I've completed the solution below in parts. If you have never done differentiation under the integral sign before, then try to study this solution perhaps; otherwise, use the spoilers to finish it yourself (asking any questions you might have in either case). But first let's get the differentiation out of the way.

Unparseable latex formula:

\displaystyle \frac{d}{d\lambda}\left( \frac{x^{\lambda}-1}{\log{x}} \right) = \frac{d}{d\lambda}\left( \frac{x^{\lambda}}{\log{x}} \right) = \frac{1}{\log{x}}\frac{d}{d \lambda}\left x^{ \lambda}

-- if we let

y = x^{\lambda}

then

\displaystyle \log{y} = \lambda \log{x} \implies \frac{1}{y} \frac{dy}{d\lambda} = \log{x} \implies \frac{dy}{d\lambda} = y \log(x) = x^{\lambda}\log{x}.

That's we have

Unparseable latex formula:

\displaystyle \frac{d}{d \lambda}\left x^{ \lambda} = x^{\lambda}\log{x}

\displaystyle \frac{d}{d\lambda}\left( \frac{x^{\lambda}-1}{\log{x}} \right) =\frac{1}{\log{x}}\cdot x^{\lambda}\log{x} =x^{\lambda}.

Spoiler

If anyone is interested, the double integral way also gets the same general value

Spoiler

Just got to this now, mate.
I totally get it, this is a cool way of integrating.

Reply 3903

Lord of the Flies

Original post by Mladenov

I would like to share the following problem, which, recently, appeared on high school olympiad.

Let f be a continuous function

f:[0,1] \rightarrow \mathbb{R}

such that:

\int_{x}^{1} f(t) \, dt \ge \frac{1-x^2}{2}

, for any

x \in [0,1]

. Prove that

\int_{0}^{1} f^2(t) \, dt \ge \frac{1}{3}

\displaystyle \int_{x}^{1} f(t) \, dt \ge \frac{1-x^2}{2}\iff \int_x^1 f(t)-t\,dt\geq 0

Let

\displaystyle F(x)=\int_x^1 f(t) -t\,dt

and consider

\displaystyle\xi (x)=x \int_x^1 F(t)\,dt

\displaystyle\xi'(x)=\int_x^1 F(t)\,dt-x F(x)\Rightarrow \xi'(0)\geq 0\quad(\star)

Partial integration gives:

\displaystyle\xi'(x)= t F(t)\bigg|_x^1+\int_x^1 t\Big(f(t)-t\Big)dt-x F(x)

Noting that

F(1)=0:

\displaystyle\xi'(0)=\int_0^1 t\Big(f(t)-t\Big)dt\geq 0

(\star)

Everything falls into place from here:

\displaystyle\int_0^1 \bigg(f(t)-t\bigg)^2\,dt\geq 0\Rightarrow \int_0^1 f^2(t)-t^2\,dt\geq 2\int_0^1 t\Big(f(t)-t\Big)dt\geq 0

Hence

\displaystyle\int_0^1 f^2(t)\,dt\geq \int_0^1 t^2\,dt=\frac{1}{3}

(edited 11 years ago)

Reply 3904

Mladenov

Spoiler

My idea was very similar to yours.

Here is my solution:

Spoiler

Permit me to propose another intriguing problem.

Let

f:[0,1]\to\mathbb{R}

be a convex and twice differentiable function so that

f(0)=0

. Prove that :

2\int_0^1f(x)\ \text{d}x\, \le\, f(1)

.
Try to obtain a more general result.

Reply 3905

L'art pour l'art

Follows

\displaystyle f\left( \frac{a+b}{2}\right) \le \frac{1}{b - a}\int_a^b f(x)\,dx \le \frac{f(a) + f(b)}{2} \implies 2\int_0^1f(x)\ dx\, \le\, f(1).

Reply 3906

I am Ace

Sorry for clogging,
but I need a university Physics student for advice on an experiment about wave-particle duality, not all at once now :rolleyes:

Reply 3907

Mladenov

Original post by L'art pour l'art

Spoiler

This solution is an overkill. Bravo.

