Get Help

Turn on suggestions

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

- Community
- :
- PTC Mathcad
- :
- PTC Mathcad
- :
- Give me a break!

Options

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Mute
- Printer Friendly Page

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Notify Moderator

03-26-2011
01:43 AM

03-26-2011
01:43 AM

Give me a break!

Is this for real?

Mathcad 14

Labels:

5 REPLIES 5

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Notify Moderator

03-26-2011
03:47 AM

03-26-2011
03:47 AM

Re: Give me a break!

Kurt Tate wrote:

Is this for real?

Mathcad 14

Yes, Indefinitely.

You have two indefinite integrals so they have a 'constant' of integration. Depending on how you formulate the problem depends how Mathcad's procedural substitution gets to the result, including the different expansions.

Both answers are correct.

Enjoy and learn the fun of maths.

Philip

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Notify Moderator

03-26-2011
01:34 PM

03-26-2011
01:34 PM

Re: Give me a break!

Mathcad has provided its own constant of integration without referance to what the boundary conditions or initial conditions are. If I want to set this indefinite integral equal to a function with my own boundary conditions so that I can set any value of x and get my desired value for the function I first have to undo what mathcad has done.

In this case it is adding the absolute value of the usually negative area to the left of the y axis (where the x-intercept is negative) or the negative area to the right of the y axis (where the y-intercept is negative). This is for a>0, for a<0 corresponding areas would be added.

What I want from an indefinite integral is the strict antiderivative (like previous versions provided) and let me provide the constant of integration. For more complicated functions does mathcad provide its own constant and how can I trust the answers now?

Thanks,

Kurt

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Notify Moderator

03-27-2011
05:52 AM

03-27-2011
05:52 AM

Re: Give me a break!

Kurt Tate wrote:

...

What I want from an indefinite integral is the strict antiderivative (like previous versions provided) and let me provide the constant of integration. ...

Thanks,

Kurt

That's exactly what Mathcad has provided! See attached.

Alan

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Notify Moderator

03-27-2011
05:58 AM

03-27-2011
05:58 AM

Re: Give me a break!

Kurt,

Mathcad has provided the 'right' answers in each case, one of which was then expanded, giving you the answer you didn't expect.

The presence of your boundary condition means that you didn't want (at the end of the day) an indefinite integral. You definitely wanted a definite integral that, at the boundary, gave your defined edge value.

The new MuPad engine does cause some difficulties because it doesn't respond to 'custom' but follows the (its) rules precisely. There have been many cases where the text books are 'wrong'. (the usual example is integral(1/x, -1,+1) where an exp(i.pi) turns up for the -1 term)

I tried googling "Strict Indefinite Integral" to see if it was a common term and this thread is already #3! It may be that you desire a "strict" keyword that simply drops those terms which do not contain the dependent variables (i.e. make the 'constants' effectively zero).

MuPad appears to like to take the (a.x+b) expression as a particular rule, which caused the effect you are seeing.

For any symbolic evaluation it is always advantageous to include all the steps you need (i.e. the 'determine constant of integration' step) rather than assume that it will get the same as the blackboard results (ie.e don't assume it will be zero). It (unexpected symbolic expressions) has caught me out before.

Philip

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Email to a Friend
- Notify Moderator

03-27-2011
11:45 AM

03-27-2011
11:45 AM

Re: Give me a break!

You can include the real boundary conditions, or get rid of the consant, see the attached