is a constant as far as the sum is concerned, so it can be pulled out front: $\sum_{i=1}^{n} c \cdot a_i = c\sum_{i=1}^{n} a_i$. This is what lets $\Delta x$ leave the sum when every rectangle has the same width, and it is exactly why an unequal-width sum cannot be written that way.

Put the pieces of Topic 6.2 into that notation. On $[a, b]$ with $n$ equal subintervals,

$$\Delta x = \frac{b - a}{n}, \qquad x_i = a + i\,\Delta x = a + i\,\frac{b - a}{n},$$

and the right Riemann sum is

$$R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x = \sum_{i=1}^{n} f\!\left(a + i\,\frac{b-a}{n}\right)\frac{b-a}{n}.$$

On $[1, 3]$ that reads $\sum_{i=1}^{n} f\!\left(1 + \frac{2i}{n}\right)\frac{2}{n}$. The $\frac{2}{n}$ appears twice and means two different things: once it is the width of a rectangle, and once it is the spacing between sample points. They are the same number because the subintervals are equal.

Three places this goes wrong.

1. Dropping the $a$. Writing $f\!\left(\frac{2i}{n}\right)$ samples from $[0, 2]$ instead of $[1, 3]$. The sample point starts at the left end of the interval and walks right; it does not start at zero.
2. Running the index from $0$. Starting at $i = 0$ makes it a left sum, since $x_0 = a$. That is a legitimate sum, and it is a different one.
3. Losing $\Delta x$. Without it the expression totals heights, which is not an area and has the wrong units.

More rectangles means thinner rectangles and a closer fit. If the limit exists as $n \to \infty$, and it does for every continuous function on a closed interval, that limit is the definite integral:

$$\int_{a}^{b} f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta x.$$

The notation is designed to remind you of the sum. The $\int$ is an elongated S for sum, $f(x)$ is the height, and $dx$ is what the width $\Delta x$ became. Whether the sample points are left, right or midpoints stops mattering in the limit, which is why the definition does not name one.

Reading a given limit backwards into an integral is a standard exam question, and it is a matching exercise on three items:

1. The factor that plays $\Delta x$ tells you $\frac{b - a}{n}$, and so the length of the interval.
2. The expression inside $f$ tells you the left endpoint: whatever constant sits beside the $i$ term is $a$.
3. Then $b = a + (\text{length})$, and $f$ is whatever the expression does to its argument.

For $\lim_{n\to\infty}\sum_{i=1}^{n} \frac{3}{n}\left(1 + \frac{3i}{n}\right)^{2}$: the width is $\frac{3}{n}$, so the interval has length $3$; the sample point is $1 + \frac{3i}{n}$, so $a = 1$ and $b = 4$; and the function squares its argument. The limit is $\int_{1}^{4} x^{2}\,dx$.

Four things about $\int_{a}^{b} f(x)\,dx$ are worth saying explicitly, because each one is a mark somewhere.

1. It is a number. Not a function, not a family of functions. $\int_{1}^{4} x^{2}\,dx = 21$, and there is nothing left to substitute into.
2. The variable is a placeholder. $\int_{0}^{2} f(x)\,dx$ and $\int_{0}^{2} f(t)\,dt$ are the same number, because the letter is consumed by the integration. That is why the variable is called a dummy, and it is what makes Topic 6.4's accumulation functions readable.
3. $a$ and $b$ are inputs. $\int_{2}^{7} f(x)\,dx$ integrates from $x = 2$ to $x = 7$. The $2$ is a place on the horizontal axis, not a value of $f$, and the integral is not "starting at a height of $2$".
4. The $dx$ is not decoration. It names which variable is being integrated, which matters the moment a second letter appears, and it carries the width that gives the answer its units.

The notation for an indefinite integral, $\int f(x)\,dx$ with no limits, looks almost identical and means something categorically different: a family of functions rather than a number. Topic 6.8 takes that up. For now, notice that the limits are the entire difference, and that a stray pair of them, or a missing pair, changes what kind of object you have written down.