Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

radnan11 · « **Reply #1710 on:** February 15, 2018, 03:40:23 pm »

Ahh i got most of RuiAce's solution but couldn't get the second last line of working. Thanks

RuiAce · « **Reply #1711 on:** February 15, 2018, 04:25:14 pm »

Quote from: radnan11 on February 15, 2018, 03:40:23 pm

Ahh i got most of RuiAce's solution but couldn't get the second last line of working. Thanks

If you meant the harder-to-see red marker, basically the first integral represents the integral of an odd function and the second represents the area of a semi-circle with radius 1. (Line above is just splitting the integral up in case I misinterpreted). Let me know if anything else was hard to see or etc

kaustubh.patel · « **Reply #1712 on:** February 17, 2018, 03:41:50 pm »

Hey Rui and all maths lovers i saw this integral online and im really trying to crack it but its just too complicated. Please give it a try if you can.
Intergrate [ (ln (x) + 1/x) e^x ] dx

RuiAce · « **Reply #1713 on:** February 17, 2018, 03:50:40 pm »

Quote from: kaustubh.patel on February 17, 2018, 03:41:50 pm

Hey Rui and all maths lovers i saw this integral online and im really trying to crack it but its just too complicated. Please give it a try if you can.
Intergrate [ (ln (x) + 1/x) e^x ] dx

$\text{It's a bit beyond the expected ability of a 4U student}\\ \text{but the idea is that it can instantly be done}\\ \text{from reversing the product rule.}$
\begin{align*}\frac{d}{dx} e^x\ln x &= e^x \ln x + \frac{e^x}{x}\\ \therefore \int \left( \ln x + \frac{1}{x} \right)e^x\, dx &= e^x \ln x + C\end{align*}
________________________________________________________
$\text{Alternatively, one could be really clever with integration by parts.}\\ \begin{align*}\int \frac{1}{x} e^x\,dx &= e^x \ln x - \int e^x \ln x\,dx\\ \therefore \int \frac{1}{x}e^x\,dx + \int e^x\ln \,dx &= e^x\ln x + C\\ \int \left(\frac{1}{x}+\ln x\right)e^x\, dx &= e^x\ln x + C\end{align*}$
$\text{Note that the idea used is similar to the one used}\\ \text{to evaluate the integral }\int e^x \sin x\,d x$
I think this particular integral popped up quite recently in the MIT integration bee?

kaustubh.patel · « **Reply #1714 on:** February 19, 2018, 03:17:07 am »

Oh wow reversing from product rule made it look really simple. I'll be honest this question was from a qualifiers test for the mit intergration bee a friend showed me. I also noticed ,when looking at your 2nd working backwards, that expanding and rearranging is more insightful. Thanks heaps and you wont mind me asking for a solution for integral of (e^x sinx)

RuiAce · « **Reply #1715 on:** February 19, 2018, 07:42:37 am »

Sounds about right; I think I did all of those qualifier questions recently. Some MIT questions are a bit too far but this particular one could be done by a 4U student. Anyway, as for $ \int e^x\sin x\,dx $ you're gonna be expected to know how to do that one.
\begin{align*}\text{Let }I&=\int e^x\sin x\,dx \\ &= -e^x\cos x + \int e^x\cos x\,dx\\ &= -e^x\cos x + e^x\sin x - \int e^x\sin x\,dx\\ &= e^x(\sin x - \cos x) -
I.\\ \therefore 2I &= e^x (\cos x - \sin x) + 2C\tag{*}\\ I &= e^x(\cos x - \sin x ) + C\end{align*}
$\text{The one thing to be wary of is how the }C\text{ (or in this case }2C)\\ \text{just magically appears there.}\\ \text{Essentially, this is only allowed because we're dealing with indefinite integrals.}\\ \text{If we have one indefinite integral on one side and one on the other,}\\ \text{when we move the integrals both to one side, a }+C\text{ is allowed to appear.}$
Conventionally, the formal justification is just to say "statement is true to within a constant"

arii · « **Reply #1716 on:** February 22, 2018, 10:34:21 am »

I need help with this question..

RuiAce · « **Reply #1717 on:** February 22, 2018, 10:43:30 am »

Quote from: arii on February 22, 2018, 10:34:21 am

I need help with this question..

