Question

find the area bounded by y1=0, y2=-x+2, y3=sqrt(x)

Answer 1

y = 0 is the the x axis....let me draw a rudemintary pic of this to get an idea....

Answer 2

is that y squared? or just the second equation of y?

Answer 3

we need to determine that value of 'c' in order to piece this thing together

Answer 4

c equals the point where y = -x+2 and y = sqrt(x) are the same value:

-x+2 = sqrt(x)

x^2 -4x +4 = x^2

-4x +4 = 0

x = 1 = c

Answer 5

\[\int\limits_{0}^{1} (-x^2+2 - \sqrt{x})dx + \int\limits_{1}^{2} (\sqrt(x)+x^2-2) dx\]

Answer 6

\[{-x^2 \over 2} +2x -\frac{\sqrt{x}}{2}\] right? for the first one that is :)

Answer 7

thank you very much!

Answer 8

-1/2 - 1/2 + 2 = 1 for the area up to 'c'

Answer 9

thats an error.... -x^2 integrates to x^3/3...so ill have to adjust for it :)

Answer 10

-1/3 - 1/2 +2 = A

Answer 11

2 - 3/2 = 2 - 1' 1/2 = 1/2 right?

Answer 12

wait i dont understand how you got x^2 from the square root of x

Answer 13

youre right; trying to do this in my head :) i put thought of (-x+2) as (-x^2 +2) when i did that :)

Answer 14

ok

Answer 15

[S] (-x+2 - sqrt(x)) dx ; [0,1]

-x^2/2 +2x - sqrt(x)/2..... right?

at 0 it equals 0

at 1 it equals 1 right?

Answer 16

for the second run; we have the top function - bottom function

sqrt(x) - (-x+2)

[S] (sqrt(x) +x -2) dx ; [1,2]

sqrt(x)/2 + x^2/2 -2x right?

Answer 17

ok

Answer 18

1/2 + 1/2 - 2 = -1 .... remember that number well need it soon enough :)

sqrt(2) 4
----- + -- - 2(2)
2 2

sqrt(2)
----- + 2 - 4
2


sqrt(2)
------ - 2 = abt. -1.29
2

-1.29 - - 1 = 1-1.29 = |-.29| = .29 as an area for the next part

add the areas togher to get:

1.29; right?

Answer 19

right

Answer 20

now that i look at it again :)

the second part is simply the area under the line (-x+2); i wonder if that changes anything :)

Answer 21

ok :)

Answer 22

F(x) = [S] -x+2 dx ; [1,2]

F(x) = -x^2/2 + 2x

F(2) = -4/2 + 4 = 2
-F(1) = -1/2 + 2 = - 1' 1/2
--------
1/2

The area in the first part is the area under sqrt(x) from 0 to 1

F(x) = [S] sqrt(x) dx ; [0,1]

Answer 23

F(x) = sqrt(x)/2

F(0) = 0

F(1) = 1/2

Add the areas together to get:

1/2 + 1/2 = 1 .... that sound more logical :)

Answer 24

yes it does! thank you again!

Answer 25

did I integrate that sqrt(x) correctly? lol...let me check.

x^(1/2)

x^(1+2/2) x^(3/2) 2 sqrt(x^3)
---------- = -------- = ----------
1+2/2 3/2 3



Dont thank me yet lol.... im just an idiot in disguise till i get this right :)

Answer 26

F(1) = 2/3..... now we can add them up :)

1 2
-- + ---
2 3


3 4
-- + --- = 7/6 for the total area :) and that my final offer
6 6

Answer 27

haha. well you are not an idiot in disguise anymore.

Answer 28

I got the area = 1.167 area unit. Not entirely sure though.

Answer 29

yay!! 7/6 = 1.166.....

Answer 30

I think the method you used is a little bit more complicated than it should be. Consider the graph you posted above, you can see that the line x=2-y is on the right of the graph of x=y^2, then the area a is:
\[A=\int\limits_{0}^{1}(2-y-y^2)dy=(2y-{y^2 \over 2}-{y^3 \over 3})_{0}^{1}=2-{1 \over 2}-{1 \over 3}={7 \over 6}\]

Answer 31

i used many methods; and eventually got smart enough to see an easier way ;)

Answer 32

thank you very much