Definite Integral

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  • #1
Saitama
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Homework Statement


Consider ##f(x)=4x^4-24x^3+31x^2+6x-8## be a polynomial function and ##\alpha, \beta, \gamma, \delta## are the roots of the equation ##f(x)=0##, where ##\alpha < \beta < \gamma < \delta##. Let sum of two roots of the equation f(x)=0 vanishes. Then the value of
[tex]\int_{2\alpha}^{2\beta} \frac{x^{\delta+1}-5x^{\gamma+1}+2\beta |x|+1}{x^2+4\beta |x|+1}dx[/tex]


Homework Equations





The Attempt at a Solution


I succeeded in finding the roots. ##\alpha=\frac{-1}{2}, \beta=\frac{1}{2}, \gamma=2## and ##\delta=4##.
Therefore, the integral is
[tex]\int_{-1}^{1} \frac{x^5-5x^3+|x|+1}{x^2+2|x|+1}dx[/tex]
I don't know how to proceed further from here. The only thing I can think of is to write the integral in two parts,
[tex]\int_{-1}^{0} \frac{x^5-5x^3-x+1}{(x-1)^2}dx +\int_{0}^{1} \frac{x^5-5x^3+x+1}{(x+1)^2}dx [/tex]

Any help is appreciated. Thanks!
 

Answers and Replies

  • #2
haruspex
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The only thing I can think of is to write the integral in two parts,
[tex]\int_{-1}^{0} \frac{x^5-5x^3-x+1}{(x-1)^2}dx +\int_{0}^{1} \frac{x^5-5x^3+x+1}{(x+1)^2}dx [/tex]
That's a very sensible step. Next, can you recombine them, by change of variable? You should get some cancellation.
 
  • #3
Saitama
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That's a very sensible step. Next, can you recombine them, by change of variable? You should get some cancellation.

Great! Thanks a lot. That thing never came into my mind. :smile:

I substituted t=-x in the first integral and the terms cancelled. :)

Is the answer 2ln2?
 
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  • #4
Dick
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Is the answer 2ln2?

Yes. But what a strange question. Why did they tell you to suppose two of the roots sum to zero? If you can solve the equation as you did you don't need to be told that.
 
  • #5
Saitama
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Yes. But what a strange question. Why did they tell you to suppose two of the roots sum to zero? If you can solve the equation as you did you don't need to be told that.

I am not sure about why did they give that piece of information. This question is a part of paragraph based questions. There are three questions. I was able to solve the other two so I did not post them here. Do you want me to post the other two questions too?

Can you help me in this thread: https://www.physicsforums.com/showthread.php?t=668114
 
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  • #6
Dick
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I am not sure about why they did give that piece of information. This question is a part of paragraph based questions. There are three questions. I was able to solve the other two so I did not post them here. Do you want me to post the other two questions too?

Can you help me in this thread: https://www.physicsforums.com/showthread.php?t=668114

Sure, post them. Maybe that will give us clue about the strange assumption. I did look at the other thread. And I don't know. g'(1/2)=f(1/2). Haven't figured that one out yet either.
 
  • #7
Saitama
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Sure, post them. Maybe that will give us clue about the strange assumption.

Here goes the questions:

Q.1) The value of the expression
[tex]\delta^{\beta}+\frac{1}{\delta^{\alpha}}+\delta^{ \gamma }+\gamma^{\delta}[/tex]
is:

Q.2) [tex]\int \left(\frac{x-\delta}{x-\gamma}\right)^{\alpha+\beta+\delta}dx[/tex]
 
  • #8
jbunniii
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Yes. But what a strange question. Why did they tell you to suppose two of the roots sum to zero? If you can solve the equation as you did you don't need to be told that.
I thought that was pretty strange, too. But I guess it does help you find the roots: if [itex]c[/itex] and [itex]-c[/itex] are both roots, then we have
$$0 = f(c)=4c^4-24c^3+31c^2+6c-8 = 4c^4 + 24c^3 + 31c^2 - 6c - 8 = f(-c)$$
so
$$48c^3 - 12c = 0$$
As ##c = 0## is not a root of ##f##, we must have
$$4c^2 - 1 = 0$$
and so ##c = \pm 1/2##
 
  • #9
Saitama
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I thought that was pretty strange, too. But I guess it does help you find the roots: if [itex]c[/itex] and [itex]-c[/itex] are both roots, then we have
$$0 = f(c)=4c^4-24c^3+31c^2+6c-8 = 4c^4 + 24c^3 + 31c^2 - 6c - 8 = f(-c)$$
so
$$48c^3 - 12c = 0$$
As ##c = 0## is not a root of ##f##, we must have
$$4c^2 - 1 = 0$$
and so ##c = \pm 1/2##

Through observation, I was able to find one root. Then figuring out the other roots was easy. But thanks for the alternative way, if all the roots would have been fractional, I would never be able to find the roots through observation.
 
  • #10
jbunniii
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Through observation, I was able to find one root. Then figuring out the other roots was easy. But thanks for the alternative way, if all the roots would have been fractional, I would never be able to find the roots through observation.
I bet you could, if you use the rational root theorem:
http://en.wikipedia.org/wiki/Rational_root_theorem
 
  • #11
Saitama
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Ah, I always forget about that.
 

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