if the integral from1-3 of f(x)dx=5 and the integral from 5-1 or f(x)dx=2 find the integral from 3-5 of f(x) dx
\[\int\limits_{1}^{3} f(x) dx =5 and \int\limits_{1}^{5}f(x)dx =2 find \int\limits_{3}^{5}f(x)dx\]
Remember that: \[ \int\limits_{a}^{b} f(x) dx =\int\limits_{a}^{c} f(x) dx + \int\limits_{c}^{b} f(x) dx \]
IN this case, a is 1, c is 3, and b is 5
so =1 b=3 c=5? maybe?
Just be consistent
so wouldn't it just be 5-3? lol I don't get it
Okay, pick what you want a, b, and c to be... Then plug them into that formula I wrote.
\[\int\limits_{1}^{5} f(x0 dx = \int\limits_{1}^{3}f(x)dx +\int\limits_{3}^{5}f(x)dx\]
You might have to use \[ \int\limits_{b}^{a}f(x)dx = -\int\limits_{a}^{b}f(x)dx \]
what I get from this is \[-3=\int\limits_{3}^{5}f(x)dx\]
this is just like 2=x+5 you need to find x ! \[\Large \int\limits_{1}^{5} f(x)=\int\limits_{1}^{3}f(x)+\int\limits_{3}^{5}f(x)\]
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