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August 1, 2014

August 1, 2014

Posted by **jasmineT** on Saturday, April 28, 2012 at 6:48pm.

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=3x^(1/2) , y=4 and 2y+1x=4

y=3(x^(1/2))/2, y=4 , y=2-(1/2)x

Intersection points = -4,1,64/9

Integral from -4 to 1

4x-2x-(x^2)/4

Integral from 1 to 64/9

4x-3(x^(5/2))/5

please help i've been working on this for hours

- Calculus: Area -
**Steve**, Saturday, April 28, 2012 at 7:29pmgo to wolframalpha . com and enter

graph x=4y^2/9,x+2y=4,y=4,y=3/2,x=1

to see the curves involved.

The area of interest is a little v-shaped chunk from

(-4,4) down to (1,3/2) and back up to (64/9,4)

If you integrate on x, then the integrands use the limits you have, but

the height in each case is 4-y, for the appropriate f(x)

4 - (4-x)/2 from -4 to 1 = 25/4

4 - 3/2 sqrt(x) from 1 to 64/9 = 175/27

match anything you got?

- Calculus: Area -
**jasmineT**, Sunday, April 29, 2012 at 9:54pmThank you soooo much steve i finally got it right!!!! :D

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