- #1

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Given this, I'm trying to prove that

1. ##\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx##

2. ##\int_{a}^{a}f(x)dx=0##.

Here are my attempts:

1. Using the definition of the Riemann integral, we have that

$$\int_{a}^{b}f(x)dx:=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\Delta x=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\frac{b-a}{n}=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\frac{-(a-b)}{n}\\ =-\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\frac{a-b}{n}=-\int_{b}^{a}f(x)dx\qquad\qquad\qquad\quad$$

2. This follows from 1. for ##a=b##, we have $$\int_{a}^{a}f(x)dx=-\int_{a}^{a}f(x)dx\Rightarrow \int_{a}^{a}f(x)dx=0$$ or, from the definition of the Riemann integral $$\int_{a}^{b}f(x)dx:=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\Delta x=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\frac{a-a}{n}=\lim_{n\rightarrow\infty}0=0$$

Are these valid proofs?

Furthermore, how can one prove these using the alternative definition of the Riemann integral: $$\int_{a}^{b}f(x)dx:=\lim_{n\rightarrow\infty}\lim_{\text{Max}\,\Delta x_{i}\rightarrow 0}\sum_{i=1}^{n}f(\zeta_{i})\Delta x_{i}$$ where ##\Delta x_{i}=x_{i}-x_{i-1}## is the width of the ##i^{th}## sub-interval and ##\zeta_{i}\in [x_{i},x_{i-1}]## is an arbitrary point within this interval, with ##\text{Max}\,\Delta x_{i}=\lbrace x_{2}-x_{1},\ldots , x_{i}-x_{i-1},\ldots , x_{n}-x_{n-1}\rbrace##?!

Finally, is there an easy way to prove the additivity property of integrals, i.e. $$\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx\;?$$