Help! Integration

If the value of

f(k)=0xk2x6+4x5+3x4+5x3+3x2+4x+2dx\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

101k101k, (k is an integer)

Note by Krishna Sharma
5 years, 2 months ago

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1 vote

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

Differentiating both sides with respect to k we get :

f(k)=0ln(x).xk(2x6+4x5+3x4+5x3+3x2+4x+2)dx=I\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 :

I=f(k)=01ln(x).xk(2x6+4x5+3x4+5x3+3x2+4x+2)dx+1ln(x).xk(2x6+4x5+3x4+5x3+3x2+4x+2)dx\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=1yx=\frac{1}{y} to get :

f(k)=01ln(x).xk(2x6+4x5+3x4+5x3+3x2+4x+2)dx+01ln(y).y4k(2y6+4y5+3y4+5y3+3y2+4y+2)dy\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 :

f(k)=01ln(x).(xkx4k)(2x6+4x5+3x4+5x3+3x2+4x+2)dx\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 00 we get :

k=4kk=4-k

k=2\boxed{\Rightarrow k=2}

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

Ronak Agarwal - 5 years, 2 months ago

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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 - 5 years, 2 months ago

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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 - 5 years, 2 months ago

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

f(k)=0xk+x4xk(2x6+4x5+3x4+5x3+3x2+4x+2)dxf(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 PalindromePalindrome 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 xx)

So Apply

AM=GMAM\quad =\quad GM. in numerator Keeping xx constant .

Or

Also thinks that in Numerator as 'k' increases First term is Increases and Second term is decreases
So Minimum Value of Numerator occurs when firstfirst termterm=SecondSecond TermTerm. in Numerator

Or

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

This Gives k=2k=2.

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

Deepanshu Gupta - 5 years, 2 months ago

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@Deepanshu Gupta Very beautiful approach , really very nice

U Z - 4 years, 11 months ago

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@U Z thank you [ feeling blessed :) ]

Deepanshu Gupta - 4 years, 11 months ago

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@Deepanshu Gupta This was really great, hats off

U Z - 4 years, 11 months ago

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

The answer is 2.

Krishna Sharma - 5 years, 2 months ago

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