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# Maple Antiderivative watch

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1. I'm being asked to do this in Maple:

Find an anti-derivative F(x) of f(x)=(2x^5−5x^3+x)(e^(x^2)) which satisfies F(0)=1.

I know how to find the antiderivative, but I don't know what to do with the F(0)=1.

2. well the antiderivate will be F(x)=somethingsomething+C
put F(0)=1 and determine the unknown C
3. isn't the anti-derivative just the integral?
4. (Original post by funkiichiicka)
I'm being asked to do this in Maple:

Find an anti-derivative F(x) of f(x)=(2x^5−5x^3+x)(e^(x^2)) which satisfies F(0)=1.

I know how to find the antiderivative, but I don't know what to do with the F(0)=1.

To my knowledge you cannot get maple to calculate the extra constant which makes this so. You can get it to calculate the antiderivative (the integral) with the int command. Then I suppose you could define this as, say, 'a' and then eval 'a' at f(0) and define this as 'b' now you would be able to put this together for the whole antiderivative

Code:
`c:= a + 1 - b`
which I think gets you the integral you want.

I got a little bored and so knocked a quick procedure which should do the job. If you don't know about procedures then you can just take the code from the middle and use your values.

It's pretty self explanatory

Spoiler:
Show

Code:
```antiderivative:=proc(f,a,b) local g,k,p;  #f is the function (in terms of x), and F(a)=b
g:=int(f,x): #this is the general solution
k:= b - eval(g,x=a):  # this works out what our constant
p:= g + k;  #this puts them both together

end proc:```
5. (Original post by tengil)
well the antiderivate will be F(x)=somethingsomething+C
put F(0)=1 and determine the unknown C
(Original post by The Muon)
To my knowledge you cannot get maple to calculate the extra constant which makes this so. You can get it to calculate the antiderivative (the integral) with the int command. Then I suppose you could define this as, say, 'a' and then eval 'a' at f(0) and define this as 'b' now you would be able to put this together for the whole antiderivative

Code:
`c:= a + 1 - b`
which I think gets you the integral you want.

I got a little bored and so knocked a quick procedure which should do the job. If you don't know about procedures then you can just take the code from the middle and use your values.

It's pretty self explanatory

Spoiler:
Show

Code:
```antiderivative:=proc(f,a,b) local g,k,p;  #f is the function (in terms of x), and F(a)=b
g:=int(f,x): #this is the general solution
k:= b - eval(g,x=a):  # this works out what our constant
p:= g + k;  #this puts them both together

end proc:```
Thank you! I keep forgetting about that +C. Sounds easy enough.

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Updated: January 30, 2010
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