Science:Math Exam Resources/Courses/MATH103/April 2010/Question 05 (b)

MATH103 April 2010

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Question 05 (b)

In this problem, you are asked to find a value for the integral $I=\int _{0}^{1}x\cos(x)\,dx$ in two different ways. (b) Use the Taylor series for the function $\cos(x)$ to write down a Taylor Series approximation for $x\cos(x)$ and find an approximation to the integral $I$ using the first 3 terms of that series. (Leave your answer in terms of the three fractions, i.e. you need not compute a simplified fraction nor a decimal approximation. Some factorial values that may be useful: $2!=2$, $3!=6$, $4!=24$, $5!=120$, $6!=720$, $7!=5040$.

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
The Taylor series expansion of cosine is $\cos(x)=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+...=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n}}{(2n)!}}$

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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. Via the hint we have that $\cos(x)=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+...=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n}}{(2n)!}}$ multiplying by x gives $x\cos(x)=x-{\frac {x^{3}}{2!}}+{\frac {x^{5}}{4!}}-{\frac {x^{7}}{6!}}+...=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{(2n)!}}$ and then integrating gives $\int x\cos(x)\,dx={\frac {x^{2}}{2}}-{\frac {x^{4}}{8}}+{\frac {x^{6}}{144}}-{\frac {x^{8}}{5760}}+...=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+2}}{(2n+2)(2n)!}}$ Taking into account the end points, we have $\int _{0}^{1}x\cos(x)\,dx={\frac {1^{2}}{2}}-{\frac {1^{4}}{8}}+{\frac {1^{6}}{144}}-{\frac {1^{8}}{5760}}+...=\sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{(2n+2)(2n)!}}$ and simplifying and using only the first three terms, we have $\int _{0}^{1}x\cos(x)\,dx\approx {\frac {1}{2}}-{\frac {1}{8}}+{\frac {1}{144}}={\frac {55}{144}}$ completing the question

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