The limit of Riemann Sums is so important that it gets its own name and symbol. It measures area, but with a sign: parts below the axis count as negative.
Key idea
The definite integral of $f$ from $a$ to $b$ is $$\int_a^b f(x)\,dx = \lim_{n \to \infty}\sum_{i=1}^{n} f(x_i^*)\,\Delta x,$$ when the limit exists (it does for every continuous $f$). It equals the net signed area: area above the $x$-axis minus area below it.
Why it works
In a Riemann sum each term is $f(x_i^*)\,\Delta x$. Where $f$ is negative the term is negative, so rectangles below the axis subtract. The limit keeps that bookkeeping. The notation records the recipe: $\int$ is a stretched S for "sum", $f(x)\,dx$ is a height times a tiny width, and $a$, $b$ mark the interval.
Worked example: Triangles above and below
Evaluate $\displaystyle\int_0^3 (x - 1)\,dx$ using geometry.
- The line crosses the axis at $x = 1$.
- On $[0, 1]$: a triangle below the axis with base 1 and height 1, area $\frac{1}{2}$, counted as $-\frac{1}{2}$.
- On $[1, 3]$: a triangle above with base 2 and height 2, area 2.
- Net: $-\frac{1}{2} + 2 = \frac{3}{2}$.
Answer: $\frac{3}{2}$
Worked example: A semicircle
Evaluate $\displaystyle\int_{-2}^{2}\sqrt{4 - x^2}\,dx$.
- $y = \sqrt{4 - x^2}$ is the top half of the circle $x^2 + y^2 = 4$, radius 2.
- From $-2$ to $2$ the region is a half disk: $\frac{1}{2}\pi(2)^2$.
Answer: $2\pi$
Worked example: Symmetry
Evaluate $\displaystyle\int_{-3}^{3}(x^3 - 5x)\,dx$.
- $x^3 - 5x$ is odd: $f(-x) = -f(x)$, so its graph is symmetric through the origin.
- The signed area on $[-3, 0]$ exactly cancels the signed area on $[0, 3]$.
Answer: $0$
Common mistake
Treating every integral as total area and adding the part below the axis as positive. In the first example that gives $\frac{5}{2}$ instead of $\frac{3}{2}$. Total area is a different question; for it you integrate $\lvert f \rvert$. Check: sketch the graph and mark which pieces lie below the axis.