# Simpsonâ€™s Rule

Simpson's Rule is a numerical method that approximates the value of a definite integral by using quadratic functions.

This method is named after the English mathematician Thomas Simpson (1710−1761).

Simpson's Rule is based on the fact that given three points, we can find the equation of a quadratic through those points.

To obtain an approximation of the definite integral $$\int\limits_a^b {f\left( x \right)dx}$$ using Simpson's Rule, we partition the interval [a, b] into an even number n of subintervals, each of width

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

On each pair of consecutive subintervals $$\left[ {{x_{i - 1}},{x_i}} \right],$$ $$\left[ {{x_i},{x_{i + 1}}} \right],$$ we consider a quadratic function $$y = a{x^2} + bx + c$$ such that it passes through the points $$\left( {{x_{i - 1}},f\left( {{x_{i - 1}}} \right)} \right),$$ $$\left( {{x_i},f\left( {{x_i}} \right)} \right),$$ $$\left( {{x_{i + 1}},f\left( {{x_{i + 1}}} \right)} \right).$$

If the function $$f\left( x \right)$$ is continuous on $$\left[ {a,b} \right],$$ then

$\int\limits_a^b {f\left( x \right)dx} \approx {\frac{{\Delta x}}{3}}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + 2f\left( {{x_4}} \right) + \cdots + 4f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right].$

The coefficients in Simpson's Rule have the following pattern:

$\underbrace {1,4,2,4,2, \ldots ,4,2,4,1}_{{n + 1}\;\text{points}}.$

## Solved Problems

### Example 1.

Use Simpson's Rule with $$n = 4$$ to approximate the integral $\int\limits_0^8 {\sqrt x dx}.$

Solution.

It is easy to see that the width of each subinterval is

$\Delta x = \frac{{b - a}}{n} = \frac{{8 - 0}}{4} = 2,$

and the endpoints $${x_i}$$ have coordinates

${x_i} = \left\{ {0,2,4,6,8} \right\}.$

Calculate the function values at the points $${x_i}:$$

$f\left( {{x_0}} \right) = f\left( 0 \right) = \sqrt 0 = 0;$
$f\left( {{x_1}} \right) = f\left( 2 \right) = \sqrt 2 ;$
$f\left( {{x_2}} \right) = f\left( 4 \right) = \sqrt 4 = 2;$
$f\left( {{x_3}} \right) = f\left( 6 \right) = \sqrt 6 ;$
$f\left( {{x_4}} \right) = f\left( 8 \right) = \sqrt 8 = 2\sqrt 2 .$

Substitute all these values into the Simpson's Rule formula:

$\int\limits_0^8 {\sqrt x dx} \approx \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + f\left( {{x_4}} \right)} \right] = \frac{2}{3}\left[ {0 + 4 \cdot \sqrt 2 + 2 \cdot 2 + 4 \cdot \sqrt 6 + 2\sqrt 2 } \right] = \frac{2}{3}\left[ {6\sqrt 2 + 4 + 4\sqrt 6 } \right] \approx 14.86$

The true solution for the integral is

$\int\limits_0^8 {\sqrt x dx} = \int\limits_0^8 {{x^{\frac{1}{2}}}dx} = \left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^8 = \frac{2}{3}\left[ {\sqrt {{x^3}} } \right]_0^8 = \frac{2}{3}\sqrt {{8^3}} = \frac{2}{3}\sqrt {{2^9}} = \frac{2}{3} \cdot 16\sqrt 2 = \frac{{32\sqrt 2 }}{3} \approx 15.08$

Hence, the error in approximating the integral is

$\left| \varepsilon \right| = \left| {\frac{{15.08 - 14.86}}{{15.08}}} \right| \approx 0.015 = 1.5\%$

### Example 2.

A function $$f\left( x \right)$$ is given by the table of values. Approximate the area under the curve $$y = f\left( x \right)$$ between $$x = 0$$ and $$x = 4$$ using Simpson's Rule with $$n = 4$$ subintervals.

Solution.

For $$n= 4$$ subintervals, Simpson's rule is given by the following equation:

${S_4} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + f\left( {{x_4}} \right)} \right].$

The width of the subinterval is

$\Delta x = \frac{{b - a}}{n} = \frac{{4 - 0}}{4} = 1.$

Substitute the values of the function from the table and calculate the approximate value of the area under the curve:

$A = {S_4} \approx \frac{1}{3}\left[ {2 + 4 \cdot 7 + 2 \cdot 12 + 4 \cdot 10 + 5} \right] = \frac{1}{3}\left[ {2 + 28 + 24 + 40 + 5} \right] = \frac{1}{3} \cdot 99 = 33$

### Example 3.

A function $$f\left( x \right)$$ is given as a table of values. Approximate the area under the curve $$y = f\left( x \right)$$ from $$x = -4$$ and $$x = 8$$ using Simpson's Rule with $$n = 6$$ subintervals.

Solution.

