# Solving Definite Integral Homework: f(x)=4x^4-24x^3+31x^2+6x-8

• Saitama
In summary: It's a neat theorem.I bet you could, if you use the rational root theorem:Yes, you could use the rational root theorem.
Saitama

## 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
$$\int_{2\alpha}^{2\beta} \frac{x^{\delta+1}-5x^{\gamma+1}+2\beta |x|+1}{x^2+4\beta |x|+1}dx$$

## 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
$$\int_{-1}^{1} \frac{x^5-5x^3+|x|+1}{x^2+2|x|+1}dx$$
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,
$$\int_{-1}^{0} \frac{x^5-5x^3-x+1}{(x-1)^2}dx +\int_{0}^{1} \frac{x^5-5x^3+x+1}{(x+1)^2}dx$$

Any help is appreciated. Thanks!

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

haruspex said:
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.

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

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Pranav-Arora said:

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.

Dick said:
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?

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Pranav-Arora said:
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?

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.

Dick said:
Sure, post them. Maybe that will give us clue about the strange assumption.

Here goes the questions:

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

Q.2) $$\int \left(\frac{x-\delta}{x-\gamma}\right)^{\alpha+\beta+\delta}dx$$

Dick said:
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 $c$ and $-c$ 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##

jbunniii said:
I thought that was pretty strange, too. But I guess it does help you find the roots: if $c$ and $-c$ 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.

Pranav-Arora said:
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

Ah, I always forget about that.

## What is a definite integral?

A definite integral is a mathematical concept that represents the area under a curve between two specific points on a graph. It is denoted by ∫f(x)dx, where f(x) is the function and dx represents the infinitesimal width of the rectangles used to approximate the area.

## What is the process for solving a definite integral?

The process for solving a definite integral involves finding the antiderivative of the function, plugging in the upper and lower limits of integration, and subtracting the resulting values to find the area under the curve.

## How do you find the antiderivative of a polynomial function?

The antiderivative of a polynomial function can be found by using the power rule, which states that the antiderivative of x^n is (x^(n+1))/(n+1). For the given function f(x)=4x^4-24x^3+31x^2+6x-8, the antiderivative is (4x^5)/5 - (6x^4)/4 + (31x^3)/3 + (3x^2)/2 - 8x + C, where C is the constant of integration.

## What are the steps for solving a definite integral involving a polynomial function?

The steps for solving a definite integral involving a polynomial function are:

1. Find the antiderivative of the function using the power rule.
2. Plug in the upper and lower limits of integration into the antiderivative.
3. Subtract the resulting values to find the area under the curve.
4. If necessary, simplify the final answer.

## What is the importance of using correct notation when solving definite integrals?

Using correct notation when solving definite integrals is crucial because it helps to clearly communicate the solution and avoid any confusion. It also follows standard mathematical conventions and allows for easy verification of the solution.

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