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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

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

effectively returning the additive inverse?

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?

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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:

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

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

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

Which yields a negative answer

Compare this to

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

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

Which gives a positive answer

Further note that $\mid \frac{1}{3}(1)^3 -\frac{1}{3}(5)^3\mid = \mid \frac{1}{3}(5)^3 -\frac{1}{3}(1)^3 \mid$

This is known as the physical area, compared to a negative area which only makes sense mathematically as a number, since you can't really have a "negative area".

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We usually define $$\int_a^b f(x)\,\mathrm{d} x = -\int_b^a f(x)\,\mathrm{d}x.$$

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