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Definite Integral with symbolic upper and lower limits and a constant

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Alexandros
Alexandros 2017-9-15
评论： Star Strider 2017-9-16
Hi! I have the following integration: on both sides is -1/k*int(1/v)dv=int(1ds) where int is the symbol of integration The limits for left integral is from v0 to v The limits for right integral is from 0 to s
I have tried the following in code:
>> g=inline('(1/-k)*(1/v)')
g =
Inline function:
g(k,v) = (1/-k)*(1/v)
>> int(g(v))
Error using inline/subsref (line 12)
Not enough inputs to inline function.
Also I have tried this:
>> syms k
>> g=inline('(1/-k)*(1/v)')
g =
Inline function:
g(k,v) = (1/-k)*(1/v)
>> int(g(k,v))
ans =
-log(v)/k
Is there a way to take into account the v0 to v integration limits? The result should be: -1/k*(log(v)-log(v0))
Also k is a constant.
This is a Dynamics Mechanics sample problem by Meriam & Kraige

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Star Strider
Star Strider 2017-9-15
Try this:
syms k v v0
assume(v > 0)
assume(v0 > 0)
g(k,v) = (1/-k)*(1/v); % Symbolic Function
Ig = int(g(k,v), v, v0, v)
Result:
Ig =
-(log(v) - log(v0))/k
You can define symbolic functions in R2012a and later.
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Alexandros
Alexandros 2017-9-16
Hi. I get a four decimal place accuracy. Is there a way to get what you got for k?
Star Strider
Star Strider 2017-9-16
Yes.
format long g
k = fzero(@(k) 1.1*k-log(1+k*8*(10/60)), 1)
See the documentation on the format (link) function for details.

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