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Topic: Calculus Problem - Can You Help?
Replies: 2   Last Post: Oct 15, 2006 9:29 PM

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Lynn Kurtz

Posts: 1,278
Registered: 12/6/04
Re: Calculus Problem - Can You Help?
Posted: Oct 15, 2006 9:29 PM
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On Mon, 16 Oct 2006 01:16:18 GMT, "[Mr.] Lynn Kurtz"
<kurtzDELETE-THIS@asu.edu> wrote:

>
>On 15 Oct 2006 17:27:06 -0700, "Bill" <bill@afplaza.com> wrote:
>

>>Hello,
>>
>>I am looking for help with a calculus problem. Can you take a look and
>>see if you can help?
>>
>>Since I have a hard time getting all the little characters straight in
>>the newsgroups, I just posted the problem on a webpage I have space on.
>> It is located at www.afplaza.com.
>>
>>I am clueless when it comes to Calculus so I am hoping someone alot
>>smarter then me can help us out. Problem 13 only. She has shown her
>>work and knows the answer - she is confused about how to handle the
>>middle of the problem.
>>
>>Thanks in advance!
>>
>>Bill
>>www.afplaza.com

>
>It doesn't require integration by parts. She has u = e^(x^3) and du =
>3x^2 dx so she should just substitute u for the e^(x^3) and (1/3)du
>for the x^2 dx. It is a straight substitution.
>
>--Lynn


Correction, I read it too quickly. With her choice of u her du should
be du = 3x^2e(x^3) dx so her integrand is just (1/3)du.

Alternatively she could try u = x^3, du = 3x^2 dx so her integrand
becomes (1/3) e^u du.

Both methods lead to the same answer.

--Lynn



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