K KEYWEST17 New member Joined Jan 19, 2011 Messages 46 Apr 21, 2011 #1 I plugged the equations into the graphing calculator and can see the areas to evaluate but I am stuck. Attachments b828591ac0ea6717d99a40d72930821.png 616 bytes · Views: 75
I plugged the equations into the graphing calculator and can see the areas to evaluate but I am stuck.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Apr 21, 2011 #2 I assume the shaded region is the area in question?. Find the area under \(\displaystyle e^{2x}\) from 0 to 1: \(\displaystyle \int_{0}^{1}e^{2x}dx\) Find the area under \(\displaystyle e^{6x}\) from \(\displaystyle -\infty \;\ \text{to} \;\ 0\): \(\displaystyle \int_{-\infty}^{0}e^{6x}dx\) Add them up. This is assuming my assessment is correct and I have the correct region as in the diagram. But, if you need the region between the two e functions: \(\displaystyle \int_{-\infty}^{0}\left[e^{2x}-e^{6x}\right]dx\) Attachments e.jpg 65.6 KB · Views: 70
I assume the shaded region is the area in question?. Find the area under \(\displaystyle e^{2x}\) from 0 to 1: \(\displaystyle \int_{0}^{1}e^{2x}dx\) Find the area under \(\displaystyle e^{6x}\) from \(\displaystyle -\infty \;\ \text{to} \;\ 0\): \(\displaystyle \int_{-\infty}^{0}e^{6x}dx\) Add them up. This is assuming my assessment is correct and I have the correct region as in the diagram. But, if you need the region between the two e functions: \(\displaystyle \int_{-\infty}^{0}\left[e^{2x}-e^{6x}\right]dx\)
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,250 Apr 21, 2011 #3 KEYWEST17 said: I plugged the equations into the graphing calculator and can see the areas to evaluate but I am stuck. Click to expand... KEYWEST17, you are also to include y = 0, as another curve to form the region, if it is supposed to be the figure drawn by galactus.
KEYWEST17 said: I plugged the equations into the graphing calculator and can see the areas to evaluate but I am stuck. Click to expand... KEYWEST17, you are also to include y = 0, as another curve to form the region, if it is supposed to be the figure drawn by galactus.
K KEYWEST17 New member Joined Jan 19, 2011 Messages 46 Apr 21, 2011 #4 I come up with -47.209. Does this seem accurate.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Apr 21, 2011 #5 No, I am sorry to say, it does not. Where did the negative come from?. The region is above the x-axis.
No, I am sorry to say, it does not. Where did the negative come from?. The region is above the x-axis.
K KEYWEST17 New member Joined Jan 19, 2011 Messages 46 Apr 21, 2011 #6 Tried the positive value and it is still incorrect.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Apr 21, 2011 #7 I am unsure of what region you are looking for. Is it \(\displaystyle \int_{0}^{1}e^{2x}dx\)?. Or \(\displaystyle \int_{-\infty}^{0}\left(e^{2x}-e^{6x}\right)dx\)?. Or both?. Or something else?. The problem may indicate both. \(\displaystyle \int_{-\infty}^{0}\left(e^{2x}-e^{6x}\right)dx+\int_{0}^{1}e^{2x}dx\) Show us what you're doing so we can pinpoint the trouble.
I am unsure of what region you are looking for. Is it \(\displaystyle \int_{0}^{1}e^{2x}dx\)?. Or \(\displaystyle \int_{-\infty}^{0}\left(e^{2x}-e^{6x}\right)dx\)?. Or both?. Or something else?. The problem may indicate both. \(\displaystyle \int_{-\infty}^{0}\left(e^{2x}-e^{6x}\right)dx+\int_{0}^{1}e^{2x}dx\) Show us what you're doing so we can pinpoint the trouble.