Use Maclaurin series to approximate the integral to three decimal-place accuracy. ∫0^0.2 √(1+x^4

Use Maclaurin series to approximate the integral to three...

Key Concepts

Maclaurin Series

A Maclaurin series is a type of Taylor series expansion centered at zero. It represents a function as an infinite sum of its derivatives at zero multiplied by powers of the variable. This series is particularly useful for approximating functions where direct evaluation is difficult, by replacing them with polynomial sums that are easier to integrate or differentiate.

Term-by-Term Integration

Once a function is expressed as a power series, it is often practical to integrate the series term by term over the interval of interest. This method relies on the uniform convergence of the series within the range, allowing the integration of each individual term, which simplifies the process of finding an approximate value of the integral.

Error Estimation

Error estimation in series approximations is crucial when truncating an infinite series to a finite number of terms to achieve a desired accuracy. By analyzing the truncation error using remainder terms from Taylor’s theorem or bounding the series terms, one can ensure that the approximation, such as up to three decimal places, is within acceptable limits of the true value.

Transcript

00:01 You have this integral between 0 up to 0 .2 of the 4th the 3rd square root of 1 plus x to the 4th x and you want to use you want to use mclaurian series to approximate to approximate up to 3 decimates up to 3 places so that would mean that you are to use a taylor point taylor expansion so that if you call this f of x well that value f so that you'd like to find the approximation by the polynomial so that f minus approximation of two degree n this has an error that is more than 10 to the minus 3 so smaller than 0 .0 .1 so also we approximate what we can do is we can approximate this function we're integrating and we can integrate the polynomials of approximation of this so for this one you do an approximation around zero because these points are or whatever we get would be evaluated at zero so zero is a good point to start with also for that we need what is the function what is the derivative of the function so if we differentiate that this is 1 plus x to the 4th power and that raised to the 1 third so the derivative of that would be 1 3rd 1 plus x to the 4th then this to the 1 third minus 1 which is minus 2 thirds and then by the inner derivative times of x to 4th which is 4 x cubed so times 4 x cube that is going to be the derivative the second derivative is going to be so 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 and then we'll differentiate first that so times minus two thirds then one plus x to the fourth and then minus two thirds minus minus one will be minus five thirds and then times all times we have our times x to the third and then by the inner derivative differentiating this term differentiating this term we get again a 4x times 4x cube and now we differentiate this term so it's going to be 3 times that so the 3 will cancel and then we will get 4 and then this power lowers by 1 so 2 4x squared and then 1 plus x to 4th minus 2 thirds so you have that well this this is gonna be this term is four square third squared and then we have a minus two so minus two times that and then x to the six power plus six fourth so you have you have that and then flux plus 4 x squared so well if we differentiate that again the so this one differentiating this one we get a x to the fifth that all times of another function and then if we differentiate this term we would get another power like x to the 6 times another constant times x to the 3rd by the the derivative of that times that and then differentiating this term you'll get four so eight x and then plus that and then by doing the derivative of this term we'll get four x squared and then times four x cube one plus our term but what matters is that this derivative would also be zero because if we want to evaluate at zero, all these terms, well, this is going to be zero, zero, zero, zero.

07:08 The first derivative that is not zero would be, well, the derivative of x to the fourth, so it's going to have, all these terms will produce zero, but this one, so it would have, uh, will differentiate.

07:27 That you get mine so 8 times 1 plus x to the 4th minus 2 thirds so this is the only term well the rest is going to be times x times another function h of x so that while this is a polynomial so that when we evaluate this fourth derivative at 0 is going to be well 8 so so we see that if we go up to the third term, so t3, the value of f minus the polynomial at t3, this would be integrating, integrating that.

08:38 So if you have this function, if we approximate this function, this function, so f minus t3 this is going to be bounded the function that we're integrating so this is going to be bounded bounded by 8 and then divided by 4 factorial times for what we're integrating there would be 0 .2 today 4 power and this will give us so zero four zeros...

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