Question

Area between two curves? Not sure I am doing this correctly?

Answer 1

I am being asked to find the area between the\[y^2=4x\] curve and the line \[x=3\].

Answer 2

Am I correct by using \[A=\int\limits_{a}^{b}g(x)-f(x) dx\] where g(x) is the funtion with the higher x values?

Answer 3

for get finding higher and lower, its pointless in this case

Answer 4

Okay then how would I set up the integral to find the area?

Answer 5

define each setup in terms of x(y)

Answer 6

\[x= \frac{ y^2 }{ 4 }\] and \[x=3\]?

Answer 7

yep, and what are the points of intersection?

Answer 8

\[\pm2\sqrt{3}\]?

Answer 9

yep

Answer 10

so lets add up them up ...

\[\int_{-2\sqrt{3}}^{2\sqrt{3}}x_1-x_2~dy\]

Answer 11

3 - y^2/4 or y^2/4 - 3

if you do it both ways youll see why it doesnt matter the order ....

Answer 12

So I just chose the positive value then between the 2?

Answer 13

just work it out ... what do we get in both cases?

Answer 14

or rather, lets say the area is K

one way gets us K the other way gets us -K

but area is not negative so the sign is irrelevant

Answer 15

all it means is the same as this:

5 - 2 = 3
2 - 5 = -3

|3| = |-3| = 3

Answer 16

I am getting 9/4

Answer 17

3 - y^2/4

3y - y^3/12

2(3(2rt3) - (2rt3)^3/12)

2(6rt3 - (2rt3)(2rt3)(2rt3)/12)

2(6rt3 - 12(2rt3)/12)

2(6rt3 - 2rt3)

2(4rt3)

Answer 18

Oops I forgot about the lower limit. But yeah then I would get 8sqr(3)

Answer 19

\[\int_{0}^{3}4x^{1/2}~dx\]
\[\frac234(3)^{3/2}\]
\[8(3)^{3/2-2/2}\]
\[8(3)^{1/2}\]

Answer 20

Thank you for the attachment! That's super helpful!

Answer 21

youre welcome