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anonymous
 5 years ago
find the area bounded by y1=0, y2=x+2, y3=sqrt(x)
anonymous
 5 years ago
find the area bounded by y1=0, y2=x+2, y3=sqrt(x)

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1y = 0 is the the x axis....let me draw a rudemintary pic of this to get an idea....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1is that y squared? or just the second equation of y?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we need to determine that value of 'c' in order to piece this thing together

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1c 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
 5 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{0}^{1} (x^2+2  \sqrt{x})dx + \int\limits_{1}^{2} (\sqrt(x)+x^22) dx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[{x^2 \over 2} +2x \frac{\sqrt{x}}{2}\] right? for the first one that is :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.11/2  1/2 + 2 = 1 for the area up to 'c'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1thats an error.... x^2 integrates to x^3/3...so ill have to adjust for it :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.12  3/2 = 2  1' 1/2 = 1/2 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait i dont understand how you got x^2 from the square root of x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1youre right; trying to do this in my head :) i put thought of (x+2) as (x^2 +2) when i did that :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1[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
 5 years ago
Best ResponseYou've already chosen the best response.1for 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.11/2 + 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 = 11.29 = .29 = .29 as an area for the next part add the areas togher to get: 1.29; right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1now that i look at it again :) the second part is simply the area under the line (x+2); i wonder if that changes anything :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1F(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
 5 years ago
Best ResponseYou've already chosen the best response.1F(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
 5 years ago
Best ResponseYou've already chosen the best response.0yes it does! thank you again!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1did 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
 5 years ago
Best ResponseYou've already chosen the best response.1F(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
 5 years ago
Best ResponseYou've already chosen the best response.0haha. well you are not an idiot in disguise anymore.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got the area = 1.167 area unit. Not entirely sure though.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1yay!! 7/6 = 1.166.....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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=2y is on the right of the graph of x=y^2, then the area a is: \[A=\int\limits_{0}^{1}(2yy^2)dy=(2y{y^2 \over 2}{y^3 \over 3})_{0}^{1}=2{1 \over 2}{1 \over 3}={7 \over 6}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i used many methods; and eventually got smart enough to see an easier way ;)
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