How does one prove the following:
[tex] \int^{c}_{a} f\left(x\right)dx = \int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx [/tex]
where [itex] f\left(x\right)[/itex] is continuous in the interval [itex] x\in \left[a, b\right][/itex], and differentiable on [itex] x\in \left(a, b\right)[/itex].
My approach was the following:
Given that [itex] \int^{b}_{a} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) [/itex], where [itex] x^{\ast}_{i} \in\left[x_{i},x_{i+1}\right] [/itex], we have that
[tex] \int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) + \lim_{n\rightarrow \infty} \frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) [/tex]
[tex]\qquad\qquad\qquad\qquad\qquad\quad= \lim_{n\rightarrow \infty} \left[\frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)+\frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)\right] = \lim_{n\rightarrow \infty} \frac{1}{n}\left[\left(b-a\right) +\left(c-b\right) \right]\sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)[/tex]
[tex]\qquad\qquad\qquad\qquad\qquad\quad= \lim_{n\rightarrow \infty} \frac{\left(c-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) = \int^{c}_{a} f\left(x\right)dx[/tex]
However, I have a feeling that this isn't quite correct?!
[tex] \int^{c}_{a} f\left(x\right)dx = \int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx [/tex]
where [itex] f\left(x\right)[/itex] is continuous in the interval [itex] x\in \left[a, b\right][/itex], and differentiable on [itex] x\in \left(a, b\right)[/itex].
My approach was the following:
Given that [itex] \int^{b}_{a} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) [/itex], where [itex] x^{\ast}_{i} \in\left[x_{i},x_{i+1}\right] [/itex], we have that
[tex] \int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) + \lim_{n\rightarrow \infty} \frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) [/tex]
[tex]\qquad\qquad\qquad\qquad\qquad\quad= \lim_{n\rightarrow \infty} \left[\frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)+\frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)\right] = \lim_{n\rightarrow \infty} \frac{1}{n}\left[\left(b-a\right) +\left(c-b\right) \right]\sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)[/tex]
[tex]\qquad\qquad\qquad\qquad\qquad\quad= \lim_{n\rightarrow \infty} \frac{\left(c-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) = \int^{c}_{a} f\left(x\right)dx[/tex]
However, I have a feeling that this isn't quite correct?!
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