Have any meaning $\int_0^{x} x dt$?

Question

If $x$ and $t$ are independent variables, does this expression have any meaning?

$$\int_0^{x} x \hspace{2pt}\mathrm{d}t$$

I admit that I'm confused about this!

Answer 1

It does. Say $x=2$. That would amount to

$$\int\limits_0^2 2 \, dt = 2 \cdot \int\limits_0^2 1\,dt=2\cdot\left[t\right]_0^2=2\cdot(2-0)=2^2=4$$

More generally, we have:

$$\int\limits_0^x x \, dt = x \cdot \int\limits_0^x 1\,dt=x\cdot\left[t\right]_0^x=x\cdot(x-0)=x^2$$

Answer 2

Yes, from the point of view of the integral $x$ is a constant, so we have $$\int_0^xx\;dt=x\int_0^x1\;dt=x\left(\left.t\right|_{t=0}^{t=x}\right)=x(x-0)=x^2$$

Answer 3

If, $x$ is a variable, then you mean to have something like

$$f(x)=\int_0^x xdt.$$

This notation is not common, however. Usually, the integrand and bounds do not use the same variable name, unless one is a constant, as is not the case here.

For examples, see here. This page shows various trigonemtric integrals, such as $$\mathrm{Si}(x)=\displaystyle\int_0^x \frac{\sin t}{t}dt$$

Answer 4

The confusion was you stated them as independent variables which implies there is no relationship between $x$ and $t$ and as far as the integral over $t$ is concerned $x$ is a constant that can be taken outside.

If $x$ and $t$ are dependent variables, say $x = x(t)$ then the correct way (that leaves no room for confusion) to right your integral is as $\int_0^{x(t)} x(t') dt'$.