Exponential integration question (1 Viewer)

[aanthnnyyy]

Hey guys I just ran into a question seems pretty easy: y=e^2x with y=1 and x=2 find area enclosed between.

I'm sorta confused.. I've never see a. Question with a y and x limit put together - someone help me pleeees

Answers: 1/2(e^4-5) where the hell did 5 come from !!

 

[InteGrand]

Hey guys I just ran into a question seems pretty easy: y=e^2x with y=1 and x=2 find area enclosed between.

I'm sorta confused.. I've never see a. Question with a y and x limit put together - someone help me pleeees

Answers: 1/2(e^4-5) where the hell did 5 come from !!

Once you draw a diagram, you'll see that the area they want you to find is the area of a curvilinear triangle. It's the area bounded by the horizontal line y = 1, x = 2, and the curve y = e^2x. If you draw a diagram, the region of the area they want you to find is easily seen. Note that x = 0 when y = 1.

Then you should see that the area will be given by .

 

[aanthnnyyy]

Once you draw a diagram, you'll see that the area they want you to find is the area of a curvilinear triangle. It's the area bounded by the horizontal line y = 1, x = 2, and the curve y = e^2x. If you draw a diagram, the region of the area they want you to find is easily seen. Note that x = 0 when y = 1.

Then you should see that the area will be given by .

omg! Thank you so much!
I drew the diagram out and all but didn't really know how to manipulate it :/

 

[InteGrand]

Shouldn't the area be given by:


As the original graph itself is e^2x, so the square of the radius should give e^4x

We're not doing volumes for this Q, we're doing areas. You'd be right for the volume of that area rotated about the x-axis though.

 

[Ekman]

We're not doing volumes for this Q, we're doing areas. You'd be right for the volume of that area rotated about the x-axis though.

Yeah I just realized, doing all those volumes questions got to my head, I deleted the post so no one else would get confused.