Integrals

Forum|Forum|14 years ago
November 2, 2011
3 replies
9861 views

It would be nice if the constant C is always equal to zero, because for some cases I have to take into account the particular value of this constant, or am I picking on too much?

Other

F

Fred_Kohlhepp

23-Emerald I

Beginning calculus: Every indefinite integral has a constant term (since the derivative of a constant is zero) and there's no way to tell what it is. So the integrals above (without a constant term given) are assuming that the constant is zero (C = 0.)

Your individual problem needs to decide what C is equal to.

1_indefinite-integrals-xmcd.zip

A

andmail

Author

1-Visitor

I understand, there are no mathematical errors, but

Is this my individual problem?

Fred Kohlhepp wrote:
So the integrals above (without a constant term given) are assuming that the constant is zero (C = 0.)

Are you sure?

F

Fred_Kohlhepp

23-Emerald I

I've tried to create a simple example, and ran into a constant of integration problem. See attached.

1_integrals-xmcd.zip

A

andmail

Author

1-Visitor

So what we decided? Symbolic notation does not match the displayed result when C equals to zero?

R

rag

1-Visitor

Maybe I am missing the point, but it seems to me from your prior postings, that you assume, that every Integral has a "solution" running through (0/0) and you would like Mathcad to show you that "main"-Integral. At least thats what your setting x to zero implies. Think of INT(1/x,x)-->ln(x) - no way to apply your method.

I agree, that Mathcad ever so often simplifies Integrals not the same way I would do and Maple in Mathcad 11 did a better job. But one can argue if (x-3)^3 really is so bad compared with the polynomial expression without the constant 1.

What is the "right" result for INT(-2*sin(x)*cos(x),x)? -sin^2(x) or cos^2(x)?

Won't call it a bug but if you want to say that Mathcad is doing somethings wrong it could be its inability to explicitely give the constant of integration in its results.

As you seem to deal mainly with polynomial function, maybe substituting the indefinite integral for INT(f(t),t,0,x) will do the job better for you.

int.03.11.png