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Express the following sum of two integrals as one:

$I=\int_{5}^1 f(x) dx+\int_{-3}^5 f(x) dx$

My professor said it would be this way:

$I= \int_{-3}^5 f(x) dx+\int_{5}^1 f(x) dx=\int_{-3}^{1} f(x) dx$

But I am not convinced, I feel like there is a mistake in the question, and $5$ should be replaced with a number in between $-3$ and $1$ for me to be able to do the trick he did. What do you think?

orangeskid
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Dima
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1 Answers1

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It adds up like vectors do:

$$\int_a^b f(x)\, dx + \int_b^c f(x)\, dx = \int_a^c f(x)\, dx$$

It helps to think this way: take $F(x)$ an antiderivative of $f(x)$ then

$$(F(b) - F(a)) + (F(c) - F(b)) = F(c) - F(a)$$

Like for vectors ( on a line), $b$ does not have to be between $a$, $c$.

orangeskid
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  • Better speak here of indefinite integral rather than antiderivative, which might not exist. Anyway, the question is a duplicate. – Anne Bauval Feb 14 '23 at 08:02
  • @AnneBauval: That is problematic as well: If $F$ is defined as $F(x) = \int_{x_0}^x f(t) dt$ then using $\int_a^b f(x) dx = F(b) - F(a)$ leads to a circular reasoning. – Martin R Feb 14 '23 at 08:04
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    the latter approach is not a good way of saying it because integrals are not defined as "difference at endpoints of anti-derivatives". That is the conclusion of the FTC, under suitable hypotheses. The original equation itself is easy enough to prove/visualize/think of geometrically (for $a\leq b\leq c$, and for all other cases by definition and case work). – peek-a-boo Feb 14 '23 at 08:05
  • @MartinR sort of, but why not for that kind of basic question – Anne Bauval Feb 14 '23 at 08:10