- #1
"Don't panic!"
- 601
- 8
How does one prove the following relation?
[tex]\int_{a}^{b}f(x)dx= \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx [/tex]
Initially, I attempted to do this by writing the definite integral as the limit of a Riemann sum, i.e.
[tex] \int_{a}^{b}f(x)dx= \lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k})[/tex]
Where [itex] x^{\ast}_{k}\in\left[x_{k}, x_{k+1} \right] [/itex].
Then,
[tex] \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx= \\ = \lim_{n\rightarrow\infty}\frac{(c-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) +\lim_{n\rightarrow\infty}\frac{(b-c)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) \\ = \lim_{n\rightarrow\infty}\frac{(c-a)+ (b-c)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) \\ = \lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k} = \int_{a}^{b}f(x)dx[/tex]
But I have a feeling that this isn't correct?!
[tex]\int_{a}^{b}f(x)dx= \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx [/tex]
Initially, I attempted to do this by writing the definite integral as the limit of a Riemann sum, i.e.
[tex] \int_{a}^{b}f(x)dx= \lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k})[/tex]
Where [itex] x^{\ast}_{k}\in\left[x_{k}, x_{k+1} \right] [/itex].
Then,
[tex] \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx= \\ = \lim_{n\rightarrow\infty}\frac{(c-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) +\lim_{n\rightarrow\infty}\frac{(b-c)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) \\ = \lim_{n\rightarrow\infty}\frac{(c-a)+ (b-c)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) \\ = \lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k} = \int_{a}^{b}f(x)dx[/tex]
But I have a feeling that this isn't correct?!