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anonymous
 one year ago
The area enclosed by the curve y^2 = x(3 − x) is given by
https://gyazo.com/d848359e382f1fcf7369e66b148fc534
anonymous
 one year ago
The area enclosed by the curve y^2 = x(3 − x) is given by https://gyazo.com/d848359e382f1fcf7369e66b148fc534

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got C. Can someone confirm it for me?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would you solve for y?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Take the square root of both sides.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I agree, you solve for y to get \[\Large y = \sqrt{x(3x)} = \sqrt{3xx^2}\] you double the area because the integral alone is just figuring out the top portion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y^2=x \left( 3x \right)=3xx^2=\left( x^23x \right)=\left( x^23x+\left( \frac{ 3 }{ 2 } \right)^2\left( \frac{ 3 }{ 2 } \right)^2 \right)\] \[y^2=\left( x^23x+\left( \frac{ 3 }{ 2 } \right)^2\frac{ 9 }{ 4 } \right)\] \[y^2=\left( x\frac{ 3 }{ 2 } \right)^2+\frac{ 9 }{ 4 }\] \[y^2+\left( x\frac{ 3 }{ 2 } \right)^2=\left( \frac{ 3 }{ 2 } \right)^2\] it is a circle with center (3/2,0) and radius 3/2 dw:1444436581363:dw area\[=2\int\limits_{0}^{3}\sqrt{x \left( 3x \right)}dx\] \[area=2\int\limits_{0}^{3}\sqrt{3xx^2}dx\]
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