×

# 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)

Note by Krishna Sharma
2 years, 3 months ago

Sort by:

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. · 2 years, 3 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.!! · 2 years, 3 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 · 2 years, 3 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 .

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 $$first$$ $$term$$=$$Second$$ $$Term$$. in Numerator

Or

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. · 2 years, 3 months ago

This was really great, hats off

· 2 years, 1 month ago

Very beautiful approach , really very nice · 2 years, 1 month ago

thank you [ feeling blessed :) ] · 2 years, 1 month ago

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. · 2 years, 3 months ago

@Michael Mendrin · 2 years, 3 months ago