Riemann Sums and Definite Integrals

The shaded region below is bounded by the curve $y = x^2$, the $x$-axis, and the vertical lines $x = 2$ and $x = 6$. We want to find its area.

There is no area formula for a region with a curved boundary, as there is for a rectangle or a triangle. We approximate the region with rectangles, whose areas we can compute, and then take a limit to make the approximation exact.

Divide the interval $[2, 6]$ into four subintervals of equal width. The width of each subinterval is

The four subintervals are $[2, 3]$, $[3, 4]$, $[4, 5]$ and $[5, 6]$. On each subinterval we draw a rectangle whose height is the value of $f$ at the right endpoint: $f(3) = 9$, $f(4) = 16$, $f(5) = 25$ and $f(6) = 36$.

The area of each rectangle is height times width. The sum of the four areas is the right Riemann sum with four rectangles, written $R_4$:

Since $f$ is increasing on $[2, 6]$, each rectangle extends above the curve. Thus $R_4 = 86$ is an overestimate of the area.

We use the same four subintervals, so again $\Delta x = 1$. Now the height of each rectangle is the value of $f$ at the left endpoint: $f(2) = 4$, $f(3) = 9$, $f(4) = 16$ and $f(5) = 25$.

Every rectangle lies below the curve, so $L_4 = 54$ is an underestimate. Thus the exact area $A$ satisfies

With four rectangles the approximation is rough. Use the slider to increase the number of rectangles $n$. The readout shows the Riemann sum, the exact area, and the error. Compare the left and right sums: at $n = 4$ they differ by $32$, and both approach the same value as $n$ increases.

The same construction works for any number of rectangles. Take an interval $[a, b]$ and divide it into $n$ subintervals of equal width. We use right endpoints.

The width of each rectangle is

and the right endpoints are

The $i$-th rectangle has height $f(x_i)$ and width $\Delta x$. The sum of the $n$ areas is the approximate area,

and the exact area is the limit of this sum as $n \to \infty$:

It can be proved that this limit always exists when $f$ is continuous. It can also be shown that we get the same value using left endpoints, where $x_0 = a$:

For $f(x) = x^2$ on $[2, 6]$, $R_n$ approaches $A$ from above and $L_n$ from below, as the explorer shows.

We return to $f(x) = x^2$ on $[2, 6]$, now with $n$ rectangles and right endpoints. Here $a = 2$ and $b = 6$, so

The height of the $i$-th rectangle is

Thus the right Riemann sum is

We cannot finish this computation yet. The sum has $n$ terms, and $n$ is a variable, so the terms cannot be written out and added. To take the limit we need closed-form formulas for $\sum_{i=1}{n} 1$, $\sum_{i=1}{n} i$ and $\sum_{i=1}{n} i2$. These formulas come from the sequences and series unit of Algebra 2 or Precalculus.

The properties let us split a sum term by term and factor out constants. Note that $n$ is a constant with respect to the index $i$, so any power of $n$ can be factored out of the sum.

We split the sum into three sums and factor out the constants:

Now we apply the summation formulas:

This is a formula for $R_n$ for any $n$. As a check, $n = 4$ gives $16 + 40 + 30 = 86$, which is $R_4$. Now we take the limit. Since $\frac{n+1}{n} \to 1$ and $\frac{(n+1)(2n+1)}{n^2} \to 2$ as $n \to \infty$,

The exact area is $\frac{208}{3} \approx 69.33$. Note that $L_4 = 54 < \frac{208}{3} < R_4 = 86$, as expected.