Posted by **Mishaka** on Wednesday, February 29, 2012 at 3:44pm.

Find the area of the region bounded by the curves y=x^(-1/2), y=x^(-2), y=1 and y=3.

You get:

a.) 1/2(sqrt(3)) + 4/3

b.) 2(sqrt(3)) - 8/3

c.) 1/2(sqrt(3) - 32/3

d.) 2(sqrt(3)) - 32/3

e.) 8/3 - 2(sqrt(3))

- Calculus (Area Between Curves) -
**Nade**, Wednesday, February 29, 2012 at 4:09pm
The definite integral of y^-.5-y^-2 evaluated from 1 to 3. I forgot 2sqrt(3)-8/3 choice b

- Calculus (Area Between Curves) -
**Mishaka**, Wednesday, February 29, 2012 at 4:11pm
I got the same thing. Working out the other answers, they were either negative or obviously too large of an area for the given bounds, thank you Nade!

## Answer this Question

## Related Questions

Calculus - Find the centroid of the area bounded by the curves: sqrt(x)+sqrt(y)=...

Calculus-Area between curves - Sketch the region enclosed by the given curves. ...

Calculus - Find the volume of the solid whose base is the region in the xy-plane...

Math Help please!! - Could someone show me how to solve these problems step by ...

Calculus - Please look at my work below: Solve the initial-value problem. y'' + ...

Math - How do you find a square root of a number that's not a perfect square? I'...

Mathematics - sqrt 6 * sqrt 8 also sqrt 7 * sqrt 5 6.92820323 and 5.916079783 So...

Math(Roots) - sqrt(24) *I don't really get this stuff.Can somebody please help ...

math calculus please help! - l = lim as x approaches 0 of x/(the square root of...

Math/Calculus - Solve the initial-value problem. Am I using the wrong value for ...