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continuity

suvadip

Member
Feb 21, 2013
69
If \(\displaystyle f(x,y)\) be a continuous function of \(\displaystyle (x,y)\) in the rectangle \(\displaystyle R:{a \leq x \leq b, c \leq y \leq d}\) , then \(\displaystyle \int_a^b f(x,y) dx\) is also a continuous function of \(\displaystyle y\) in \(\displaystyle [c,d]\)

How to proceed with the proof of the above theorem?
 

chisigma

Well-known member
Feb 13, 2012
1,704
If \(\displaystyle f(x,y)\) be a continuous function of \(\displaystyle (x,y)\) in the rectangle \(\displaystyle R:{a \leq x \leq b, c \leq y \leq d}\) , then \(\displaystyle \int_a^b f(x,y) dx\) is also a continuous function of \(\displaystyle y\) in \(\displaystyle [c,d]\)

How to proceed with the proof of the above theorem?
If an f(x,y) is continous in a closed and bounded region, then f(x,y) is also uniformly continous here, so that setting...

$\displaystyle G(y) = \int_{a}^{b} f(x,y)\ dx\ (1)$

... for any h>0 is...

$\displaystyle |G(y + h) - G(y)| = | \int_{a}^{b} \{ f(x,y+h) - f(x,y)\}\ dx| \le \int_{a}^{b} |f(x,y+h) - f(x,y)|\ d x\ (2)$

Now f(x,y) is uniformly continous so that choosing h 'small enough' You can do the last term of (2) 'small as You like' and that means that G(y) is continous...

Kind regards

$\chi$ $\sigma$