Riemann sums
The instrument covers the region under $f(x) = \sin x + 1.2$ on $[0, 5]$ with $n$ rectangles, each as tall as the curve at its left edge. As $n$ grows, what does the rectangle total do?
At $n = 2$ the left sum reads $7.4962$ — error $+0.7798$. Slide up: $6.9925$ at $n = 8$, $6.7946$ at $n = 30$, $6.7559$ at $n = 60$, against the exact $6.7163$. The estimate falls with every slice added, never dipping below the truth. At $n = 8$, Right gives $6.3932$ and Midpoint $6.7281$.
Each rectangle pretends $f$ is constant across its slice. This curve falls over most of $[0, 5]$, so the left edge is usually the tall edge: left rectangles carry spare area, and the total overshoots. Thinner slices shrink each overhang, so the estimate falls toward the truth; right edges undershoot for the mirror reason, midpoint nearly cancels. All three funnel to one number — the definition of the definite integral:
The notation remembers the construction: $\int$ is an elongated S for sum, and $dx$ is the slice width $\Delta x$ pushed to the limit. The sampling rule sets only the sign and speed of the error.
At $n = 8$ the readouts show left $6.9925$, midpoint $6.7281$, right $6.3932$ against exact $6.7163$. Why does the left sum overshoot on this curve?
Take $n = 5$ slices on $[0, 5]$, so $\Delta x = 1$. Which expression is the right Riemann sum?
Transfer: $g(x) = x$ on $[0, 4]$, right sampling. At $n = 4$ the rectangles have heights $1, 2, 3, 4$ and width $1$, so the sum is $10$; the exact area of the triangle is $8$. What does the right sum give at $n = 8$?
A definite integral is not a new kind of number — it is the one value that every rectangle scheme is squeezed toward as the slices thin. Left, right, and midpoint disagree at every finite $n$ and agree in the limit.