Sorry I was assuming that the constant was zero b/c that makes fewer things to deal with. So yes you have it except there is no requirement to be continuous. I am just asking about a concept. So if I do a definite integral then I get a value. It seems to me that a definite integral of a part of any piecewise function should give the same result as an indefinite integral of the equation defining that segment and then plugging in values, they are after all the same process, right?
It has been some time since I took calc, but I was helping someone and the answer they were given had something where they integrated from zero to the variable, not the limit, then plugged in a value later and somehow the answer said it was double what I thought it would be.
So here if I integrate
dx/dt=b*sin(a*t) integrate from 0-->t
x=-b/a*cos(a*t)+b/a then if I plug in 5 for t I get x=-b/a*cos(a*5)+b/a
If I just directly integrate I get
-b/a*cos(a*5)+b/a*cos(0) which matches
I just wanted to double-check there was no weird thing with piecewise functions where you needed to do something different. If you are between the limits it is totally irrelevant as far as I remember.
It has been some time since I took calc, but I was helping someone and the answer they were given had something where they integrated from zero to the variable, not the limit, then plugged in a value later and somehow the answer said it was double what I thought it would be.
So here if I integrate
dx/dt=b*sin(a*t) integrate from 0-->t
x=-b/a*cos(a*t)+b/a then if I plug in 5 for t I get x=-b/a*cos(a*5)+b/a
If I just directly integrate I get
-b/a*cos(a*5)+b/a*cos(0) which matches
I just wanted to double-check there was no weird thing with piecewise functions where you needed to do something different. If you are between the limits it is totally irrelevant as far as I remember.