Estimating Area with Slices

When a quantity changes, slicing it into thin rectangles turns a hard question into arithmetic.

The Question

How much area sits between a curve and the x-axis over an interval? For a constant height it is just length times width, but a curve changes height at every point. The fix is to freeze the curve for a moment, treat that height as constant over a tiny slice, and add the slices up.

A \approx \sum f(x_i^*)\,\Delta x

Riemann Sums

Cut the interval from a to b into n equal pieces, so each has width Delta x = (b-a)/n. On each piece pick a sample point, draw a rectangle of that height, and add all n rectangles. Left endpoints undershoot an increasing curve and overshoot a decreasing one; more, thinner rectangles shrink the error either way.

\Delta x = \frac{b-a}{n}

A Worked Slice

Estimate the area under f(x) = x^2 on [0, 2] using 4 rectangles with right endpoints. Delta x = 0.5 and the sample heights are 0.25, 1, 2.25, and 4, giving a sum of 3.75. The exact answer is 8/3, about 2.67, so the overestimate is already shrinking as rectangles get thinner.

S_4 = 0.5(0.25 + 1 + 2.25 + 4) = 3.75

Flashcard

\sum_{i=1}^{n} f(x_i^*)\,\Delta x

A Riemann sum approximating area

Adds n rectangle areas of width Delta x and heights sampled from the curve. As n grows the sum approaches the exact area under the curve between a and b.

Check your understanding

Using right endpoints to approximate an increasing function with rectangles always produces what kind of estimate?

Take the unit quiz

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