Question

can somone help me in trying to get the are between these curves. I have : y^2-4x=4 and 4x-y=16. I solved them for x (cause i was gonna integrate them with respect to y) and got: x=(-1)+(y^2/4) and x=(4)-(y/4). I am not sure is those are right??

Answer 1

And then i had the integral from -4 to 5 of -1+y^2/4-4+y/4

Answer 2

ok, so did you integrate it?

Answer 3

i am having problem b/c of the fractions

Answer 4

can you help

Answer 5

integrate it term by term.

Answer 6

but what about the fractions

Answer 7

\[\int f(x) + g(x) dx = \int f(x)dx + \int g(x)dx\]

Answer 8

you can pull constant multiples out of an integral

Answer 9

\[\int kf(x)dx = k\int f(x)dx\]

Answer 10

so when you get to the terms with fractions, just pull out the 1/4 before you evaluate the integral

Answer 11

so i got -5x+y^3/12+y^2/8

Answer 12

why is there an x in there?

Answer 13

oh its -5y

Answer 14

ok, now evaluate that for your limits.

Answer 15

man, i am still not getting the answer

Answer 16

what are you getting?

Answer 17

i am getting -225/8

Answer 18

the books say it should be 243/8

Answer 19

Well it certainly shouldn't be negative..

Answer 20

is this right : y^2-4x=4 : x=-1+y^2/4

Answer 21

?

Answer 22

yes

Answer 23

how about this4x-y=16: x=4-y/4

Answer 24

?

Answer 25

yes.

Answer 26

so would the f(y)-g(y) look like this : -5+y^2/4-y/4

Answer 27

?

Answer 28

I think the line is your f(y) and the quadratic is g(y) since the line is 'above' the parabola.

Answer 29

yeah your right, the right most curve is the f(y)

Answer 30

-y/4+4 - (y^2/4 - 1)
= -y/4 + 4 - y^2/4 + 1

Answer 31

can we add the 4 and the 1

Answer 32

?