Require both the trapezoidal rule and Simpson’s rule. They...

Require both the trapezoidal rule and Simpson’s rule. They can be worked without calculator programs if such programs are not available, although they require more calculation than the other problems in this exercise set. The difference between the true value of an integral and the value given by the trapezoidal rule or Simpson’s rule is known as the error. In numerical analysis, the error is studied to determine how large $n$ must be for the error to be smaller than some specified amount. For both rules, the error is inversely proportional to a power of $n$, the number of subdivisions. In other words, the error is roughly $k / n^{p},$ where $k$ is a constant that depends on the function and the interval, and $p$ is a power that depends only on the method used. With a little experimentation, you can find out what the power $p$ is for the trapezoidal rule and for Simpson's rule. a. Find the exact value of $\int_{0}^{1} x^{4} d x .$ b. Approximate the integral in part a using the trapezoidal rule with $n=4,8,16,$ and $32 .$ For each of these answers, find the absolute value of the error by subtracting the trapezoidal rule answer from the exact answer found in part a. c. If the error is $k / n^{p},$ then the error times $n^{p}$ should be approximately a constant. Multiply the errors in part b times $n^{p}$ for $p=1,2,$ etc., until you find a power $p$ yielding the same answer for all four values of $n .$

Require both the trapezoidal rule and Simpson’s rule. They can be worked without calculator programs if such programs are not available, although they require more calculation than the other problems in this exercise set.
The difference between the true value of an integral and the value given by the trapezoidal rule or Simpson’s rule is known as the error. In numerical analysis, the error is studied to determine how large $n$ must be for the error to be smaller than some specified amount. For both rules, the error is inversely proportional to a power of $n$, the number of subdivisions. In other words, the error is roughly $k / n^{p},$ where $k$ is a constant that depends on the function and the interval, and $p$ is a power that depends only on the method used. With a little experimentation, you can find out what the power $p$ is for the trapezoidal rule and for Simpson's rule.

a. Find the exact value of $\int_{0}^{1} x^{4} d x .$
b. Approximate the integral in part a using the trapezoidal rule with $n=4,8,16,$ and $32 .$ For each of these answers, find the absolute value of the error by subtracting the trapezoidal rule answer from the exact answer found in part a.
c. If the error is $k / n^{p},$ then the error times $n^{p}$ should be approximately a constant. Multiply the errors in part b times $n^{p}$ for $p=1,2,$ etc., until you find a power $p$ yielding the same answer for all four values of $n .$

Key Concepts

Order of Convergence

The order of convergence is a concept in numerical analysis that describes how quickly the error decreases as the number of subintervals increases. In the context of numerical integration methods like the trapezoidal and Simpson's rules, it indicates the power to which the subdivision count is raised in the error term, guiding the selection of an appropriate method based on the desired accuracy.

Simpson's Rule

Simpson's rule is another numerical integration technique that uses quadratic polynomials to approximate the function over subintervals. It typically provides a higher degree of accuracy compared to the trapezoidal rule, with an error term that decreases faster as the number of subintervals increases, reflecting its higher order of convergence.

Error Analysis

Error analysis in numerical integration studies the difference between the true value of an integral and the approximated value obtained via numerical methods. It helps in understanding how the choice of method and the number of subintervals affect the accuracy, and it provides insights into the rate at which the error decreases as the computational effort (number of subdivisions) increases.

Numerical Integration

Numerical integration involves approximating the definite integral of a function when an exact answer is difficult or impossible to obtain analytically. This subject area covers various techniques for estimating the area under curves and is especially important when dealing with complex or experimentally obtained functions.

Trapezoidal Rule

The trapezoidal rule is a method in numerical integration that approximates the area under a curve by breaking the interval into small segments and approximating the function with linear functions (trapezoids) over these segments. It is simple to implement and its error is generally proportional to the inverse square of the number of subintervals used, known as the order of convergence.

Transcript

00:01 Okay, so for this problem, the first thing that we're going to do is it's asking us to find the exact value when we're given the integral from 0 to 1 of x to the 4th dx.

00:12 So we're going to obviously is going to add an exponent and then do the reciprocal of that exponent and then from 1 to 0.

00:22 So it's going to be 1 5th minus 0 or 0 .2.

00:27 Okay? so part b is asking us to use a calculator of some kind.

00:32 To determine the approximation using the trapezoidal rule when n equals 4, n equals 8.

00:47 I'm going to scoot down just a little bit, so i have a little bit more space, when n equals 16 and 32.

00:55 So we don't have to calculate these by hand, and i don't recommend doing that because it would take forever.

01:03 And so we're going to do that and then calculate the error.

01:05 And so i'll walk through that formula whenever we get to it.

01:08 So t4 is going to be approximately 0 .2207.

01:15 And so to calculate the error, i'm going to take the absolute value of the exact value, which is 0 .2.

01:21 And i'm going to subtract what i got for my approximation.

01:27 So this is going to be 0 .027.

01:33 Okay.

01:34 So then going down to where n equals 8, the trapezoidal rule tells us that it's 0 .2052.

01:42 So the error is going to be 0 .2, which is the exact value, minus 0 .202, which is going to be 0 .0052.

01:58 And then for t, when in a 16, t is going to be approximately 0 .2013.

02:06 So for the error, i'm going to take that exact value, subtract to 0 .2013.

02:13 And i'm going to get 0 .0013.

02:18 And then last but not least, i'm going to do where n equals 32.

02:22 So it's going to be approximately 0 .203.

02:28 So the error is going to be 0 .2 minus 0 .203.

02:34 And so i'm going to get 0 .003.

02:40 Okay...

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