We use Simpson's rule formula which has the following form for $$n = 6$$ subintervals:

${S_6} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + 2f\left( {{x_4}} \right) + 4f\left( {{x_5}} \right) + f\left( {{x_6}} \right)} \right].$

Determine the width $$\Delta x$$ of the subinterval:

$\Delta x = \frac{{b - a}}{n} = \frac{{8 - \left( { - 4} \right)}}{6} = 2.$

Using the values of the function given in the table, we calculate the approximate value of the area under the curve.

$A = {S_6} \approx \frac{2}{3}\left[ {1 + 4 \cdot 3 + 2 \cdot 4 + 4 \cdot 4 + 2 \cdot 6 + 4 \cdot 9 + 14} \right] = \frac{2}{3}\left[ {1 + 12 + 8 + 16 + 12 + 36 + 14} \right] = \frac{2}{3} \cdot 99 = 66$

### Example 4.

Approximate the area under the curve $$y = f\left( x \right)$$ between $$x = -1$$ and $$x = 5$$ using Simpson's Rule with $$n = 6$$ subintervals.

Solution.

The Simpson's Rule formula for $$n = 6$$ subintervals is given by

${S_6} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + 2f\left( {{x_4}} \right) + 4f\left( {{x_5}} \right) + f\left( {{x_6}} \right)} \right].$

It follows from the figure that $$\Delta x = 1.$$ The function values at the endpoints of the intervals are

$f\left( {{x_0}} \right) = f\left( -1 \right) = 4;$
$f\left( {{x_1}} \right) = f\left( 0 \right) = 3;$
$f\left( {{x_2}} \right) = f\left( 1 \right) = 2;$
$f\left( {{x_3}} \right) = f\left( 2 \right) = 3;$
$f\left( {{x_4}} \right) = f\left( 3 \right) = 6;$
$f\left( {{x_5}} \right) = f\left( 4 \right) = 6;$
$f\left( {{x_6}} \right) = f\left( 5 \right) = 4.$

So, the approximate value of the area under the curve is

$A = {S_6} \approx \frac{1}{3}\left[ {4 + 4 \cdot 3 + 2 \cdot 2 + 4 \cdot 3 + 2 \cdot 6 + 4 \cdot 6 + 4} \right] = \frac{1}{3}\left[ {4 + 12 + 4 + 12 + 12 + 24 + 4} \right] = \frac{1}{3} \cdot 72 = 24$

### Example 5.

Approximate the area under the curve $y = {3^x}$ between $$x = -2$$ and $$x = 2$$ using Simpson's Rule with $$n = 4$$ subintervals.

Solution.

The Simpson's Rule formula formula for $$n = 4$$ is written in the form

${S_4} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + f\left( {{x_4}} \right)} \right].$

We calculate the function values $$f\left( {{x_i}} \right):$$

$f\left( {{x_0}} \right) = f\left( {-2} \right) = {3^{ - 2}} = \frac{1}{9};$
$f\left( {{x_1}} \right) = f\left( {-1} \right) = {3^{ - 1}} = \frac{1}{3};$
$f\left( {{x_2}} \right) = f\left( 0 \right) = {3^0} = 1;$
$f\left( {{x_3}} \right) = f\left( 1 \right) = {3^1} = 3;$
$f\left( {{x_4}} \right) = f\left( 2 \right) = {3^2} = 9.$

As $$\Delta x = 1,$$ we obtain

$A = {S_4} \approx \frac{1}{3}\left[ {\frac{1}{9} + 4 \cdot \frac{1}{3} + 2 \cdot 1 + 4 \cdot 3 + 9} \right] = \frac{1}{3}\left[ {\frac{1}{9} + \frac{4}{3} + 23} \right] = \frac{1}{3} \cdot \frac{{1 + 12 + 207}}{9} = \frac{{220}}{{27}} = 8\frac{4}{{27}}.$

### Example 6.

Approximate the integral $\int\limits_1^2 {\frac{{dx}}{x}}$ using Simpson's Rule with $$n = 2$$ subintervals.

Solution.

The Simpson's Rule formula with $$n = 2$$ subintervals is given by

${S_2} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + f\left( {{x_2}} \right)} \right].$

The width of the subinterval is

$\Delta x = \frac{{b - a}}{n} = \frac{{2 - 1}}{2} = \frac{1}{2}.$

Calculate the values of the function at the points $${x_i} = \left\{ {1,\frac{3}{2},2} \right\}:$$

$f\left( {{x_0}} \right) = f\left( 1 \right) = \frac{1}{1} = 1;$
$f\left( {{x_1}} \right) = f\left( {\frac{3}{2}} \right) = \frac{1}{{\frac{3}{2}}} = \frac{2}{3};$
$f\left( {{x_2}} \right) = f\left( 2 \right) = \frac{1}{2}.$

Then

$\int\limits_1^2 {\frac{{dx}}{x}} \approx {S_2} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + f\left( {{x_2}} \right)} \right] = \frac{1}{6}\left[ {1 + 4 \cdot \frac{2}{3} + \frac{1}{2}} \right] = \frac{1}{6} \cdot \frac{{6 + 16 + 3}}{6} = \frac{{25}}{{36}}.$

See more problems on Page 2.