Help! Integration

If the value of

$\displaystyle f(k) = \int_0^{\infty} \frac{x^{k}}{2x^{6} + 4x^{5} + 3x^{4} + 5x^{3} + 3x^{2} + 4x + 2} dx $

Is minimum

Then find

$101k$ , (k is an integer)

#Calculus #Integration

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

I don't know whether I am right or not but I am getting $k=2$ , here's a solution.

Differentiating both sides with respect to k we get :

$\large { f }^{ ' }(k)=\int _{ 0 }^{ \infty }{ \frac { ln(x){ .x }^{ k } }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx }=I$

I differentiated here using newton leibnitz rule.

Splitting the integral into two parts we get :

$\large I={ f }^{ ' }(k)=\int _{ 0 }^{ 1 }{ \frac { ln(x){ .x }^{ k } }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx } +\int _{ 1 }^{ \infty }{ \frac { ln(x){ .x }^{ k } }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx }$

In the second part put $x=\frac{1}{y}$ to get :

$\large { f }^{ ' }(k)=\int _{ 0 }^{ 1 }{ \frac { ln(x){ .x }^{ k } }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx } +\int _{ 0 }^{ 1 }{ \frac { -ln(y).y^{ 4-k } }{ (2y^{ 6 }+4y^{ 5 }+3y^{ 4 }+5y^{ 3 }+3y^{ 2 }+4y+2) } dy }$

Combining both integrals we get :

$\large { f }^{ ' }(k)=\int _{ 0 }^{ 1 }{ \frac { ln(x){ .(x }^{ k }-{ x }^{ 4-k }) }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx }$

Equating the derivative equal to $0$ we get :

$k=4-k$

$\boxed{\Rightarrow k=2}$

I won't do the second derivative test as I know this has the minimum possible value.

Ronak Agarwal - 6 years, 8 months ago

Is The answer is 202 means K=2 ??
( I have feeling That it might not be correct)

Since You Didn't Specify That 'K' is any real number or an integer So I have Assumed That
K is integer.. If it is real then please mention it.!!

Deepanshu Gupta - 6 years, 8 months ago

Question was integer type and we need to find K...I guess so it is integer. How do got 2?

The answer is 2

Krishna Sharma - 6 years, 8 months ago

Well I first substitute $x=\frac { 1 }{ t }$ . and then using Properties of integration I get :

$f(k)=\int _{ 0 }^{ \infty }{ \frac { { x }^{ k }+{ \frac { { x }^{ 4 } }{ { x }^{ k } } } }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx }$ .

Since Denominator is an $Palindrome$ Then it must be Products of Perfect squares. So Denominator is always Positive.

And To Minimize integral then We have To minimize The numerator. (For a Particular $x$ )

So Apply

$AM\quad =\quad GM$ . in numerator Keeping $x$ constant .

Also thinks that in Numerator as 'k' increases First term is Increases and Second term is decreases
So Minimum Value of Numerator occurs when $first$ $term$ = $Second$ $Term$ . in Numerator

You can also Prove it by using calculus That numerator is minimum when $K=2$ . Hence The Integral also minimum.

This Gives $k=2$ .

But I'am not sure that this method is correct or not.

Deepanshu Gupta - 6 years, 8 months ago

@Deepanshu Gupta – Very beautiful approach , really very nice

U Z - 6 years, 5 months ago

@U Z – thank you [ feeling blessed :) ]

Deepanshu Gupta - 6 years, 5 months ago

@Deepanshu Gupta – This was really great, hats off

U Z - 6 years, 5 months ago

@Deepanshu Gupta – Yes this is correct! My teacher also said to substitute $x = \frac {1}{y}$ You should have differentiated the function. $(f'(x) = 0$ ) would be more correct.

The answer is 2.

Krishna Sharma - 6 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$