Section 9.4: Approximation of Definite Integrals
Review of Riemann Sums

If $a < b$, $f(x)$ is a function on $[a,b]$, and $a = x_0 \leq a_0 \leq x_1 \leq a_1 \leq \cdots a_{n-1} \leq x_n = b$, then the Riemann sum associated to these data is

$$\sum_{i=0}^{n-1} f(a_i) (x_{i+1} - x_i)$$

By definition, the integral, $\int_a^b f(x) dx$ is the limit (if it exists) of these Riemann sums as maximum of $\vert x_{i+1} - x_i\vert$ tends to zero.

(Applet illustrating Riemann sums)

Uniform divisions

For the sake of convenience, we often assume that the interval $[a,b]$ has been decompose into $N$ pieces of equal length, for some positive integer $N$. The length of each piece is then $\Delta := \frac{b-a}{N}$.

So, $x_i = a + i \Delta$ and $x_i \leq a_i \leq x_{i+1}$ and for such a uniform decomposition, the Riemann sum is

$$\begin{eqnarray*} \sum_{i=0}^{N-1} (f(a_i) (x_{i+1} - x_i) ) & = & \sum_{i=0}^{N... ...elta - a - i \Delta)) \\ & = & \sum_{i=0}^{N-1} f(a_i) \Delta \end{eqnarray*}$$

Left- and Right-hand rules

Let $a \leq b$, $N$, and $f(x)$ be given. Then the lefthand Riemann sum approximating $\int_a^b f(x) dx$ with $N$ subdivisions is given by setting $a_i := x_i = a + i \Delta$ where $\Delta = \frac{b-a}{N}$.

$$L = \sum_{i=0}^{N-1} f(a + i \Delta) \Delta$$

The righthand Riemann sum is given by setting $a_i := x_{i+1} = a+ (i+1) \Delta$.

$$R = \sum_{i=0}^{N-1} f(a+(i+1)\Delta) \Delta$$

Example

Compute the left and the right approximations to $\int_1^9 x^2 dx$ with $N = 4$ subdivisions.

Solution

In this case $\Delta = \frac{9 - 1}{4} = 2$. So,

$$\begin{eqnarray*} L & = & \sum_{i=0}^3 (1 + i 2)^2 2 \\ & = & (1^2 + 3^2 + 5^2 + 7^2) 2 \\ & = & (1 + 9 + 25 + 49) 2 \\ & = & 168 \end{eqnarray*}$$

Solution, continued

$$\begin{eqnarray*} R & = & \sum_{i=0}^3 (1 + (i+1)2)^2 2 \\ & = & (3^2 + 5^2 + 7^2 + 9^2) 2 \\ & = & (9 + 25 + 49 + 81) 2 \\ & = & 328 \end{eqnarray*}$$

Midpoint rule

The midpoint approximation (to the integral $\int_a^b f(x) dx$ with $N$ subdivisions) is given by taking $a_i$ to be the midpoint of the interval $[x_i,x_{i+1}]$ where $x_i = a + i \Delta$ and $\Delta = \frac{b-a}{N}$. Thus, $a_i = a + (i + \frac{1}{2}) \Delta$ and

$$M = \sum_{i=0}^{N-1} f(a_i) \Delta = \sum_{i=0}^{N-1} f(a + (i + \frac{1}{2}) \Delta) \Delta$$

Example

Approximate $\int_1^9 x^2 dx$ using the midpoint rule and $N = 4$ subdivisions.

Solution

$$\begin{eqnarray*} M & = & \sum_{i = 0}^{3} (1 + ( (i + \frac{1}{2})2 ))^2 2 \\ ... ...2 + 6^2 + 8^2) 2 \\ & = & (4 + 16 + 36 + 64) 2 \\ & = & 240 \end{eqnarray*}$$

Trapezoidal rule

Rather than approximating the area bounded by a function by rectangles, one may use other shapes. For example, one may use trapezoids. The area of the trapezoid with corners at $(a,0)$, $(b,0)$, $(a,f(a))$, and $(b,f(b))$ is $\frac{1}{2} (f(a) + f(b)) (b-a)$.

The trapezoidal approximation to $\int_a^b f(x) dx$ with $N$ subdivisions is

$$T = \sum_{i=0}^{N-1} \frac{1}{2} (f(x_i) + f(x_{i+1})) \Delta$$

where $\Delta = \frac{b-a}{N}$ and $x_i = a + i \Delta$.

Example

Compute the trapezoidal approximation to $\int_1^9 x^2 dx$ with $4$ subdivisions.

