# What is Simpson’s Rule? – Example & Formula

In this lesson, you’ll learn how to approximate the integration of a function using a numerical method called Simpson’s Rule. This method is particularly useful when integration is difficult or even impossible to do using standard techniques.

Simpson’s Rule

Integration, or anti-differentiation, is a fascinating math idea. We have methods and rules for integrating that work for most f(x) functions we encounter. There are some functions, however, that are difficult if not impossible to integrate using the usual techniques. One application in the real world is calculating the moment of inertia of a Gaussian shaped part. There is no closed-form solution for the integral of a Gaussian curve between two values (other than -∞ to +∞). So what do we do? We use numerical methods like Simpson’s Rule, named after the English mathematician, Thomas Simpson.

Parabolas & Area

The big picture idea behind Simpson’s Rule is finding the area under a parabola between two points. A parabola is a curve resembling the letter U or an upside-down U.

We start by fitting a parabola to the curve between x = 1 and x = 3

For the area between x = 3 and x = 5, we again fit a parabola to the curve:

But._it_is_a_different_parabola

Origins of the Rule

Simpson’s Rule is an algorithm for finding area. Taking the mystery out of the algorithm can be fun. We’ll do this step by step:

In this first Simpson equation:

integral_f(x)_from_xo_to_x2=(h/3)(yo+4y1+y2)

The integral f(x), from xo to x2 equals h over 3 times yo plus 4 times y1 plus y2. But what are all these variables?

First of all, f(x) is a parabola.

the_variables_are_three_points_on_the_curve

The separation between the x values is h. Do you see how

x1 = xo + h

x2 = xo + 2h

It’s time for some algebra:

x2-x1=x0+2h-xo=2h

Here, we substituted for x2 and simplified.

Here’s another algebra challenge:

From the difference of the squares.

x2^2-xo^2=(x2-xo)(x2+xo)

We’ll use the previous result for x2 – xo and substitute xo + 2h for x2:

simplify

Then, we’ll simplify and factor the 2:

factor_out_the_2

Using the difference of the cubes:

(x_2^3-x_o^3

We’ll substitute just as we did before:

eliminate_x2_and_x1

and expand the bracketed terms:

expanding

Grouping and simplifying gives us:

grouping

The point of this algebra is to come up with some expressions for proving the first Simpson equation. In the left-hand side (or LHS) of the equation:

LHS:int_x0^x2_f(x)dx

f(x), is a parabola:

f(x)=alpha_x^2+beta_x+gamma

where α, β and γ are constants defining the parabola curve.

Integrating we get:

int_f(x)dx=alpha_x^3/3+beta_x^2/2+gamma

plus a constant of integration.

Integrating from xo to x2 gives us:

int_xo^x2_f(x)dx

Now, we’ll substitute the limits:

insert_limits_of_integration

The parentheses hold the algebra we just worked out. Substituting and simplifying, we get this:

done_with_LHS

For the right-hand side (or RHS) of the equation, we have this:

LHS:2alpha/3_h(3xo^2+6hxo+4h^2)+_2betah(xo+h)+2gamma_h

Using the expression for the parabola and substituting xo for x gives us yo. We’ll use the same process to get y1 and y2:

integrate

Factoring α, β and γ:

factor_the_alpha_beta_gamma_terms

Substituting xo + h for x1 and xo + 2h for x2:

eliminate_x2_x1

Simplifying, we get this:

the_RHS

Which is the same as the LHS! We’ve just shown that the integral of a parabola can be expressed in terms of y values and the separation, h, between the x values.

Now, let’s integrate a function from a to b:

Using_three_parabola_areas

Here’s the first area, Ao:

area_of_Ao

We write ≈ because the parabola does not fit the curve exactly.

Here’s the second area, A1:

area_of_A1

And, the third area, A2:

area_of_A2

Adding the areas together, we get the total area of A:

total_area

This last line with the alternating 4s and 2s is Simpson’s Rule. In general, we write

integral_f(x)_from_a_to_b=(h/3)(yo+4y1+y2+..)

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