Definite integral with negative difference of bounds

Not sure the title has the correct terminology but I'd like to evaluate the following definite integral:

$$\int_5^1{x^2}dx$$

Does this require the same approach as

$$\int_1^5{x^2}dx$$

Or is there some kind of voodoo that must be done, such as perhaps using $[-∞, ∞]$ as limits of integration and then subtracting using $[1,5]$?

Also, what does it mean? Does it represent an area as well?

If you swap the bounds, the area will be exactly the same magnitude, but of the opposite sign.

If you graph $y=x^2$, in both cases you are still taking the area under the curve, just in different directions.

For example:

$\int_5^1{x^2}dx$

= $[\frac{1}{3}x^3]_5^1$ =

= $\frac{1}{3}(1)^3 -\frac{1}{3}(5)^3$

Compare this to

$$\int_1^5{x^2}dx$$ = $[\frac{1}{3}x^3]_1^5$ =

= $\frac{1}{3}(5)^3 -\frac{1}{3}(1)^3$

Further note that $\mid \frac{1}{3}(1)^3 -\frac{1}{3}(5)^3\mid = \mid \frac{1}{3}(5)^3 -\frac{1}{3}(1)^3 \mid$
We usually define $$\int_a^b f(x)\,\mathrm{d} x = -\int_b^a f(x)\,\mathrm{d}x.$$