Integration by Parts - Choice of variables

In summary, the conversation discusses a problem with choosing the correct u and dv values for Integration by Parts on a given integral. The correct solution is found by setting u = x^2 and dv = 2x e^x^2 dx, while the incorrect solution is found by setting u = 2x^3 and dv = e^x^2 dx. The error is in the calculation of the integral of e^{x^2}, which leads to a different final solution.
  • #1
takarin
2
0

Homework Statement



I'm getting different results when choosing my u & dv for Integration by Parts on the following integral:

[tex]\int 2x^3 e^x^2 dx [/tex]
(Note, the exponent on 'e' is x^2)

This yields the correct solution:
u = [tex]x^2[/tex]
dv = [tex]2x e^x^2 dx [/tex]

du = [tex]2xdx[/tex]
v = [tex]e^x^2[/tex]

However, I have tried using this instead (*)

u = [tex]2x^3[/tex]
dv = [tex]e^x^2 dx [/tex]

du = [tex]6x^2 dx[/tex]
v = [tex](e^x^2) / 2x[/tex]

and this is yielding the incorrect solution (see 3.)

Homework Equations


Integration by Parts:
[tex]\int udv = uv - \int vdu [/tex]

The Attempt at a Solution



The correct solution turns out to be
[tex] x^2 e^x^2 - e^x^2 + C[/tex]

When I use my other choice of variables (*), I get (using IBP)
[tex] \int 2x^3 e^x^2 dx = 2x^3 e^x^2 / 2x - \int e^x^2 / 2x * 6x^2 dx [/tex]
which leads to:
[tex]x^2 e^x^2 - 3/2 e^x^2 + C [/tex]

which is different from the other choice of variables.

I've looked over both choices of variables, and I don't know why the second choice (*) comes up with a different solution.

Thanks for the help!
 
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  • #2
Your error is that the integral of [tex]e^{x^2}[/tex] isn't [tex]e^{x^2}/(2x)[/tex]. (I'm assuming you meant to have parentheses on the bottom, but perhaps not. Regardless, the answer is incorrect either way.)
 
  • #3
Ah! Silly me, I've been staring at it and completely overlooked that. Thanks for your quick reply, I'll avoid such carelessness in the future :)
 

Related to Integration by Parts - Choice of variables

1. What is the concept of integration by parts?

The concept of integration by parts is a technique used to find the integral of a product of two functions. It is based on the product rule of differentiation, where the derivative of the product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. This technique is useful when the integral cannot be easily evaluated using other methods.

2. How do I choose the variables when using integration by parts?

When using integration by parts, the choice of variables can greatly affect the ease of integration. Typically, the choice of variables should be such that the integral of the second function becomes simpler or easier to integrate after differentiating it. A common rule of thumb is to choose the second function as the one that becomes simpler when differentiated.

3. Can integration by parts be used for definite integrals?

Yes, integration by parts can also be used for definite integrals. In this case, the integration by parts formula is modified to include the limits of integration and the resulting integral is evaluated using the fundamental theorem of calculus.

4. What are some common examples where integration by parts is used?

Integration by parts is commonly used in various fields of science and mathematics, such as physics, engineering, and statistics. It is often used to solve integrals involving trigonometric functions, logarithmic functions, and exponential functions. It is also used in probability and statistics to find the expected value of a continuous random variable.

5. Are there any other techniques for evaluating integrals?

Yes, there are several other techniques for evaluating integrals, such as substitution, trigonometric substitution, partial fractions, and using trigonometric identities. Each of these techniques is useful for different types of integrals and it is important to understand when to apply each method.

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