$\text{When this question was examined in 2007 it had previous parts}\\ \text{that I will be assuming.}$
$\text{Since it was previously proved that }\sin x < x\\ \text{upon rearranging we have }-\sin x + x > 0\\ \text{i.e. }g^{\prime\prime}(x) > 0$
$\text{Hence, }g^\prime (x) \text{ is monotonic increasing.}$
Of course, this also means that $ g(x) $ is concave up, but we don't really need this.
$\text{Now that we now that }g^\prime(x)\text{ is monotonic increasing, we take note that}\\ g^\prime(0) = 0.\\ \text{Due to what it means to be a 'monotonic increasing function', this suggests}\\ \text{that }g^\prime(x)\text{ is constantly increasing further and further away from 0, for all }x > 0$
$\text{i.e. }g^\prime(x) > 0\text{ for all }x > 0\\ \text{Of course, this now means that }g(x)\text{ is in fact, also increasing.}$
$\text{So we then note that }g(0) = 0\text{ as well.}\\ \text{Given this, we see that }g(x)\text{ must also increase further and further}\\ \text{away from 0, for all }x > 0$
$\text{i.e. }g(x) > 0\text{ for all }x > 0.\\ \text{From here, just plugging our formula in gives }\sin x - x + \frac{x^3}{6} > 0\\ \text{which rearranges to produce the desired result.}$

arii · « **Reply #1718 on:** February 22, 2018, 11:14:39 am »

Quote from: RuiAce on February 22, 2018, 10:43:30 am

$\text{When this question was examined in 2007 it had previous parts}\\ \text{that I will be assuming.}$
$\text{Since it was previously proved that }\sin x > x\\ \text{upon rearranging we have }-\sin x + x > 0\\ \text{i.e. }g^{\prime\prime}(x) > 0$
$\text{Hence, }g^\prime (x) \text{ is monotonic increasing.}$
Of course, this also means that $ g(x) $ is concave up, but we don't really need this.
$\text{Now that we now that }g^\prime(x)\text{ is monotonic increasing, we take note that}\\ g^\prime(0) = 0.\\ \text{Due to what it means to be a 'monotonic increasing function', this suggests}\\ \text{that }g^\prime(x)\text{ is constantly increasing further and further away from 0, for all }x > 0$
$\text{i.e. }g^\prime(x) > 0\text{ for all }x > 0\\ \text{Of course, this now means that }g(x)\text{ is in fact, also increasing.}$
$\text{So we then note that }g(0) = 0\text{ as well.}\\ \text{Given this, we see that }g(x)\text{ must also increase further and further}\\ \text{away from 0, for all }x > 0$
$\text{i.e. }g(x) > 0\text{ for all }x > 0.\\ \text{From here, just plugging our formula in gives }\sin x - x + \frac{x^3}{6} > 0\\ \text{which rearranges to produce the desired result.}$

Yeah there was a part before which I weirdly proved but I think your method might be better if I can get my head around the first bit.

Proved:
sinx > x
but as soon as you start rearranging...
sinx - x > 0
-(x-sinx) > 0
x-sinx < 0 (sign flips because you divide both sides by -1?)

RuiAce · « **Reply #1719 on:** February 22, 2018, 11:17:38 am »

Quote from: arii on February 22, 2018, 11:14:39 am

Yeah there was a part before which I weirdly proved but I think your method might be better if I can get my head around the first bit.

Proved:
sinx > x
but as soon as you start rearranging...
sinx - x > 0
-(x-sinx) > 0
x-sinx < 0 (sign flips because you divide both sides by -1?)

Typo (I'll fix it now), sin x is definitely less than x.

arii · « **Reply #1720 on:** February 22, 2018, 07:17:33 pm »

Hey Rui (and other math students),

Can you check if I did (a - first box) correctly and help me with (b - second box)?

Consider the graph of the function y=sqrtx
Spelling typo in picture: Curve not Cruve...

RuiAce · « **Reply #1721 on:** February 22, 2018, 08:08:43 pm »

Quote from: arii on February 22, 2018, 07:17:33 pm

Hey Rui (and other math students),

Can you check if I did (a - first box) correctly and help me with (b - second box)?

Consider the graph of the function y=sqrtx
Spelling typo in picture: Curve not Cruve...

$ y$ increasing means $ y^\prime > 0$. $y^{\prime\prime} > 0$ means $y^\prime$ is increasing and $y$ is concave up. So what you had to solve was $\frac12 x^{-1/2} > 0 $.
___________________________________________

Firstly, two useful results.

