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quackensack

Member

Registered: 2007-02-27

Posts: 47

Maclaurin Series problem

If anyone could explain how the following problem is done, it would be greatly appreciated.

Write the Maclaurin series for cos(x^2).  By antidifferentiating its first three (nonzero) terms, obtain an approximation for the integral of cos(x^2) dx from .6 to 1.  Is this approximation larger than or smaller than the actual value of the integral of cos(x^2) from .6 to 1?  How does the series itself tell you the answer to this question?

I've done the first part which is write out the Maclaurin series for cos(x^2):
cos(x^2) = 1 - x^4/2! +x^8/4! - x^12/6!...
And I've antidifferentiated the first three terms to get:
x - x^5/10 + x^9/216...

However, I'm confused about how to obtain an approximation for the integral of cos(x^2) dx from .6 to 1.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Maclaurin Series problem

That's the easy bit! All you need to do from there (assuming the rest of it is right) is find the value of x - x^5/10 + x^9/216 when x=1, and then take away the value for when x=0.6.

That should get you an answer of ≈ 0.312.

Edit: And to answer the other part, this approximation will be larger than the actual value because the first term in the Maclaurin expansion that you ignore is negative.

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