U-sub integral: Not sure where i went wrong.

• dwdoyle8854
In summary, the conversation discusses an issue with evaluating an integral and producing two different answers. The person took pictures of the integral evaluated two ways and noticed that even when evaluating from 0 to 1, the numbers were different. The incorrect way of solving the integral involves a mistake in basic algebra with a rational expression. The correct way is shown in the provided images.
dwdoyle8854

Homework Statement

Im not sure what i did wrong, every operation i took to evaluate the integral seems valid. I took pictures of the integral evaluated two different ways, producing two different answers. What bugs me is even if i evaluate (as i did) from say 0 to 1 the numbers are different, so they are not equivalent. I don't see any errors in the first one, but the textbook lists the latter as a correct answer.

The incorrect way:

Correct:

The Attempt at a Solution

The incorrect way:
<a href="[PLAIN]http://imgur.com/2QfhdVs"><img src="http://i.imgur.com/2QfhdVs.jpg?1" title="Hosted by imgur.com" alt="" /></a>[/PLAIN]

Correct:
<a href="[PLAIN]http://imgur.com/24E2jat"><img src="http://i.imgur.com/24E2jat.jpg" title="Hosted by imgur.com" alt="" /></a>[/PLAIN]

Last edited by a moderator:
Saying u/(2+u^2-2u) equals u/2+u^2/u-2u/u is seriously bad algebra. (a+b)/c=a/c+b/c. c/(a+b) IS NOT equal to c/a+c/b. Try some numbers.

dwdoyle8854 said:

Homework Statement

I'm not sure what i did wrong, every operation i took to evaluate the integral seems valid. I took pictures of the integral evaluated two different ways, producing two different answers. What bugs me is even if i evaluate (as i did) from say 0 to 1 the numbers are different, so they are not equivalent. I don't see any errors in the first one, but the textbook lists the latter as a correct answer.

The incorrect way:
[ IMG]http://i.imgur.com/2QfhdVs.jpg?1[/PLAIN]
Correct:
[ IMG]http://i.imgur.com/24E2jat.jpg[/PLAIN]

The Attempt at a Solution

The incorrect way:
<a href="[PLAIN]http://imgur.com/2QfhdVs"><img src="http://i.imgur.com/2QfhdVs.jpg?1" title="Hosted by imgur.com" alt="" /></a>[/PLAIN]

Correct:
<a href="[PLAIN]http://imgur.com/24E2jat"><img src="http://i.imgur.com/24E2jat.jpg" title="Hosted by imgur.com" alt="" /></a>[/PLAIN]
Your error is in doing basic algebra with a rational expression.

$\displaystyle \frac{u}{u^2-2u+2}\ne\frac{u}{u^2}-\frac{u}{2u}+\frac{u}{2}$

After all, is $\displaystyle \ \ \frac{1}{1+2+3}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\ ? \ \$

Last edited by a moderator:
oh christ. I am dumb.

Apologies.

1. What is a "U-sub integral"?

A "U-sub integral" is a type of integration technique that involves substituting a variable in an integral to simplify the expression and make it easier to solve.

2. How do I know when to use a U-sub integral?

You can use a U-sub integral when the integrand (the expression inside the integral) contains a function and its derivative. The function and derivative must also be in the same order, for example, the function must be the first term and its derivative must be the second term.

3. What is the process for solving a U-sub integral?

The process for solving a U-sub integral involves choosing a substitution, computing the derivative of that substitution, substituting the values into the integral, and then solving the new integral.

4. Why is my U-sub integral not working?

There could be several reasons why your U-sub integral is not working. Some common mistakes include choosing the wrong substitution, not correctly computing the derivative, or making a mistake while substituting the values into the integral.

5. How can I check if my U-sub integral is correct?

You can check if your U-sub integral is correct by differentiating the solution and comparing it to the original integrand. If they are equal, then your U-sub integral is correct.

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