ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: tiffany88 on June 13, 2018, 01:27:09 pm

Title: Reimann integrals
Post by: tiffany88 on June 13, 2018, 01:27:09 pm: Hello,
I was wondering how to approach part b of this question
Title: Re: Reimann integrals
Post by: RuiAce on June 13, 2018, 01:56:12 pm: $\text{Note that over }[1,6]\text{, }f(x) = 2x^2-1\text{ is monotonic increasing.}$
$\text{This allowed us to note that the lower Riemann sums were drawn from the left endpoint}\\ \text{and that the upper Riemann sums were drawn from the right endpoint.}$
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$\text{The relevant Riemann sums for part a) were}\\ \begin{align*}L_5 &= \frac{1}{5} \big( f(1) + f(2) + f(3) + f(4) + f(5) \big) \\ &= \frac15 \left( f\left(1 \right) + f \left( 1 + \frac{5}{5}\right) + f \left(1 + \frac{10}{5} \right) + f \left(1 + \frac{15}{5} \right) + f \left( 1 + \frac{20}{5} \right)\right)\\ &= \frac15 \sum_{k=0}^4 f\left(1 + \frac{5k}{5}\right)\\ U_5 &= \frac{1}{5} \big( f(2) + f(3) + f(4) + f(5) + f(6) \big) \\ &= \frac15 \left(f \left( 1 + \frac{5}{5}\right) + f \left(1 + \frac{10}{5} \right) + f \left(1 + \frac{15}{5} \right) + f \left( 1 + \frac{20}{5} \right) + f \left(1 + \frac{25}{5} \right)\right)\\ &= \frac15 \sum_{k=1}^5 f\left(1 + \frac{5k}{5}\right) \end{align*}$
Of course, this isn't the only representation that gets to the answer.
$\text{In general,}\\ \begin{align*}L_n &= \frac1n\sum_{k=0}^{n-1} f \left(1+\frac{5k}{n} \right) \\ U_n &=\frac1n \sum_{k=1}^n f \left(1 + \frac{5k}{n} \right) \end{align*}$
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$\text{We require }U_n\text{ and }L_n\text{ to not be more than 0.001 apart}\\ \text{i.e. }|U_n - L_n| < 0.001.\\ \text{Because }U_n \geq L_n\text{ always, we can take off the absolute values and simply consider}\\ \boxed{U_n - L_n < 0.001}$
$\text{Subbing this in gives}\\ \frac1n \sum_{k=1}^n f \left(1 + \frac{5k}{n} \right) - \frac1n\sum_{k=0}^{n-1} f \left(1 + \frac{5k}{n} \right) < 0.001\\\text{which is just a telescoping sum, and simplifies to}\\ \boxed{\frac1n\left( f(6) - f(0) \right) < 0.001}$
$\text{From the calculator, }f(6) - f(0) = 70\\ \text{so }\frac1n < \frac{0.001}{70} \implies \boxed{n >70000 }$
Subject to minor computational error