# Perpetual integration

I have to integrate the following:

https://www.physicsforums.com/attachment.php?attachmentid=24485&d=1268939055

as you can see I have an answer - this is such a long problem - it's like an integration of an integration of an...
and the final answer gives a math error in the calculator.

I also get a value out of calculator of -7.169337845

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I am unsure of what the problem is. What are you trying to integrate? I see "I have to integrate the following:" and nothing following.

you cant see the problem on the image?

gabbagabbahey
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you cant see the problem on the image?
No. I used the quote button on your original post and it looks like you are trying to post an image just by putting image tags around an http address for an attachment. I don't think that works even if the http address links to an image in jpeg form. To make things worse, it looks like the attachment http address is invalid.

Try uploading your image to a free image hosting site like imageshack.us and then just posting the forum hotlink url that they give you.

Either that or take the time to type out your problem statement and the key lines in your attempt like everybody else does.

This was typed out properly in mathType.

gabbagabbahey
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I wasn't being sarcastic. This forum supports $\LaTeX$ and it is often quicker for you to type it out here than to type it up in a separate document and then post it as an attachment or upload it to an image hosting site.

sorry - it's been a long week. I will look into it

$$\begin{array}{l} \int_0^3 {2x^3 e^{ - x} dx} \\ = \frac{{13\left( { - 12 - e^3 } \right)}}{{e^3 }} \\ \end{array}$$

tricky

gabbagabbahey
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there are two different types of tags you can use...the ones I used were "itex" which is short for inline LaTeX and is only good for short expression that you want to appear inline with the rest of your text. For larger expressions, just use "tex" tags.

\begin{aligned}\int_0^3 2x^3 e^{ - x} dx & = \left.-6x^2e^{-x}\right|_0^3 +\int_0^3 6x^2 e^{ - x} dx\\ & = \left.-6x^2e^{-x}\right|_0^3+ \left.12xe^{-x}\right|_0^3 + \int_0^3 12x e^{ - x} dx \\ & = \frac{{13\left( { - 12 - e^3 } \right)}}{e^3}\end{aligned}

Were you trying to post something like this? You can click on the image to see the code that generated it (or quote my post to see it without all kinds of extra spaces and line-breaks).

thank you gabbagabbahey
I will take a look at it over the weekend.

Is that the correct finished solution you posted above?

gabbagabbahey
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Is that the correct finished solution you posted above?
No.

Is that the problem you are trying to calculate?

yes
it is the problem
but it went on and on and...

gabbagabbahey
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Gold Member
It shouldn't go too long. Just use integration by parts (IBP) 3 times.

Why not show me your first iteration of IBP...

i have attached a .pdf file of my written workings
hope you can read my scribbles

#### Attachments

• 24.4 KB Views: 63
it is all part of this attached .pdf file.
I have succeeded to complete part one of the question.

#### Attachments

• 3.9 KB Views: 56
Mark44
Mentor
Changed your [ itex] tags, and got rid or your array stuff.
$$\int_0^3 {2x^3 e^{ - x} dx} \\ = \frac{{13\left( { - 12 - e^3 } \right)}}{{e^3 }}$$

gabbagabbahey
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i have attached a .pdf file of my written workings
hope you can read my scribbles
It looks like you've done all the individual iterations of IPB properly (although you've neglected to write the integration limits in at each step).

You're problem is with the arithmetic after you substitute in your limits of integration:

\begin{aligned}\int_0^3 2x^3 e^{ - x} dx & = \left.-2x^3e^{-x}\right|_0^3- \left.6x^2e^{-x}\right|_0^3 - \left.12xe^{-x}\right|_0^3-\left.12e^{-x}\right|_0^3 \\ & = \left.-2e^{-x}\left(x^3+3x^2+6x+6\right)\right|_0^3 \\ & = -2e^{-3}\left(27+27+18+6\right)+2e^{0}\left(0+0+0+6\right) \\ & \neq \frac{{13\left( { - 12 - e^3 } \right)}}{e^3}\end{aligned}

Mark44
Mentor
It looks like you've done all the individual iterations of IPB properly
IPB - integration py barts?

gabbagabbahey
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Gold Member
IPB - integration py Barts?
My calculus teacher's name was Ms. Edna Krabappel

That's the approximate numerical value of the answer, yes. The exact answer involves $e^3$.