Asked by Natalie

Sketch the region enclosed by 2y=5x^(1/2), y=3, 2y+4x=9. Decide whether to integrate with respect to x or y, and then find the area of the region.
Answered by Reiny
your sketch should look like this:
a small "triangular shape"

I would take horizontal slices, that is with respect to y

you first need the intersection of 2y=5x^(1/2) and 2y+4x=9

5√x = 9-4x
25x = 81 - 72x + 16x^2
16x^2 - 97x + 81 = 0
(16x - 81)(x - 1) = 0 , wow it factored
x = 81/16 , or x = 1
as can be seen from the graph, 81/16 is an extraneous root introduced by squaring, so
x = 1, y = 2.5

so you want to integrate with respect to y from 2.5 to 3
Answered by Ayo
Could u integrate it for us
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