What is the area under the curves of y=3, 2y=5(x)^1/2 and 2y+x=6

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What is the area under the curves of y=3, 2y=5(x)^1/2 and 2y+x=6

Mathematics
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is there a picture attached?
Ah well it's just that graphing this is going to be a pain.. So let's see what we can do
no but I can attach one

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if u can, please do so haha
k here it is
1 Attachment
omg lol more of these webworks... i'm so sick of them haha
sorry
nah it's fine.. actually give the wolframalpha link, they also list the intercepts. it's easier.
http://www.wolframalpha.com/input/?i=Graph+y%3D3%2C+2y%3D5x^%281%2F2%29%2C+2y%2Bx%3D6
all right great thanks. Here we go..
So here I'm going to split the graph in two at the point where x+2y = 6 and 2y = 5sqrt(x) intersect
okay we did that
That's the upper limit of our first integral
okay
Sorry my browser's having trouble, anyway the upper intersect was 1. Now the lower intersect, which occurs when y =3 and 2y+x=6 intersect.
6+x=6, x = 0. So the first half of that area is: Integral (upper curve - lower curve, 0,1)
how did you get one I got 1.36 doesnt it turn into a quadratic
x+5sqrt(x) = 6? Well I mean take a look, what number plus 5 times it's square root is 6?
okay i get it i never thought of it like that
Keep in mind, we're still calculating the first half. So now, the upper curve minus the lower curve (well they're lines, but whatever) is: 3 - (6-x)/2
So so far, the first half of our area is: Integral (3 - (6-x)/2, 0,1)
Now let's find the 2nd half: This time the lower limit is 1, because we are starting at one, and the upper limit is the intersection of y=3 and y = 5/2sqrt(x)
(5/2)*sqrt(x) = 3 sqrt(x) = 6/5 => x = 36/25
So 2nd integral is again, upper curve minus the lower curve from 1 to 36/25, or Integral (3 - (5/2)*sqrt(x), 1, 36/25)
Calculate each integral and then add them to get 107/300 (lol I cheated, used wolf to compute)
so it is .356666667
Yea that looks right.
k i will try it and tell you if it works
it works thanks sooooo much you have no idea
np.

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