Integrating $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$: U-Substitution or Parts?

In summary, the conversation discusses the difficulty of integrating the function \int_0^\sqrt{6}}e^{-x^2}\frac{x^2}{2}, with suggestions of using u-substitution or integration by parts. However, it is concluded that the integral cannot be solved in terms of elementary functions and instead can be expressed in terms of the Error function. The conversation also mentions using infinite series to approximate the integral, but it is argued that an answer in terms of the Error function is more informative and elegant. The conversation ends with a humorous suggestion of creating a new function, "easyanswer(t)", to simplify integrals.
  • #1
RadiationX
256
0
[tex]\int_0^\sqrt{6}}e^{-x^2}\frac{x^2}{2}[/tex]

should i use a u-substitution or integration by parts?
 
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  • #2
You can not solve this in terms of elementary functions. It is possible to express the answer in terms of the Error function, that is:

[tex]\text{Erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt[/tex]

You might want to give a shot at it yourself knowing that?
 
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  • #3
I don't think that I was being explicit enough in my first post: This is the double integral[tex]\int_0^3\int_{\sqrt{2y}}^{\sqrt{6}}e^{-x^2}dxdy[/tex] that reduces to my original post integral

[tex]\int_0^\sqrt{6}}e^{-x^2}\frac{x^2}{2}dx[/tex]

If I reduced correctly then how can i possibly integrate the integral with elementary functions?
 
  • #4
Did I at least change the order of integration correctly?
 
  • #5
Yes, you have changed it correctly.

It is not possible to integrate with elementary functions no matter how you look at it.
 
  • #6
[tex]\int \left( e^{-x^2} \frac{x^2}{2} \right) dx = \frac{1}{4} \left( \int_0^x e^{t^2} dt - e^{-x^2} x \right) + C[/tex]
 
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  • #7
RadiationX said:
[tex]\int_0^\sqrt{6}}e^{-x^2}\frac{x^2}{2}[/tex]

should i use a u-substitution or integration by parts?


These a-hole intergral disgust me greatly when i worked on calaulus..

It is better if you just express e^t as a infinite serie. Substitude t=-x^2
in to the series. After that, multiple the entire series by x/2. intergrat it term by term, and plug numbers. This function can only be tame;not solve.
 
  • #8
kant said:
These a-hole intergral disgust me greatly when i worked on calaulus..

It is better if you just express e^t as a infinite serie. Substitude t=-x^2
in to the series. After that, multiple the entire series by x/2. intergrat it term by term, and plug numbers. This function can only be tame;not solve.
An answer in terms of erf is no worse than one in terms of sin or exp. There are table and computer programs to find values. Infinite series are helpful for some purposes, but unless one is going to compute an approximation by hand, an expression in terms of erf looks nicer and is more informative. Were you also discusted by integrals like
[tex]\int_1^2\frac{dx}{x}[/tex]
What good is an answer like log(2) or sin(exp(sqrt(2))) anyway.
End special treatment for elementary function.
Equality for special functions.
Equal rights for all functions.
 
  • #9
lurflurf said:
End special treatment for elementary function.
Equality for special functions.
Equal rights for all functions.
Mathematical constipation :biggrin:
I mean constitution o:)

-- AI
 
  • #10
lurflurf said:
What good is an answer like log(2) or sin(exp(sqrt(2))) anyway.
End special treatment for elementary function.
Equality for special functions.
Equal rights for all functions.
Well in that case my new function is called easyanswer(t), easyanswer(t) is defined such that where t is some real number of my choice it is the solution to the given numerical integral in front of me. Much easier exams now :biggrin:
 

FAQ: Integrating $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$: U-Substitution or Parts?

What is the formula for integrating $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$?

The formula for integrating $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$ is $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}dx = \frac{\sqrt{\pi}}{4}erf(\sqrt{6}) - \frac{1}{4}e^{-6}$.

What is the difference between using u-substitution and integration by parts for $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$?

U-substitution involves substituting a variable, u, for part of the integrand in order to simplify the integral. Integration by parts involves identifying a part of the integrand to be the differential of another function, and using the product rule to simplify the integral. In this case, u-substitution is more straightforward and efficient.

What is the benefit of using u-substitution for $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$?

The benefit of using u-substitution for $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$ is that it simplifies the integral and makes it easier to solve. This method is especially useful when the integrand involves a complicated function, such as an exponential or trigonometric function.

What is the general process for using u-substitution to solve $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$?

The general process for using u-substitution to solve $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$ is as follows:

  1. Identify the variable to be substituted, u.
  2. Find the derivative of u, du.
  3. Substitute u and du into the integrand, simplifying the integral.
  4. Solve the new integral with respect to u.
  5. Substitute back in the original variable, x, and solve for the final answer.

What are some common mistakes to avoid when using u-substitution to solve $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$?

Some common mistakes to avoid when using u-substitution for $\int_0^{\sqrt{6}}e^{-x^2}\frac{x^2}{2}$ include:

  • Choosing the wrong variable to substitute.
  • Forgetting to include the "dx" at the end of the integral when substituting.
  • Forgetting to change the bounds of integration when substituting.
  • Not simplifying the integral after substituting.

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