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July 24, 2014

July 24, 2014

Posted by **Brit** on Tuesday, August 16, 2011 at 8:18pm.

2y=3sqrtx

and

y=3

and

2y+2x=5

- CALCULUS:) -
**Brit**, Tuesday, August 16, 2011 at 8:18pmits not 8

- CALCULUS:) -
**Reiny**, Tuesday, August 16, 2011 at 8:54pmI 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:) -
**Brit**, Tuesday, August 16, 2011 at 11:29pm-4?? that is incorrect..hmm, what am i doing wrong?

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