# Thread: find the area bounded by the y-axis

1. ## find the area bounded by the y-axis

from the graphs

$y = e^x$

$y = 7-e^x$

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.

2. 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.

$(y=e^x) \rightarrow (x=\ln(y))$
$(y=7-e^x) \rightarrow (x=\ln(7-y))$

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 $x=\ln(y)$ and after they intersect until y=6 the x-values are defined by $x=\ln(7-y)$

So let's figure out where they intersect.

$\ln(y)=\ln(7-y)$
$y=7-y$
$y=\frac{7}{2}$

So I think the whole are is:

$\int_{1}^{\frac{7}{2}} \ln(y)dy + \int_{\frac{7}{2}}^{6} \ln(7-y)dy$

Does that make sense? Hopefully I don't have any huge errors.

3. Originally Posted by Jameson
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=7.

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.

$(y=e^x) \rightarrow (x=\ln(y))$
$(y=7-e^x) \rightarrow (x=\ln(7-y))$

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 $x=\ln(y)$ and after they intersect until y=7 the x-values are defined by $x=\ln(7-y)$

So let's figure out where they intersect.

$\ln(y)=\ln(7-y)$
$y=7-y$
$y=\frac{7}{2}$

So I think the whole are is:

$\int_{1}^{\frac{7}{2}} \ln(y)dy + \int_{\frac{7}{2}}^{7} \ln(7-y)dy$

Does that make sense? Hopefully I don't have any huge errors.
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!

4. Doh! e^0 = 1, not 0. Brain fade. That shouldn't change anything though.