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April 24, 2014

April 24, 2014

Posted by **Candice** on Tuesday, November 1, 2011 at 4:24am.

(I've worked out the x intercepts to be 0 and 1 in the first quadrant, but how would I find the area?)

thanks :)

- Math- integration -
**Reiny**, Tuesday, November 1, 2011 at 8:43amIs your first function y = x^(2/3) ?

then your intersection is indeed at x-0 and x=1

in that domain x = y^2 or y = x^(1/2) is above y = x^(2/3)

so the effective height is x^(1/2) - x^(2/3)

area = ∫( x^(1/2) - x^(2/3) ) dx

= (2/3)x^(3/2) - (3/5)x^(5/3) | from 0 to 1

= (2/3 - 3/5 - 0)

= 1/15

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