A more general result:

Spoiler

Here is another problem:

Let

\displaystyle f: [0,1]\to\mathbb R

be a derivable function, with a continuous derivative

\displaystyle f'

\displaystyle [0,1]

. Prove that if

Unparseable latex formula:

\displaystyle f\left( \frac 12\right) \equal{} 0

, then

Unparseable latex formula:

\displaystyle \[ \int^1_0 \left( f'(x) \right)^2 dx \geq 12 \left( \int^1_0 f(x) dx \right)^2.\]

\displaystyle c=12

the best possible constant?

Reply 3908

Dadeyemi

Original post by Mladenov

This solution is an overkill. Bravo.

A more general result:

Spoiler

Here is another problem:

Let

\displaystyle f: [0,1]\to\mathbb R

be a derivable function, with a continuous derivative

\displaystyle f'

\displaystyle [0,1]

. Prove that if

Unparseable latex formula:

\displaystyle f\left( \frac 12\right) \equal{} 0

, then

Unparseable latex formula:

\displaystyle \[ \int^1_0 \left( f'(x) \right)^2 dx \geq 12 \left( \int^1_0 f(x) dx \right)^2.\]

\displaystyle c=12

the best possible constant?

Differentiable

Reply 3909

I am Ace

Can someone who had an interview this year for Maths and Physics at Pembroke explain to me how to find the volume of that squiggly shape that came up like a cone?
I can never remember how it was done - although I (sort of) got it in the interview.
Or if someone knows what I'm talking about, that'd be nice

Reply 3910

Farhan.Hanif93

Original post by I am Ace

Afraid not; interview questions aren't to be discussed on TSR as they're often recycled by the Colleges. If you know anyone that applied personally, perhaps ask them directly instead through a private medium.

Reply 3911

I am Ace

Original post by Farhan.Hanif93

Is this so?
I thought that interview questions could be discussed post-interview?

Reply 3912

Farhan.Hanif93

Original post by I am Ace

Is this so?
I thought that interview questions could be discussed post-interview?

Yup, the reason for that is that Colleges often reuse questions in future cycles so any specific discussion on here may give an unfair advantage to some/render the question ineffective from an admissions standpoint.

Reply 3913

I am Ace

Has any reapplicant ever been asked the same question twice?

Reply 3914

Lord of the Flies

Original post by I am Ace

Has any reapplicant ever been asked the same question twice?

Each college deals with admissions separately, and a reapplicant will usually have enough common sense not to apply to the same college twice: so even though colleges may recycle questions, getting the exact same question from two different colleges seems incredibly unlikely, unless it's something very basic like "sketch sin(x)".

(edited 11 years ago)

Reply 3915

I am Ace

How do you prefer solving differential equations:
With limits from y to y nought and same for x
or with constant of integration?

Reply 3916

I am Ace

Why is the constant, e used for growths?

Reply 3917

Mladenov

Original post by Mladenov

Let

\displaystyle f: [0,1]\to\mathbb R

be a derivable function, with a continuous derivative

\displaystyle f'

\displaystyle [0,1]

. Prove that if

Unparseable latex formula:

\displaystyle f\left( \frac 12\right) \equal{} 0

, then

Unparseable latex formula:

\displaystyle \[ \int^1_0 \left( f'(x) \right)^2 dx \geq 12 \left( \int^1_0 f(x) dx \right)^2.\]

\displaystyle c=12

the best possible constant?

Hint:

Spoiler

Reply 3918

number23

Is anyone doing maths at uni? I have 5 weeks over easter before my exams and was wondering how best to organize my time/revise as i really want over a 2:1. I was thinking spending the first week or 2 going over lecture notes/making summary notes and the rest of time practicing questions... what do u think?

Reply 3919

Kallisto

Entertainment Forum Helper

I need help to solve an integral. To be exactly I was able to solve the integral to a term. And know I'm wonder what should I do to get the figure of x to a figure of y. Anyone who want to help me?