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How to show $\int_a^b f(x)dx = - \int_b^a f(x)dx$ using the definition of Riemann integration?

$\int_a^b f(x)dx := \lim_{||P(x,t)|| \rightarrow 0} \sum_{i = 1}^{n-1} f(t_i) (x_i - x_{i-1}) $, where $P(x,t)$ is a tagged partition of $[a,b]$. But the definition of a partition assumes $ a = x_0 \leq x_1 ... \leq x_{n-1} \leq x_n = b \Rightarrow a \leq b$, so then how is $\int_b^a f(x)dx$ defined if $ b > a$?

Anon
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    This is usually a definition (for $a<b$, once then defines $\int_b^af$ as $-\int_a^bf$), since intervals $[x_{i-1},x_i]$ imply $x_{i-1}\le x_i$. – StackTD May 09 '18 at 07:47
  • So are you saying that $\int_b^a f(x)dx :=- \int_a^b f(x)dx$ for $b > a$? – Anon May 09 '18 at 07:49
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    Exactly. This definition, together with $\int_a^a f = 0$, is useful as it ensures that another important property, $\int_a^bf = \int_a^cf + \int_c^bf$, is valid in general and not only for $a<c<b$. – StackTD May 09 '18 at 07:50
  • Thank you for the help. – Anon May 09 '18 at 07:54
  • You're welcome; I turned this into an answer. – StackTD May 09 '18 at 08:08
  • Please check my answer about it .https://math.stackexchange.com/questions/261244/is-there-a-fundamental-reason-that-int-ba-int-ab/261328#261328 – Mathlover May 09 '18 at 08:22

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For the usual definition to make sense, you need a non-empty interval $[a,b]$ (which implies $a<b$) and then you have a definition for the definite integral of $f$ over $[a,b]$, denoted as: $$\int_a^bf$$ Notice that this definition does not make sense for $b=a$ (since there is no interval to partition) and not for $b<a$ (since the partitioning requires $a<b$; that's how intervals work).

Using this definition, you can prove the following important property: $$\int_a^bf = \int_a^cf + \int_c^bf \tag{$*$}$$ With the usual definition, this property holds only for $a<c<b$. To allow more flexibility, one could want to give meaning to integrals of the form $\int_a^af$ and $\int_b^af$ (with $a<b$).

If we define them as: $$\int_a^af := 0 \quad \mbox{and} \quad \int_b^a f:=-\int_a^b f$$ then the formula $(*)$ remains valid and holds in general, for all $a,b,c$.

How to show $\int_a^b f(x)dx = - \int_b^a f(x)dx$ using the definition of Riemann integration?

Long story short: this is usually a definition, not a property you prove. We define it in such a way that another important property remains valid in general.

StackTD
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