A concept from calculus that has always bothered me

1. Jan 22, 2010

ronaldor9

When one writes $$\int p(t) \, dt$$ is the constant of integration implied? I have always thought that it wasn't necessary to write $$\int p(t) \, dt +k$$. However, in my diff. eq. book the constant is ussually written out, why is this so?

In addition the book also writes $$\int_{t_o}^t p(t) \, dt +k$$ isn't the constant here unnecessary since we now have the limits of integration included?

2. Jan 22, 2010

snipez90

3. Jan 22, 2010

l'Hôpital

It's usually written because it's definite.

Suppose

$$y ' = f(x, y(x)) \ , y(x_0) = y_0$$

It does seem logical to say that (assuming f is integratable)

$$\int y' dx = \int f(x,y(x)) dx \rightarrow y = \int f(x,y(x)) dx + k$$

but since we have initial conditions, we can do better.

$$\int_{x_0}^{x} y'(t) dt = \int_{x_0}^{x} f(t,y(t)) dt \rightarrow y - y_0 = \int_{x_0}^{x} f(t,y(t)) dt$$
Thus,
$$y = \int_{x_0}^{x} f(t,y(t)) dt + y_0$$

Better?

Last edited: Jan 23, 2010
4. Jan 23, 2010

ronaldor9

Thanks l'Hopital
Why is the later form preferred over the first form?

5. Jan 23, 2010

l'Hôpital

Simply because it actually involves the initial conditions.

6. Jan 27, 2010

ronaldor9

There is one part in my book where it writes $$R'(y)=Q(x_0,y)$$
and then by integration $$R(y)=\int_{y_0}^y Q(x_0,y)\,dy$$.

Shouldn't theauthor here have included a constant at the end of the integral as you have written in you example?

7. Jan 28, 2010

HallsofIvy

Yes, because with that notation R(y0)= 0 which is not true for all anti-derivatives. I would also object to using "y" both as a limit of integration (and so outside the integral) and as the variable of integration. Much better would be either
$$R(y)= \int_{y_0}^y Q(x_0,t)dt+ C$$
or
$$R(y)= \int^y Q(x_0, t)dt$$
where we don't need the "C" because the lower limit of integration is left open.

If you write just
$$R(y)= \int Q(x_0,y)dy$$
the usual notation for the "anti-derivative", the constant C is implied. You do not write it there. Of course, if you wrote
$$R(y)= \int y^2 dy= \frac{1}{3}y^2+ C$$
The constant on the right is necessary.