from the graphs

i got the lines in a online function grapher but not sure where the area is bounded by the y-axis and which intervals i should use.

here is the graph with both functions.

http://img267.imageshack.us/img267/3584/graphzv1.jpg

Printable View

- Nov 3rd 2008, 07:49 PMjvignaciofind the area bounded by the y-axis
from the graphs

i got the lines in a online function grapher but not sure where the area is bounded by the y-axis and which intervals i should use.

here is the graph with both functions.

http://img267.imageshack.us/img267/3584/graphzv1.jpg - Nov 3rd 2008, 08:43 PMJameson
I think the region you're supposed to evaluate starts at y=1, has a right edge where the green and red graphs intersect, and an upper bound where the green graph crosses the y-axis again, which should be at y=6.

So because of the way the region is set up, we should integrate with respect to y instead of x. That means we need to solve both equations for x.

So looking at the region it looks like it's good to break it up into two smaller regions. From y=1 to where the graphs intersect, the x-values of the region are defined by and after they intersect until y=6 the x-values are defined by

So let's figure out where they intersect.

So I think the whole are is:

Does that make sense? Hopefully I don't have any huge errors. - Nov 3rd 2008, 09:06 PMjvignacio
just letting you know i shrunk the graph and the green line crosses the y axis on 6, would that make any major changes to the working out? just change the interval number yeah? so from 7/2 till 6? makes more sence now man. thanks alot!

http://img267.imageshack.us/img267/7772/graphfixbn9.jpg - Nov 3rd 2008, 09:11 PMJameson
Doh! e^0 = 1, not 0. Brain fade. That shouldn't change anything though.