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Given function $y(t)$, I am confused by the notation

$$x(t) = \int\limits_0^t y(\tau) d\tau$$

Can someone explain to me if $x(t)$ is a function of $t$, or the value evaluated at some $t$. Logically speaking, it should be a function, because when you plug in different $t$, you get different $x(t)$. But an integral is roughly a summation over infinitesimal intervals, so it is a number.

How do you distinguish between these two, notion-wise?

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As an objection to $x(t)$ be a function you said:

But an integral is roughly a summation over infinitesimal intervals, so it is a number.

However, as you pointed out before, this number depends on $t$. Note that it does not depend on the function $y$; the function $y$ is fixed. For example, if $y(t)=t^2$ the function $x(t)$ is given by

$$x(t)=\int_0^t\tau^2\;d\tau=\frac{\tau^3}{3}\Bigg|_{\tau=0}^{\tau=t}=\frac{t^3}{3}.$$

For each $t$, the expression $\sin(t)$ is a number and we can define $x(t)=\sin(t)$. Analogously: for each $t$, the expression $\int_0^ty(\tau)d\tau$ is a number and we can define $x(t)=\int_0^ty(\tau)d\tau$.

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$\int_0^ty(\tau)d\tau$ is a value. However for different $t$ you have a different value. So $t\mapsto \int_0^ty(\tau)d\tau$ is a function.

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  • $\begingroup$ Why is that the domain of the function is $t$, instead the function $y$? If a function is a machine, then this function is taking in a function and spitting out a number, no? $\endgroup$ – Shamisen Expert May 30 '17 at 23:21
  • $\begingroup$ Sorry I do not understand your question. The domain of a function is independent of the variable you use. So $y(t)$, $y(\tau)$, $y(x)$ they are all the same things. A function is an object that maps a value to another value. $\endgroup$ – Yining Wang May 30 '17 at 23:22
  • $\begingroup$ I see your confusion. $x$ does not map a function to a value. $x$ maps a value ($t$) to another value $\int_0^t f(\tau)d\tau$. $\endgroup$ – Yining Wang May 30 '17 at 23:23
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$x$ is the name of the function and its value at some $t$ is defined by $$ x(t) = \int_0^t y(\tau) \,d\tau. $$

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