ChetIs the Integral Convergent?

In summary, the original function is convergent after completing the square and using a trigonometric substitution. The antiderivative of the function is not necessary to find for the purpose of determining convergence.
  • #1
Neil21
2
0

Homework Statement


Find whether the integral is convergent or not, and evaluate if convergent.

Homework Equations


integral 1/sqrt(x^4+x^2+1) from 1 to infinity

The Attempt at a Solution


1/sqrt(x^4+x^2+1)<1/sqrt(x^4)
1/sqrt(x^4)=1/x^2 which is convergent for 1 to infinity and is 1
therefore, the original function is convergent
I can't seem to find the antiderivative of the original function after this.
 
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  • #2
Try completing the square
 
  • #3
Ok, so I complete the square and got [tex] \int \frac{1}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}} [/tex] . I am still not seeing a way find an antiderivative though.
 
  • #4
Neil21 said:
Ok, so I complete the square and got [tex] \int \frac{1}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}} [/tex] . I am still not seeing a way find an antiderivative though.
Don't forget the dx.
$$\int \frac{dx}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}}$$

I haven't worked this through, yet, but here's what I would do, FWIW.
Let u = x2 + 1/2
So ##x = \sqrt{u - 1/2}##
Then du = 2xdx
So ##dx = \frac{du}{2x} = \frac{du}{2\sqrt{u - 1/2}}##

After making the substitution and getting everything in terms of u and du, I would try for a trig substitution. I think that might work.
 
  • #5
You don't need to integrate the exact expression. If the 3/4 wasn't in there, would the integral be convergent?

Chet
 
Last edited:
  • #6
Trig substitution is always my first instinct when I see a fraction with higher powers of x.
 
  • #7
Chestermiller said:
You don't need to integrate the exact expression. If the 3/4 wasn't in there, would the integral be convergent?

Chet
@chet, I think he already established that in the opening post.
 
  • #8
Mark44 said:
Don't forget the dx.
$$\int \frac{dx}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}}$$

I haven't worked this through, yet, but here's what I would do, FWIW.
Let u = x2 + 1/2
So ##x = \sqrt{u - 1/2}##
Then du = 2xdx
So ##dx = \frac{du}{2x} = \frac{du}{2\sqrt{u - 1/2}}##

After making the substitution and getting everything in terms of u and du, I would try for a trig substitution. I think that might work.
What I first saw was rewriting the expression as $${\frac{1}{(x^4+x^2+1)^{1/2}}}={\frac{1}{((x^2+({\frac{1}{2}))^2}+(\frac{\sqrt{3}}{2})^2)^{1/2}}}$$.
Which I'm pretty sure can be solved with a trig sub.
 
  • #9
Mark44 said:
@chet, I think he already established that in the opening post.
Ooops. Missed that. Sorry guys.

Chet
 

Related to ChetIs the Integral Convergent?

1. What is the definition of convergence of an integral?

The convergence of an integral refers to whether or not the value of the integral exists or is finite. In other words, it is a measure of whether the function being integrated approaches a finite limit as the interval of integration becomes smaller.

2. How is the convergence of an integral determined?

The convergence of an integral is determined by evaluating the integral using various methods such as the comparison test, the limit comparison test, or the integral test. These methods help determine if the integral converges or diverges.

3. What is the difference between absolute and conditional convergence of an integral?

Absolute convergence of an integral means that the integral converges regardless of the order of integration. Conditional convergence, on the other hand, means that the integral only converges when integrated in a specific order.

4. Can an integral converge to a negative value?

Yes, an integral can converge to a negative value. The convergence of an integral is determined by the limit of the function being integrated, so it is possible for the limit to be negative, resulting in a negative value for the integral.

5. What are some real-world applications of the convergence of an integral?

The convergence of an integral has many applications in physics, engineering, and economics. For example, it is used to calculate the area under a curve, which is important in determining work done, finding the center of mass, and calculating probabilities in economics.

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