Solution

$$\begin{eqnarray*} T & = & \sum_{i=0}^3 \frac{1}{2} ( (1 + i 2)^2 + (1 + (i+1)2)^... ...2 \\ & = & (1 + 9 + 9 + 25 + 25 + 49 + 49 + 81) \\ & = & 248 \end{eqnarray*}$$

Other formulae for the trapezoidal rule

$$\begin{eqnarray*} T & = & \sum_{i=0}^{N-1} \frac{1}{2} (f(x_i) + f(x_{i+1}) \Del... ...um_{i=0}^{N-1} f(x_{i+1}) \Delta)] \\ & = & \frac{1}{2} [L + R] \end{eqnarray*}$$

Another formula for $T$

$$\begin{eqnarray*} T & = & \sum_{i=0}^{N-1} \frac{1}{2} (f(x_i) + f(x_{i+1}) ) \D... ...rac{1}{2} [f(x_0) + 2 (\sum_{i=1}^{N-1} f(x_i)) + f(x_N)] \Delta \end{eqnarray*}$$

Simpson's Rule

Simpson's rule for approximating integrals is based on approximating $\int_{x_i}^{x_{i+1}} f(x) dx$ by the area bounded by the parabola passing through $(x_i,f(x_i))$, $(\frac{x_i + x_{i+1}}{2}, f(\frac{x_i + x_{i+1}}{2}))$, and $(x_{i+1}, f(x_{i+1}))$.

Rather than deriving Simpson's rule from its geometric description, we write the Simpson's rule approximation in terms of the midpoint and trapezoidal approximations.

$$S = \frac{2}{3} M + \frac{1}{3} T$$

Another formula for Simpson's rule

Directly,

$$\begin{eqnarray*} S & = & \frac{1}{6} [f(a) + 4 f(a + \frac{1}{2} \Delta) + 2 f(... ... \\ & & + 2 (\sum_{i=1}^{N-1} f(a + i \Delta)) + f(b)] \Delta \end{eqnarray*}$$

Example

Compute the Simpson's rule approximation to $\int_1^9 x^2 dx$ with $N = 4$ subdivisions.

Solution

$$\begin{eqnarray*} S & = & \frac{1}{6} [1^2 + 4 (2^2 + 4^2 + 6^2 + 8^2) \\ & & ... ...66 + 81] \\ & = & \frac{1}{3} [656] \\ & = & 242 \frac{2}{3} \end{eqnarray*}$$

Comparison

The exact value of $\int_1^9 x^2 = \frac{1}{3} x^3 \vert_{1}^9 = \frac{1}{3}[9^3 - 1^3] = \frac{728}{3} = 242 \frac{2}{3}$.

Approximations for $N = 4$.

Left	$168$
Midpoint rule	$240$
Simpson's rule	$242 \frac{2}{3}$
Exact value	$242 \frac{2}{3}$
Trapezoidal rule	$248$
Right	$328$

Error Analysis

The error of an approximation $\alpha$ is the absolute value of the difference between $\alpha$ and the exact value of the itegral.

• If $\vert f''(x)\vert \leq A$ for all $x \in [a,b]$, then the error of a midpoint approximation with $N$ subdivisions is at most $\frac{A(b-a)^3}{24 N^2}$.
• If $\vert f''(x)\vert \leq A$ for all $x \in [a,b]$, then the error of a trapezoidal approximation with $N$ subdivisions is at most $\frac{A(b-a)^3}{12 N^2}$.
• If $\vert f''''(x)\vert \leq A$ for all $x \in [a,b]$, then the error of a Simpson's rule approximation with $N$ subdivisions is at most $\frac{A(b-a)^5}{2880 n^4}$.

Example

How many subdivisions do we need to guarantee that a midpoint approximation to $\int_1^2 \frac{dx}{x}$ approximates $\ln(2)$ to two decimal points.

Solution

We need to have an error of less than $0.005$. So, we want

$$0.005 > \frac{A(2-1)^2}{24 N^2}$$

where $A = \max \{ \vert f''(x)\vert \vert  1 \leq x \leq 2 \}$.

We compute $f'(x) = -x^{-2}$ and $f''(x) = 2 x^{-3}$. This function is decreasing between $1$ and $2$. So, $A = 2$. So, we want $N^2 > \frac{2 \times 200}{24} = 16 \frac{2}{3}$. Thus, $N = 5$ suffices.

We compute

$$\begin{eqnarray*} \ln(2) & = & \int_1^2 \frac{1}{x} dx \\ & \approx & (1/(1.1)... .../(1.3) + 1/(1.5) + 1/(1.7) + 1/(1.9)) .2 \\ & = & 0.6919\ldots \end{eqnarray*}$$