The simplest numerical approximations to the integral are the left and right Riemann sums. More efficient approximations (below) are the trapezoidal and Simpson approximations.

$Δx = (b-a)/n,$
$x_0 = a,$
$x_1 = a + Δx,$
$x_2 = a + 2Δx,$
$. . .$
$x_n = a + nΔx = b$

In Calculus Applied to the Real World, we defined the left Riemann sum as follows:

Left Riemann Sum	$= \sum_{n = 0}^{n-1} f(x_k)Δx$
	$= f(x_0)Δx + f(x_1)Δx + . . . + f(x_{n-1})Δx$
	$= [f(x_0) + f(x_1) + . . . + f(x_{n-1})]Δx$

The left Riemann sum gives the area shown below.

Notice that the left side of each rectangle matches the height of the graph -- hence the name "left sum." The right Riemann sum is defined similarly:

Right Riemann Sum	$= \sum_{n = 1}^{n} f(x_k)Δx$
	$= f(x_1)Δx + f(x_2)Δx + ... + f(x_n)Δx $
	$= [f(x_1) + f(x_2) + ... + f(x_n)]Δx$

The right Riemann sum gives the area shown below.

 
Example 1 Computing Left- and Right Riemann Sums

Solution

The left Riemann sum for the function is illustrated here:

To compute it, we use the following setup:

$Δx = (b-a)/n =$

We now calcuate the partition values in the following table.

$x_0 = a$	$x_1 = a+Δx$	$x_2 = a+2Δx$	$x_3 = a+3Δx$	$x_4 = a+4Δx$	$x_5 = a+5Δx$	$x_6 = a+6Δx$	$x_7 = a+7Δx$	$x_8 = b =$

 
 

$x$	$x_0$	$x_1$	$x_2$	$x_3$	$x_4$	$x_5$	$x_6$	$x_7$	$x_8$
Value
$\color{blue}{f(x) = 1-x^2}$

 

We can now compute the left- and right Riemann sums as follows:

Left Riemann Sum	$=$	$[f(x_0) + f(x_1) + ... + f(x_7)]Δx$
	$=$	Sum of the first $8$ terms in the bottom row above $\times Δx$
	$=$	$5.8125 \times 0.125 = 0.7265625.$
Right Riemann Sum	$=$	$[f(x_1) + f(x_2) + ... + f(x_8)]Δx$
	$=$	Sum of the last $8$ terms in the bottom row above $\times Δx$
	$=$	$4.8125 \times 0.125 = 0.6015625.$

Trapezoid Sums

A trapezoid is a four-sided region with two opposite sides parallel. In the figure below, it is the two vertical sides that are parallel.

The area of a trapezoid is the average length of the parallel sides, times the distance between them.

The areas of the individual trapezoids (from left to right) are as follows.

Leftmost trapezoid:	$\frac{1}{2} (f(x_0) + f(x_1))Δx$	Average height $\times$ width
Next trapezoid:	$\frac{1}{2} (f(x_1) + f(x_2))Δx$
. . .
Last trapezoid:	$\ \ \frac{1}{2} (f(x_{n-1}) + f(x_n))Δx$

Adding these together, gives the trapezoid sum:

Trapezoid Sum

$= \frac{1}{2} (f(x_0) + f(x_1))Δx + \frac{1}{2} (f(x_1) + f(x_2))Δx + . . . + \frac{1}{2} (f(x_{nµ1}) + f(x_n))Δx$

Simplifying gives

 
If you look carefully at the above formula, you will see that the trapezoid sum is just the average of the left- and right Riemann sums.

 
Example 2 Computing a Trapezoid Sum

Solution

In view of the comment immediately above, we have already computed it, in effect:

Trapezoid Sum	$=$ Average of Left- and Right Riemann Sums
	$=\frac{0.7265625 + 0.6015625}{2} = 0.6640625$

Q How about computing the trapezoid sum without first computing the left- and right Riemann sums?
A Here is an example.

 
Example 3 Computing a Trapezoid Sum

Solution

Now proceed as follows using the table below:

$\color{blue}{x}$	$x_0 = a$	$x_1$	$x_2$	$x_3$	$x_4$
Enter Formula Here: $f(x) =$
		$\times 2$	$\times 2$	$\times 2$
						Total:

 

We can now compute the Trapezoid sum as follows:

Trapezoid Sum	$=$	$\frac{1}{2} [f(x_0) + 2f(x_1) + ... + 2f(x_{n-1}) + f(x_n)]Δx$
	$=$	$\frac{1}{2} \times $   Total on last row above $\times 0.25$	If your total is not $1.10975...,$ check the values of $x$ you used.
	$\approx$	$0.5 \times 1.109758 \times 0.25 = 0.13871975$

Before we go on...

Simpson's Rule

Area under parabola $= \frac{b - a}{3n} [f(x_{k-1}) + 4f(x_k) + f(x_{k+1})].$

When we add the area under the parabola over the first two strips to the area under the parabola over the 3rd and 4th strips, and so on, we get Simpson's rule.

 
Example 4 Simpson's Rule

Solution

$\int_{0}^{1}\ f(x)\ dx \approx \frac{b - a}{3n} [f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4)]$

$= \frac{1}{12} [0 + 4(0.0625) + 2(0.25) + 4(0.5625) + 1] = \frac{1}{3}.$

Before we go on...

This is the exact answer. What is going on? Remember that Simpson's rule is based on approximating the graph by quadratic functions. If the function is already quadratic, as it is here, the approximation is exact.

 
Example 5 Simpson's Rule

Solution

The following table summarizes the main part of the computation. (Figures are rounded to 6 decimal places.)

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$f(x) = e^{-x^2}$	$1$	$0.367879$	$0.018316$	$0.000123$	$0$	$0$	$0$
		$\times 4$	$\times 2$	$\times 4$	$\times 2$	$\times 4$
	$1$	$1.471518$	$0.036631$	$0.000494$	$0$	$0$	$0$

Thus, Simpson approximation $= (6 - 0)/18 \times $ Sum of bottom row $\approx 0.83621$

Error = |Approximation - Exact integral|

The following formulas give bounds on the errors for the rules we have been using.

 
Example 6 Estimating the Error in the Trapezoid Sum

How accurate is the calculation in Example 3?

Solution

Calculating,

$f'(x) = -2xe^{-x^2},$ y
$f"(x) = 2(2x^2-1)e^{-x^2}.$

$f^{(3)}(x) = 4x(3-2x^2) e^{-x^2}.$

$\frac{(2 - 1)^3}{12\times 4^2}\times 0.90 \approx 0.0047$

 
Example 7 Estimating the Error in the Simpson Sum

Solution

Our formula for the error in Simpson's rule says that

$\|Error\| ≤ \frac{(b - a)^5}{180n^4} \|f^{(4)}(M)\|$

$f^{(4)}(x) = e^{-x},$

$\|f^{(4)}(M)\| = e \approx 3$

As before, we have overestimated rather than underestimated the quantity e (never underestimate an error!). This gives

$\|Error\| ≤ \frac{(b - a)^5}{180n^4} \|f^{(4)}(M)|$

$< \frac{3^5}{180n^4}3 = \frac{81}{20n^4}.$