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Use the Trapezoidal Rule with [tex]n=10[/tex] to approximate

[tex]\int _0 ^{20} \cos ( \pi x) \: dx [/tex]

Compare your result to the actual value. Can you explain the discrepancy?

My work

The actual value can be found as follows

[tex]\int _0 ^{20} \cos ( \pi x) \: dx = \left. \frac{\sin ( \pi x)}{\pi} \right] _{x=20} - \left. \frac{\sin ( \pi x)}{\pi} \right] _{x=0} = 0 [/tex]

Next, we have the Trapezoidal Rule with [tex]n=10[/tex]

[tex]T_{10} = \left( \frac{20-0}{10\cdot 2} \right) \left[ \cos (0\cdot \pi) + 2\cos (2\cdot \pi) + 2\cos (4\cdot \pi) + \dots + 2\cos (16\cdot \pi) + 2\cos (18\cdot \pi) + \cos (20\cdot \pi) \right] = 20[/tex]

Can I just say that the reason for the discrepancy is the low value of [tex]n[/tex] ? According to the error bound, we have

[tex] \left| E_T \right| \leq \frac{K(b-a)^3}{12n^2} < \left| \mbox{ error } \right| \Longrightarrow \frac{1\cdot (20-0)^3}{12 n^2} < 10^{-1} \Rightarrow n \geq 82 [/tex]

which shows that in order to achieve accuracy to within one decimal place, it is required to have [tex]n \geq 82[/tex].

Is that it?

Thank you

PS: I should have calculated that for the first digit...

[tex] \left| E_T \right| \leq \frac{K(b-a)^3}{12n^2} < \left| \mbox{ error } \right| \Longrightarrow \frac{1\cdot (20-0)^3}{12 n^2} < 10^{0} \Rightarrow n \geq 26[/tex]

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# Homework Help: Trapezoidal Rule (just need to check my answer)

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