Question : Representing the area by integral, find the area of the region bounded by the lines and .

------------------------------------------------------------------------

Solution:

Printable View

- December 29th 2009, 12:29 AMzorroArea of the region bounded by the line
Question : Representing the area by integral, find the area of the region bounded by the lines and .

------------------------------------------------------------------------

Solution: - December 29th 2009, 01:59 AMwauwaujust straight forward
- December 29th 2009, 04:02 AMHallsofIvy
I just notice that you said "representing the area by an integral". wau wau's answer is best. But if it were not for that requirement, that's a right triangle: area= (1/2)height * base. Find the points where x+ 4y= 4 and you are almost done. You can use that to check your work.

- December 29th 2009, 01:19 PMzorro
Thanks mite the answer which i am getting is 2

is that correct - December 29th 2009, 01:30 PMintegral
Find the area under the top function and subtract by the area under the bottom function.

Any vertical line restraint only represents were to stop.

x=4 then

--- - December 29th 2009, 02:12 PMzorro
dude ur are confusing me ..............the answer to that question is 2

isnt it or do i have to do something else as well - December 29th 2009, 02:13 PMintegral
Sorry, yes it is 2 :)

- December 29th 2009, 04:35 PMzorro
thanks mite for conferming

- December 30th 2009, 07:54 AMHallsofIvy
I told you earlier, in post number 3 how you could easily check this yourself. This is a right triangle. It's area is just (1/2)height times width: one leg is from y= 0 to y= 1 so has height 1, the other is from x= 0 to x= 4 so has width 4. The area is (1/2)(1)(4)= 2.