Asked by karen
Set up the integrals that can be used to find the area of the region by integrating with respect to X and Y
(the region is bounded by y=3x,y=x,y=4-x)
Anyone help me set up the integrals??
(the region is bounded by y=3x,y=x,y=4-x)
Anyone help me set up the integrals??
Answers
Answered by
Reiny
Did you make a sketch?
Here is Wolfram's take on the situation.
http://www.wolframalpha.com/input/?i=plot+y%3D3x%2Cy%3Dx%2Cy%3D4-x
You need the intersection of the lines
I assume you want the area of the enclosed region.
the intersection points are (0,0), (1,3), and (2,2)
from x = 0 to 1, y = 3x is above y = 3x
so the effective height = 3x-x = 2x
from 1 to 2, y = 4-x is above y = x
so the effective height is 4-x - x = 4 - 2x
So the area
= ∫2x dx from 0 to 1 + ∫(4-2x) dx from 1 to 2
take over from here
Here is Wolfram's take on the situation.
http://www.wolframalpha.com/input/?i=plot+y%3D3x%2Cy%3Dx%2Cy%3D4-x
You need the intersection of the lines
I assume you want the area of the enclosed region.
the intersection points are (0,0), (1,3), and (2,2)
from x = 0 to 1, y = 3x is above y = 3x
so the effective height = 3x-x = 2x
from 1 to 2, y = 4-x is above y = x
so the effective height is 4-x - x = 4 - 2x
So the area
= ∫2x dx from 0 to 1 + ∫(4-2x) dx from 1 to 2
take over from here
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