Antiderivatives answer "what function had this as its slope." A completely different-looking question asks what the exact area trapped under this curve is. Answering it turns out to need its own idea before it can even be stated precisely. Chop the region into thin rectangles, add up their areas, and ask what happens as the rectangles get thinner and more numerous. That sum is a Riemann sum, and the definite integral is simply its limit.
Think of tiling a curved swimming pool's floor with rectangular tiles to estimate its area. Wide tiles leave visible gaps and overhangs wherever the wall curves. Thinner tiles fit the curve more closely. In the limit of infinitely many, infinitesimally thin tiles, the tiling error vanishes completely. What's left is the pool's actual area, not an approximation of it anymore.
A Riemann sum is the name for the "chop into thin pieces and add" approximation: partition [a,b] into n equal pieces of width \Delta x = \frac{b-a}{n} and sum f(x_i^*)\cdot \Delta x over some sample point x_i^* in each piece, approximating \int_a^b f(x)\,dx. The LEFT endpoint, the RIGHT endpoint, and the MIDPOINT are the three standard choices for x_i^*. Each one systematically over- or under-shoots on a curve that's consistently increasing, decreasing, or curved. That direction is predictable, and it's worth being able to name before trusting any single one. The definite integral, \int_a^b f(x)\,dx, is defined as the limit of this sum as n\to\infty. It is not approximated by that limit; it is DEFINED by it, for any continuous f.
Watch left, right, and midpoint sums converge
Try left vs. right at small n, then slide n up — both close in on the same true value from opposite sides.
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Trapezoids fit a curve better than flat-topped rectangles
Compare a trapezoid sum to left/right at the same n.
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Worked example
A left-endpoint Riemann sum by hand
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1
Approximate \displaystyle\int_0^4 x^2\,dx using a left-endpoint sum with n=4 rectangles.
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\Delta x = \frac{4-0}{4} = 1, so the left endpoints are x = 0, 1, 2, 3.
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Sum f(x)\cdot\Delta x at each: f(0)\cdot1 + f(1)\cdot1 + f(2)\cdot1 + f(3)\cdot1 = 0 + 1 + 4 + 9 = 14.
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Since x^2 is increasing on [0,4], a left sum uses the SMALLEST height on each piece. This underestimates the true area. \boxed{\text{left sum} = 14 \text{ (an underestimate)}}
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Where this shows up
Numerical methods
When a function has no closed-form antiderivative at all, a Riemann-type sum is often the ONLY way to get a usable numeric answer. It gets refined into the trapezoid and Simpson's rules several chapters ahead. This isn't a stepping stone discarded once FTC arrives. It's a permanent tool.
Check your understanding (counts toward your progress)
Approximate \displaystyle\int_0^2 x^2\,dx using a midpoint sum with n=2 rectangles.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Midpoints 0.5, 1.5, each times \Delta x=1 — f(0.5)+f(1.5) = 0.25+2.25 = 2.5.
Correct answer
Also accepted:
Why
Midpoints of [0,1] and [1,2] are 0.5 and 1.5.
With \Delta x=\dfrac{2}{5} and right-endpoint values f(x_1)=17, f(x_2)=2, f(x_3)=-2, f(x_4)=13, f(x_5)=-3, f(x_6)=3, f(x_7)=3, find the Riemann sum \displaystyle\sum_{i=1}^{7}f(x_i)\Delta x.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\left(17 + 2 + -2 + 13 + -3 + 3 + 3\right)\times\dfrac{2}{5}\approx13.2.
Correct answer
Also accepted:
Why
Add all the f-values, then multiply the total by \Delta x.
For \int_0^{10} f(x)\,dx split into n=5 equal subintervals, what is \Delta x?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\Delta x = \dfrac{b-a}{n} = \dfrac{10-0}{5} = 2.
Correct answer
Also accepted:
Why
Approximate \displaystyle\int_{-5}^{-1} x^2\,dx using a LEFT-endpoint Riemann sum with 4 rectangles of width 1.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Left endpoints: x=-5, -4, -3, -2. Sum of f(x)=x^2 at each, times width 1: -5^2 + -4^2 + -3^2 + -2^2 = 54$.
Correct answer
Also accepted:
Why
The left endpoints are x = -5, -4, -3, -2 — evaluate f at each and sum (each rectangle has width 1).