September 30th, 2016, 11:20 AM  #1 
Newbie Joined: Sep 2016 From: Vancouver, Canada Posts: 6 Thanks: 0  Setting up integral
I'm trying to calculate the area between two parallel linear lines f1=4x f2=5x Integrating between x= 05 But I only want the area between y=0 and y=5 (as constraints) all in the first quadrant. Now these are easy equations to integrate but the biggest issue I am having is setting up the integral... I've done something like this: [F(5)  F(f2)]dx  [F(5)F(f1)]dx where definite integral range is 05 But I end up getting a negative value which makes no sense. The only thing I can conclude is the way I am setting up my integral which is incorrect. Anyone know how I can setup this integral given where I want my area? 
September 30th, 2016, 12:17 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,975 Thanks: 1026 
I would set this up as $\displaystyle{\int_0^5}f_2(x)  f_1(x)~dx$ $\displaystyle{\int_0^5}(5x)  (4x)~dx$ Basically you find the area under $f_2$ and subtract off the area under $f_1$ leaving the part you are interested in. $\displaystyle{\int_0^5}(5x)  (4x)~dx = $ $\displaystyle{\int_0^5}1~dx = 5$ btw your pic is a bit off. $f_1(0)=4,~f_2(0)=5$ 
September 30th, 2016, 12:25 PM  #3  
Newbie Joined: Sep 2016 From: Vancouver, Canada Posts: 6 Thanks: 0  Quote:
*Sorry I should expand on the above... My f1 and f2 functions are very general and I'm working thru this on excel where f1 and f2 will change depending on m and b values input. But based on the m and b values, I'd like to constrain that area within a box of 5x5 if that makes sense? Last edited by ihaque; September 30th, 2016 at 12:28 PM.  
September 30th, 2016, 12:32 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 1,975 Thanks: 1026  Quote:
$\displaystyle{\int_0^4}f_2f_1~dx+\displaystyle{\int_4^5}f_2~dx = \dfrac 9 2$  
September 30th, 2016, 12:45 PM  #5  
Newbie Joined: Sep 2016 From: Vancouver, Canada Posts: 6 Thanks: 0  Quote:
Assume f1 as a linear function similar to above. Assume f2 as a linear function similar to above. And now let's add f3, which is equal to f3=5 If I want the definite integral from 05 in the first quadrant only given the above functions, is it fair to write it as the following: (F2F1)(F2F3)? Last edited by skipjack; October 12th, 2016 at 11:47 AM.  
September 30th, 2016, 01:01 PM  #6  
Senior Member Joined: Sep 2015 From: USA Posts: 1,975 Thanks: 1026  Quote:
If I understand what you're trying to do, you want to find a way to express that triangular area of the integral of $f1$ that is below the xaxis, so you can subtract it off the simple integral of $f2f1$ I don't see any really clever way of doing that that isn't equivalent to what I did treating the range $x \in (4,5]$ separately. Last edited by skipjack; October 12th, 2016 at 11:48 AM.  
September 30th, 2016, 04:55 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,524 Thanks: 588 
For x < 4, the integrand is f2f1. For 4 < x < 5, the integrand is f2.

October 12th, 2016, 11:40 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,054 Thanks: 1618 
$\displaystyle \int_0^5\!f_2\,dx  \int_0^4\!f_1\,dx = \frac{25}{2}  8 = \frac92$


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