Area of the region bounded by the line

• December 28th 2009, 11:29 PM
zorro
Area of the region bounded by the line
Question : Representing the area by integral, find the area of the region bounded by the lines $x=4, y=1$ and $x+4y = 4$.
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Solution:
• December 29th 2009, 12:59 AM
wauwau
just straight forward
$
\int_0^4 \!1\,dx-\int_{0}^{4}\!\frac{4-x}{4}\,dx
$
• December 29th 2009, 03:02 AM
HallsofIvy
Quote:

Originally Posted by zorro
Question : Representing the area by integral, find the area of the region bounded by the lines $x=4, y=1$ and $x+4y = 4$.
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Solution:

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, 12:19 PM
zorro
Thanks mite the answer which i am getting is 2
is that correct
• December 29th 2009, 12:30 PM
integral
Find the area under the top function and subtract by the area under the bottom function.

$\int_{0}^{4}1dx-\left |\int_{0}^{4}\frac{x-4}{4}dx \right |=2$
Any vertical line restraint only represents were to stop.

x=4 then
$\int_{n}^{4}$

--- $\int$
• December 29th 2009, 01:12 PM
zorro
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, 01:13 PM
integral
Sorry, yes it is 2 :)
• December 29th 2009, 03:35 PM
zorro
thanks mite for conferming
• December 30th 2009, 06:54 AM
HallsofIvy
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.