What is the point in using integrals with boundary x while using variable t? I often see this:
$$  \int_{0}^{x} t^2 \,dt $$
Why is it defined in $ t$, what is the significance?
Why not just define it like this:
$$  \int_{0}^{x} x^2 \,dx $$
What is gained with using the $t$ variable? Thanks in advance
 A: To answer your question, I would first like to let us dig a little deeper as to what the symbol $$\int_a^bf(x)\,\mathrm{d}x$$ represents mathematically, and what an integral ultimately is.
In the context of calculus and beginner-level real analysis, the symbol above is known as a Riemann integral, defined rigorously in terms of partitions of the domain of the function. In this context, the integrand $f$ is interpreted to be a function $[a,b]\rightarrow\mathbb{R}.$ In fact, if the domain is already made explicit in the definition of $f,$ mathematicians will often omit specifying it, and just write $$\int{f(x)}\,\mathrm{d}x.$$ In higher level mathematics, though, specifying the variable of integration is often superfluous, and specifying it can even be confusing, so the integral is just seen as a function $f\mapsto{I}$ with $$I:=\int{f}\in\mathbb{R}.$$ In fact, specifying the variable is really a computational device, rather than a mathematically necessary thing to do, since nowhere in the definition of the Riemann integral does the choice of variable of integration become relevant. This is why $x$ in this context is known as "dummy" variable: it's a dummy. In a sense, it is not real, and is only there due to historical tradition, and because it can be useful for computational purposes. But as far as the theory of integrals is concerned, in the context of real analysis, it plays no role.
So, as far as your specific integral is concerned, the symbol $$\int_0^x{t^2}\,\mathrm{d}t$$ should be interpreted as $$\int{f}$$ where $f:[0,x]\rightarrow\mathbb{R},$ $f(t)=t^2.$
This is why $x$ and $t$ play very different roles, and this distinction is important. The variable $x$ is a free variable, and it communicates something about the domain of the function being integrated. The variable $t,$ on the other hand, is a bound variable, or a "dummy" variable, and in a sense, it is  not real. To highlight this, here is an interesting thing: if you replace $t$ with $u,$ the integral remains unchanged: $$\int{f}:=\int_0^x{t^2}\,\mathrm{d}t=\int_0^x{u^2}\,\mathrm{d}u.$$ To make things more confusing, if you replace $t$ with $t^2,$ the integral also stays the same, contrary to what you may intuitively expect: $$\int_0^x{t^2}\,\mathrm{d}t=\int_0^x{(t^2)^2}\,\mathrm{d}(t^2)=\int_0^x{t^4}\,\mathrm{d}(t^2).$$ It is in this precise sense that $t$ is not a "real" variable, because it is actually irrelevant to the value of the integral, and is merely there as part of a notational artifact. One could choose a different notation that simply got rid of it altogether, and it would not matter. Hence the name "dummy" variable.
But changing $x$ to $x^2$ does change the integral. The specific choice of $x$ changes the domain of integration, and thus, the integral itself. In fact, $$\int_0^{x^2}t^2\,\mathrm{d}t=\int_0^x{t^2}\,\mathrm{d}t+\int_x^{x^2}t^2\,\mathrm{d}t.$$ So this is why you cannot write the integral as $$\int_0^x{x^2}\,\mathrm{d}x.$$ If you did, then $x$ would both be a free variable (a "real" or "true" variable) and a bound variable (a "dummy" variable). This would lead to contradictions.
