Integration (well i think it is) (1 Viewer)

kwabon · Oct 3, 2009

help me with this question.

i couldnt understand Mr. Jim Coroneos' answer to this question, could someone be kind enough to answer it for me.

and how would tackle this sort of questions, these questions trouble me.

Trebla · Oct 3, 2009

$$Look at the limits of integration$\\0 \leq x \leq \dfrac{1}{2}\\\\$We know that $ 0 \leq x^n \leq x^2 $for $n\geq2\\\\\Rightarrow 1 \geq 1 - x^n \geq 1 - x^2\\\\\Rightarrow 1 \geq \sqrt{1 - x^n} \geq \sqrt{1 - x^2}\\\\\Rightarrow 1 \leq \dfrac{1}{\sqrt{1 - x^n}} \leq \dfrac{1}{\sqrt{1 - x^2}}\\\\$Integrate all sides from 0 to $\dfrac{1}{2}\\\\\displaystyle\int_{0}^{\frac{1}{2}}dx \leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^n}}\,dx \leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^2}}\,dx\\\\\Rightarrow \dfrac{1}{2}\leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^n}}\,dx \leq \dfrac{\pi}{6}$

Sorry about the unconventional use of inequality signs, (generally two sided inequalities are written as __ < __ < __, not __ > __ > __, but you should get my idea) lol...

kwabon · Oct 3, 2009

Trebla said:
$$Look at the limits of integration$\\0 \leq x \leq \dfrac{1}{2}\\\\$[B]We know tha[/B]t $ 0 \leq x^n \leq x^2 $for $n\geq2\\\\\Rightarrow 1 \geq 1 - x^n \geq 1 - x^2\\\\\Rightarrow 1 \geq \sqrt{1 - x^n} \geq \sqrt{1 - x^2}\\\\\Rightarrow 1 \leq \dfrac{1}{\sqrt{1 - x^n}} \leq \dfrac{1}{\sqrt{1 - x^2}}\\\\$Integrate all sides from 0 to $\dfrac{1}{2}\\\\\displaystyle\int_{0}^{\frac{1}{2}}dx \leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^n}}\,dx \leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^2}}\,dx\\\\\Rightarrow \dfrac{1}{2}\leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^n}}\,dx \leq \dfrac{\pi}{6}$

Sorry about the unconventional use of inequality signs, (generally two sided inequalities are written as __ < __ < __, not __ > __ > __, but you should get my idea) lol...

i do not get how did you get that line.(line which says "we know that") thats troubling me, how would you think of that in the exam, is there a clue or something? i knew there was something with the limits, i didnt know where to go from that, could you please clear that up for me.

thanks.

Trebla · Oct 3, 2009

How to tackle such a question? Simply look at the thing you're trying to prove and work backwards to find your starting point.

$\dfrac{1}{2}\leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^n}}\,dx \leq \dfrac{\pi}{6}\\\\$From the first part you're meant to realise you can rewrite it as$\\\\\dfrac{1}{2}\leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^n}}\,dx \leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^2}}\,dx\\\\$and a bit a thinking should help realise the RHS of the inequality can be rewritten as an integral, hence$\\\\\displaystyle\int_{0}^{\frac{1}{2}}\,dx\leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^n}}\,dx \leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^2}}\,dx\\\\$This implies$\\\\1\leq \dfrac{1}{\sqrt{1 - x^n}}\ \leq \dfrac{1}{\sqrt{1 - x^2}}\\\\\Rightarrow \sqrt{1 - x^2} \leq \sqrt{1 - x^n}\leq 1\\\\\Rightarrow 1 - x^2 \leq 1 - x^n\leq 1\\\\\Rightarrow 0 \leq x^n \leq x^2 \\\\$which easily shown to be true for $0 \leq x \leq \dfrac{1}{2}$

$\\\\$If you want to go even further you get$\\\\0 \leq x^{n-2} \leq 1\\\\$which again is easily shown to be true for $0 \leq x \leq \dfrac{1}{2}$

kwabon · Oct 3, 2009

Trebla said:
How to tackle such a question? Simply look at the thing you're trying to prove and work backwards to find your starting point.

$\dfrac{1}{2}\leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^n}}\,dx \leq \dfrac{\pi}{6}\\\\$From the first part you're meant to realise you can rewrite it as$\\\\\dfrac{1}{2}\leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^n}}\,dx \leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^2}}\,dx\\\\$and a bit a thinking should help realise the RHS of the inequality can be rewritten as an integral, hence$\\\\\displaystyle\int_{0}^{\frac{1}{2}}\,dx\leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^n}}\,dx \leq \displaystyle\int_{0}^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - x^2}}\,dx\\\\$This implies$\\\\1\leq \dfrac{1}{\sqrt{1 - x^n}}\ \leq \dfrac{1}{\sqrt{1 - x^2}}\\\\\Rightarrow \sqrt{1 - x^2} \leq \sqrt{1 - x^n}\leq 1\\\\\Rightarrow 1 - x^2 \leq 1 - x^n\leq 1\\\\\Rightarrow 0 \leq x^n \leq x^2 \\\\$which easily shown to be true for $0 \leq x \leq \dfrac{1}{2}$

legendary, thanks man.

and btw, do you know where i could find these sort of questions? Past papers :-S, couldnt find them other than this question.
will do some practice on the way of the "working backwards"

Drongoski · Oct 3, 2009

$\100dpi For\ 0\leq \ x\ \leq \frac{1}{2} \ \ and\ \ n\geq 2:\\ \\ 1\ =\ \frac{1}{\sqrt{1-0^{n}}} \ \leq \ \frac{1}{\sqrt{1-x^{n}}}\\ \\ (\frac{1}{2})^{n}\ \leq \ (\frac{1}{2})^{2} \ \ \Rightarrow \ \ \sqrt{1-(\frac{1}{2})^{n}}\ \ \geq \ \ \sqrt{1-(\frac{1}{2})^{2}}\\ \\ \Rightarrow \ \ \frac{1}{\sqrt{1-(\frac{1}{2})^{n}}}\ \leq \frac{1}{\sqrt{1-(\frac{1}{2})^{2}}}\\ \\ so\ that:\ \ \int_{0}^{\frac{1}{2}}\ 1 \ dx\ \ \leq \ \int_{0}^{\frac{1}{2}}\ \frac{1}{\sqrt{1-x^{n}}}\ dx\ \leq \int_{0}^{\frac{1}{2}}\ \frac{1}{\sqrt{1-x^{2}}}\ dx\\ \\ i.e.\ \ \frac{1}{2}\ \ \leq \ \ \int_{0}^{\frac{1}{2}}\ \frac{1}{\sqrt{1-x^{n}}}\ dx\ \ \leq \ \ \frac{\pi }{6}$

Edit: looks like I'm 10 minutes late!

Integration (well i think it is) (1 Viewer)

kwabon

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