$\text{Now, by definition, what it means for }f(x)\text{ to be increasing is that}\\ \text{if }x\le a\text{, then }f(x)\le f(a).$
The idea that $f^\prime(x) \ge 0$ is only a way to actually prove that $f(x)$ is increasing. It's not, bydefinition, what increasing actually means.
$\text{Also, integration preserves inequality.}\\ \text{Suppose on some interval }a\le x \le b\text{, we know that }h(x) \le g(x).\\ \text{Then, integrating over this interval doesn't flip the sign of the inequality, i.e.}\\ \int_a^b h(x)\,dx \le \int_a^b g(x)\,dx$

$\text{Given these two results, we can consider the following:}\\ \text{Take }f(x) = \sqrt{x}.\text{ We know that }f(x)\text{ is monotonic increasing.}\\ \text{So whenever }x \le 1, f(x) \le f(1).\\ \text{i.e. }\sqrt{x} \le \sqrt{1}.$
$\text{Now, take }g(x) = \sqrt{1}\text{ and }h(x) = \sqrt{x}.\\ \text{Since we know that }h(x) \le g(x)\text{ for all }0 \le x \le 1\\ \text{we realise that }\int_0^1 h(x)\,dx \le \int_0^1g(x)\,dx\\ \text{i.e. }\boxed{\int_0^1 \sqrt{x}\,dx \le \int_0^1 \sqrt{1}\,dx}$
$\text{Keeping the boxed result in mind, we now rinse and repeat.}\\ \text{This time, we will still take }f(x) = \sqrt{x}.\text{ But whenever }x \le 2\text{ instead,}\\ f(x) \le f(2)\text{, and thus }\sqrt{x} \le \sqrt{2}$
$\text{Now, take }g(x) = \sqrt{2}\text{ and }h(x) = \sqrt{2}.\\ \text{Since we know that }h(x) \le g(x)\text{ for all }1 \le x \le 2\\ \text{now we have }\int_1^2 h(x)\,dx \le \int_1^2 g(x)\,dx\\ \text{i.e. }\boxed{\int_1^2 \sqrt{x}\,dx \le \int_1^2 \sqrt{2}\,dx}$
$\text{Essentially, we rinse and repeat over and over again.}\\ \text{This will give us}\\ \begin{align*}\int_2^3 \sqrt{x}\,dx &\le \int_2^3 \sqrt{3}\,dx\\ \int_3^4 \sqrt{x}\,dx &\le \int_3^4 \sqrt{4}\,dx\\ &\vdots\\ \int_{n-1}^n \sqrt{x}\,dx &\le \int_{n-1}^n \sqrt{n}\,dx\end{align*}$
___________________________________________
$\text{We now have the necessary ingredients to do this question.}\\ \begin{align*}RHS &= \frac{2}{3} n\sqrt{n}\\ &= \left[\frac{2}{3}x^{3/2} \right]_0^n\\ &= \int_0^n \sqrt{x}\,dx \tag{= MHS}\\ &= \int_0^1 \sqrt{x}\,dx + \int_1^2 \sqrt{x}\,dx + \int_2^3 \sqrt{x}\, dx + \dots + \int_{n-1}^n \sqrt{x}\,dx\tag{chopping the integral up}\\ &\tag{from above}\le \int_0^1 \sqrt{1}\,dx + \int_1^2 \sqrt{2}\,dx + \int_2^3 \sqrt{3}\,dx + \dots + \int_{n-1}^n \sqrt{n}\,dx\\ &= \sqrt{1}+\sqrt{2}+\sqrt{3}+\dots +\sqrt{n}\tag{evaluating the integrals}\\ &= LHS\end{align*}\\ \text{as required.}$
Remark: This question could be done a tiny bit more quickly if it said "hence or otherwise". We can just sketch $ y = \sqrt{x} $ and draw a bunch of rectangles above it.

arii · « **Reply #1722 on:** February 22, 2018, 09:01:00 pm »

Thank you so much Rui (you're very likely going to be saving my 4U life throughout the year).

I just wanted to ask you about the last section of your working where you start off as:
RHS = (given)
= (how did you get this... the pro-numerals went from n to x a bit sudden)
= MHS (given in the question)"

RuiAce · « **Reply #1723 on:** February 22, 2018, 09:10:17 pm »

Quote from: arii on February 22, 2018, 09:01:00 pm

Thank you so much Rui (you're very likely going to be saving my 4U life throughout the year).

I just wanted to ask you about the last section of your working where you start off as:
RHS = (given)
= (how did you get this... the pro-numerals went from n to x a bit sudden)
= MHS (given in the question)"

$\text{I pretty much worked backwards there. Probably easier to think about this:}\\ \begin{align*}\int_0^n \sqrt{x}\,dx &= \int_0^n x^{1/2}\,dx\\ &= \left[ \frac{2}{3}x^{3/2}\right]_0^n\\ &= \frac{2}{3}n^{3/2} - \frac{2}{3}\times 0^{3/2}\\ &= \frac{2}{3}n \times n^{1/2}\\ &= \frac23 n\sqrt{n}\end{align*}$

arii · « **Reply #1724 on:** February 23, 2018, 05:12:37 pm »

Not sure how to continue here...

Normally I'd try to factorise out sqrt{k+1} but it seems to complicate it even more...

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