If x=0 then e^(2*0)=e^(6*0)
If 0<x<1 then e^2x<e^6x.
The area=integral from 0 to 1
(e^6x-e^2x)dx =
(1/6)e^6x-(1/2)e^2x from 0 to 1 =
(1/6)e^6-(1/2)e^2-(1/6)+(1/2)
Decide whether to integrate with respect to x or y. Find the area of the region.
y=e^2x y=e^6x x=1
I have no idea what to do
6 answers
The computer homework program says that is incorrect. Any other suggestions?
Your think your computer program is incorrect
I think in "anonymous' " last line he/she should have had
(1/6)e^6-(1/2)e^2-(1/6) - (1/2))
or
e^6/6 - e^2/2 + 1/3
(1/6)e^6-(1/2)e^2-(1/6) - (1/2))
or
e^6/6 - e^2/2 + 1/3
I wrote ...-(1/6)+(1/2)
...-(1/6)+(1/2)=...+1/3.
You mean -((1/6)-(1/2))
...-(1/6)+(1/2)=...+1/3.
You mean -((1/6)-(1/2))
Yay!! Thank you both very much! It worked.
1 answer