Re: Missunderstanding on integration

On May 8, 10:51 pm, polymedes <goo...@xxxxxxxxxxxxxxxxxx> wrote:
On May 8, 11:25 pm, leigh123 <Leighsmu...@xxxxxxxxxxxxxx> wrote:

On 8 May, 21:47, polymedes <goo...@xxxxxxxxxxxxxxxxxx> wrote:

I'm looking for help on this integral:

e^{-\int_{t_0}^{t}p(s)ds} \int_{t_0}^{t} g(r) dr = \int_{t_0}^{t} e^{-
\int_{t_0}^{t}p(s)ds} g(r) dr

why is this true? What I am missing?

Hi,

The multiple on the left hand side, e^{-\int_{t_0}^{t}p(s)ds}, depends
only on s, so you can treat it just as you would a constant and move
it inside the
integral that is dependent on r.

But it is a function of t. According to the definition of the
indefinite integral F(t) = int_a^t f(s)ds, right? I would understand
it as a constant, if

e^{-\int_{t_0}^{t}p(s)ds} \int_{t_0}^{t} g(r) dr = \int_{t_0}^{y}
e^{-
\int_{t_0}^{t}p(s)ds} g(r) dr

(i.e. the second integral was not a function of t)

Hi,

But it is a function of t. According to the definition of the
indefinite integral F(t) = int_a^t f(s)ds, right? I would understand
it as a constant, if

You're correct that it is a function of t. However, the variables of
integration are r and s and do not depend on t, so you run into no
problems when taking the integral.






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