anonymous
  • anonymous
find the area bounded by y1=0, y2=-x+2, y3=sqrt(x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
y = 0 is the the x axis....let me draw a rudemintary pic of this to get an idea....
amistre64
  • amistre64
is that y squared? or just the second equation of y?
amistre64
  • amistre64
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amistre64
  • amistre64
we need to determine that value of 'c' in order to piece this thing together
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amistre64
  • amistre64
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
amistre64
  • amistre64
\[\int\limits_{0}^{1} (-x^2+2 - \sqrt{x})dx + \int\limits_{1}^{2} (\sqrt(x)+x^2-2) dx\]
amistre64
  • amistre64
\[{-x^2 \over 2} +2x -\frac{\sqrt{x}}{2}\] right? for the first one that is :)
anonymous
  • anonymous
thank you very much!
amistre64
  • amistre64
-1/2 - 1/2 + 2 = 1 for the area up to 'c'
amistre64
  • amistre64
thats an error.... -x^2 integrates to x^3/3...so ill have to adjust for it :)
amistre64
  • amistre64
-1/3 - 1/2 +2 = A
amistre64
  • amistre64
2 - 3/2 = 2 - 1' 1/2 = 1/2 right?
anonymous
  • anonymous
wait i dont understand how you got x^2 from the square root of x
amistre64
  • amistre64
youre right; trying to do this in my head :) i put thought of (-x+2) as (-x^2 +2) when i did that :)
anonymous
  • anonymous
ok
amistre64
  • amistre64
[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?
amistre64
  • amistre64
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?
anonymous
  • anonymous
ok
amistre64
  • amistre64
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?
anonymous
  • anonymous
right
amistre64
  • amistre64
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 :)
anonymous
  • anonymous
ok :)
amistre64
  • amistre64
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]
amistre64
  • amistre64
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 :)
anonymous
  • anonymous
yes it does! thank you again!
amistre64
  • amistre64
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 :)
amistre64
  • amistre64
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
anonymous
  • anonymous
haha. well you are not an idiot in disguise anymore.
anonymous
  • anonymous
I got the area = 1.167 area unit. Not entirely sure though.
amistre64
  • amistre64
yay!! 7/6 = 1.166.....
anonymous
  • anonymous
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: \[A=\int\limits_{0}^{1}(2-y-y^2)dy=(2y-{y^2 \over 2}-{y^3 \over 3})_{0}^{1}=2-{1 \over 2}-{1 \over 3}={7 \over 6}\]
amistre64
  • amistre64
i used many methods; and eventually got smart enough to see an easier way ;)
anonymous
  • anonymous
thank you very much

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