MATH103 April 2011
• Q1 (a) • Q1 (b) • Q1 (c) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q5 (d) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q7 (c) • Q8 •
Question 07 (b)

The following definite integral cannot be solved analytically: $\int _{0}^{1}{\frac {\sin x}{x}}\,dx.$
Approximate the integral based on the first three nonzero terms in the Taylor series (around $x=0$).
Hint: Taylor series for $\sin x=x{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}{\frac {x^{7}}{7!}}+\dots =\sum _{k=0}^{\infty }(1)^{k}{\frac {x^{2k+1}}{(2k+1)!}}$

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

Divide the first three terms in series for $\sin x$ by $x$, then integrate termbyterm.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
 If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
 If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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.
Dividing the first three nonzero terms in the Taylor series for $\sin x$ by $x$,
${\begin{aligned}{\frac {\sin x}{x}}\approx 1{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}}.\end{aligned}}$
Hence,
${\begin{aligned}\int _{0}^{1}{\frac {\sin x}{x}}\,dx&\approx \int _{0}^{1}\left(1{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}}\right)\,dx\\&=\left.x{\frac {x^{3}}{3\cdot 3!}}+{\frac {x^{5}}{5\cdot 5!}}\right_{0}^{1}\\&=1{\frac {1}{18}}+{\frac {1}{5\cdot 120}}=1{\frac {1}{18}}+{\frac {1}{600}}={\frac {1800100+3}{1800}}={\frac {1703}{1800}}.\end{aligned}}$
(Note that this is a pretty good approximation. ${\frac {1703}{1800}}\approx 0.946111,$ while WolframAlpha approximates the full integral $\int _{0}^{1}{\frac {\sin x}{x}}\,dx$ with $0.946083$.)

Click here for similar questions
MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Taylor series, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag

Math Learning Centre
 A space to study math together.
 Free math graduate and undergraduate TA support.
 Mon  Fri: 12 pm  5 pm in LSK 301&302 and 5 pm  7 pm online.
Private tutor
