Oh sorry then, In haste i just read the post topic and attachment, and answered
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Finding area beneath/below x-axis; integration watch
- 24-01-2015 20:14
- Study Helper
(Original post by apronedsamurai)
- 24-01-2015 22:40
Your both right
There are three questions in total.
I was asking for help regarding the first question, so that I might be better enabled to work on the other two myself.
I was going to take a look but for some reason your attachment isn't coming up properly on my browser and will only save as a tiny image!
I presume you're trying to find the area between the line y = 2-x and the x-axis between the limits x=0 and x=3, in which case, as already explained, you need to look at two separate regions - the part from x=0 to x=2 and the part from x=2 to x=3 because your line crosses the x-axis at that point.
- 24-01-2015 22:43
- Study Helper
- 24-01-2015 22:50
Although I was hoping we might see some feedback from the OP to indicate whether he's overcome his conceptual difficulties with the assistance that people have provided already...
- Thread Starter
(Original post by Actaeon)
- 27-01-2015 11:44
This is where integrating the parts of the graph above and below the x-axis separately comes in. The section from 0 to 2 will integrate to give 2. The section from 2 to 3 will integrate to give -0.5. So if you simply integrate between 0 and 3 you'll get 1.5 which is wrong - area is always positive. So integrate them separately, add them ignoring the negative sign, and you get 2+0.5 =2.5
Ok, so I am unsure as to where/how you get the -0.5
When I integrate 2 and 3; I get 1.5...
EDIT: God, what a ****ing moron I am. I had forgotten to subtract the value of the 3 and 2; and right enough, got -0.5; as advised, disregarded the sign, from there; upon adding the two values together, got the required 2.5.
I was getting frustrated because I had managed to finally grasp the -x+2=0; but was thinking why on earth was I missing out on the 2.5
Thanks guys. I am not 100% confident on this topic, but will be working on it today so hopefully will be able to report back more definitively one way or another.
In any case, big thank you to everyone for your help.Last edited by User947387; 27-01-2015 at 11:52.