Asked by Oscar
Find the area of the region. Please show work so I can understand.
2y=5x^(1/2),y=3,2y+4x=9
2y=5x^(1/2),y=3,2y+4x=9
Answers
Answered by
Steve
The curves bound a small triangular region with vertices (left-to-right) at
(3/4,3) (1,5/2) (36/25,3)
so, using vertical strips of width dx, we have to divide the area into two regions, divided at x=3/4. The area is
โซ[3/4,1] 3-(9-4x)/2 dx + โซ[1,36/25] 3-5โx/2 dx
Using horizontal strips is a bit easier, since the area has an easier boundary. The strips have length equal to the difference between the two curves' x-coordinates:
โซ[36/25,3] ((2y/5)^2 - (9-2y)/4) dy
as always, check my algebra
(3/4,3) (1,5/2) (36/25,3)
so, using vertical strips of width dx, we have to divide the area into two regions, divided at x=3/4. The area is
โซ[3/4,1] 3-(9-4x)/2 dx + โซ[1,36/25] 3-5โx/2 dx
Using horizontal strips is a bit easier, since the area has an easier boundary. The strips have length equal to the difference between the two curves' x-coordinates:
โซ[36/25,3] ((2y/5)^2 - (9-2y)/4) dy
as always, check my algebra
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