Question : Representing the area by integral, find the area of the region bounded by the lines $\displaystyle x=4, y=1$ and $\displaystyle x+4y = 4$.

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Solution:

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- Dec 28th 2009, 11:29 PMzorroArea of the region bounded by the line
Question : Representing the area by integral, find the area of the region bounded by the lines $\displaystyle x=4, y=1$ and $\displaystyle x+4y = 4$.

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Solution: - Dec 29th 2009, 12:59 AMwauwaujust straight forward
$\displaystyle

\int_0^4 \!1\,dx-\int_{0}^{4}\!\frac{4-x}{4}\,dx

$ - Dec 29th 2009, 03:02 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.

- Dec 29th 2009, 12:19 PMzorro
Thanks mite the answer which i am getting is 2

is that correct - Dec 29th 2009, 12:30 PMintegral
Find the area under the top function and subtract by the area under the bottom function.

$\displaystyle \int_{0}^{4}1dx-\left |\int_{0}^{4}\frac{x-4}{4}dx \right |=2$

Any vertical line restraint only represents were to stop.

x=4 then

$\displaystyle \int_{n}^{4}$

---$\displaystyle \int$ - Dec 29th 2009, 01:12 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 - Dec 29th 2009, 01:13 PMintegral
Sorry, yes it is 2 :)

- Dec 29th 2009, 03:35 PMzorro
thanks mite for conferming

- Dec 30th 2009, 06:54 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.