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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I got C. Can someone confirm it for me?
How would you solve for y?
Take the square root of both sides.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Anyone?
I agree, you solve for y to get \[\Large y = \sqrt{x(3-x)} = \sqrt{3x-x^2}\] you double the area because the integral alone is just figuring out the top portion
\[y^2=x \left( 3-x \right)=3x-x^2=-\left( x^2-3x \right)=-\left( x^2-3x+\left( \frac{ -3 }{ 2 } \right)^2-\left( \frac{ -3 }{ 2 } \right)^2 \right)\] \[y^2=-\left( x^2-3x+\left( \frac{ -3 }{ 2 } \right)^2-\frac{ 9 }{ 4 } \right)\] \[y^2=-\left( x-\frac{ 3 }{ 2 } \right)^2+\frac{ 9 }{ 4 }\] \[y^2+\left( x-\frac{ 3 }{ 2 } \right)^2=\left( \frac{ 3 }{ 2 } \right)^2\] it is a circle with center (3/2,0) and radius 3/2 |dw:1444436581363:dw| area\[=2\int\limits_{0}^{3}\sqrt{x \left( 3-x \right)}dx\] \[area=2\int\limits_{0}^{3}\sqrt{3x-x^2}dx\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question