# CALCULUS:)

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Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
2y=3sqrtx
and
y=3
and
2y+2x=5

• CALCULUS:) -

its not 8

• CALCULUS:) -

I would take vertical slices, that is, integrate with respect to x
But after looking at the sketch, I realize that we have to find the intersection of the line 2x+2y=5 and 2y = 3√x
equating 3√x = 5-2x and squaring both sides I got
4x^2 - 29x + 25=0
(x-1)(4x-25) =0
x = 1 or x = 25/4

also if y=3, then 6=3√x ----> x = 4

so we need ∫(3 - 5/2 + x dx from 0 to 1 + ∫(3 - 3x^(1/2) ) dx from 1 to 4

Can you finish it?

• CALCULUS:) -

-4?? that is incorrect..hmm, what am i doing wrong?

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