Science:Math Exam Resources/Courses/MATH103/April 2012/Question 01 (g)

MATH103 April 2012

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Question 01 (g)
Answer the following multiple choice question. Check your answer very carefully. Your answer will be marked right or wrong (work will not be considered for this problem). Which of the following is the closest approximation to the value of the integral $\int _{0}^{1}{\frac {1-\cos(x)}{x}}\,dx$ (Hint: use Taylor series) A. 1 B. 1/4 C. 1/2 D. 5/6

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall the Taylor series of cos(x) and use just the first two terms for a simplification of the integrand.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The Taylor expansion of cos(x) is $\cos(x)=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+\cdots =1-{\frac {x^{2}}{2}}+{\frac {x^{4}}{24}}+\cdots$ Keeping the terms up to x2, we get ${\frac {1-\cos(x)}{x}}\approx {\frac {x^{2}/2}{x}}={\frac {x}{2}}$ Therefore $\int _{0}^{1}{\frac {1-\cos(x)}{x}}\,dx\approx \int _{0}^{1}{\frac {x}{2}}\,dx=\left.{\frac {x^{2}}{4}}\right|_{0}^{1}=1/4$ So 1/4 is a good approximation. Note that taking more terms of the Taylor expansions of cos(x) would simply yield an even better approximation.

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