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Math Help - Area of the region bounded by the line

  1. #1
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    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.
    ------------------------------------------------------------------------
    Solution:
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  2. #2
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    just straight forward

     <br />
\int_0^4 \!1\,dx-\int_{0}^{4}\!\frac{4-x}{4}\,dx<br />
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  3. #3
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    Quote Originally Posted by zorro View Post
    Question : Representing the area by integral, find the area of the region bounded by the lines x=4, y=1 and x+4y = 4.
    ------------------------------------------------------------------------
    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.
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  4. #4
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    Thanks mite the answer which i am getting is 2
    is that correct
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  5. #5
    Member integral's Avatar
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    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
    Last edited by integral; December 29th 2009 at 12:39 PM. Reason: typo :)
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  6. #6
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    dude ur are confusing me ..............the answer to that question is 2
    isnt it or do i have to do something else as well
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  7. #7
    Member integral's Avatar
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    Sorry, yes it is 2
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  8. #8
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    thanks mite for conferming
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  9. #9
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    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.
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