MATH103 April 2011
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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 $\sin x=x{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}{\frac {x^{7}}{7!}}+\dots =\sum _{k=0}^{\infty }(1)^{k}{\frac {x^{2k+1}}{(2k+1)!}}$

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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.

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Solution

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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 $0.946083$.)

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