Use the definition to find $\int_0^2f(x)dx$ where $f (x )=\cases{ 1,&$x \ne 1$\\2,& $x = 1$}$

Asked Nov 23 '23 at 18:09

Active Nov 23 '23 at 22:17

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Use the definition to find $\int_0^2f(x)dx$ where $f (x )=\cases{ 1,&$x \ne 1$\\2,& $x = 1$}$

I am stuck with this question, we studying Riemann integration, in my mind I can do the integral over $[0,1] +[1,2] =[0,2]$

the question is how I set up the integral, I mean i understand $f (x )=\cases{ 1,&$x \ne 1$\\2,& $x = 1$}$., so the first integral $\int_0^1 f(x) dx=x , 1-0 =1$, second integral $\int_1^2 f(x) dx= 2-1=1$, so $\int_0^2 f(x)dx=1+1=2$?

edited Nov 23 '23 at 22:17

asked Nov 23 '23 at 18:09

user1131328

1

This Q is hard to read. Hence, consider to visit MathJax basic tutorial and quick reference for format improvement. – Anton Vrdoljak Nov 23 '23 at 18:15
what's your definition of an integral? is it Riemann integration? Perhaps state the definition and then work from there? – Siong Thye Goh Nov 23 '23 at 18:31
The jump at $x=1$ adds no area to the function, so you can just integrate $\int_0^2 f(x)dx=\int_0^2dx$ – Тyma Gaidash Nov 23 '23 at 18:38
What does “use the definition to …” mean? – Ted Shifrin Nov 23 '23 at 18:41
It's not pair wise addition, it's called splitting into disjoint intervals, so use \cup. $$(0,1)\cup (1,2)=(0,2)\setminus {1}$$ – Nothing special Nov 23 '23 at 18:42
See this more general situation. Consider $g(x)=1$ for all $x$. $g$ and $f$ differ only at one point. – Nothing special Nov 23 '23 at 18:50

Use the definition to find $\int_0^2f(x)dx$ where $f (x )=\cases{ 1,&$x \ne 1$\\2,& $x = 1$}$

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