How much area lies under a curve? For rectangles and triangles we have formulas; for a curved boundary we do not. The plan: cover the region with thin rectangles and add their areas. Sigma Notation keeps the sums tidy, and limits will later make them exact.
Key idea
Split $[a, b]$ into $n$ equal strips of width $\Delta x = \frac{b - a}{n}$, with endpoints $x_i = a + i\,\Delta x$. The left sum uses the heights at the left ends, $L_n = \sum_{i=0}^{n-1} f(x_i)\,\Delta x$; the right sum uses the right ends, $R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x$.
Why it works
Each rectangle's area is height times width, $f(x_i)\,\Delta x$. Over a thin strip the curve barely changes, so the rectangle is close to the true slice of area. Thinner strips make the error smaller. For a rising curve, left heights are the lowest point of each strip (an underestimate) and right heights are the highest (an overestimate), so the true area is trapped between them.
Worked example: Left and right sums
Estimate the area under $f(x) = x^2$ on $[0, 2]$ with $n = 4$, using left and right endpoints.
- $\Delta x = \frac{2 - 0}{4} = 0.5$. Endpoints: $0, 0.5, 1, 1.5, 2$.
- Left (the first four): $$0.5\,(0 + 0.25 + 1 + 2.25) = 0.5 \cdot 3.5 = 1.75$$.
- Right (the last four): $$0.5\,(0.25 + 1 + 2.25 + 4) = 0.5 \cdot 7.5 = 3.75$$.
Answer: $L_4 = 1.75$ and $R_4 = 3.75$. The true area, $\frac{8}{3}$, lies between them.
Worked example: From data
A car's speed is recorded every 10 seconds: 20, 26, 30, 31, 33 ft/s at $t = 0, 10, 20, 30, 40$. Estimate the distance traveled with a left sum.
- Distance = area under the speed graph. Width 10 s.
- Left endpoints use the first four readings: $10\,(20 + 26 + 30 + 31) = 10 \cdot 107 = 1070$.
Answer: About 1070 ft (the right sum, 1200 ft, gives an upper estimate here).
Common mistake
Using the wrong endpoints, for example all five readings, or starting the right sum at $x = a$. Left sums use $x_0$ through $x_{n-1}$; right sums use $x_1$ through $x_n$. Check: count your heights. There must be exactly $n$ of them, one per strip.