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September 2, 2014

September 2, 2014

Posted by **Anonymous** on Thursday, October 28, 2010 at 11:05am.

y = 2x - x2

- calc -
**jai**, Thursday, October 28, 2010 at 11:15amis y = 2x - x^2 the function given? if it is,

area under the curve means the integral at these particular bounds,, first you have to recall how to integrate:

for ax^n (a is constant), add 1 to the power, then this n+1 must be divided:

integral (ax^n) = (a)[x^(n+1)]/(n+1)

thus for 2x - x^2, we do it separately (by term):

integral 2x = 2x^(1+1)/(1+1) = x^2

integral x^2 = x^(2+1)/(2+1) = (x^3)/3

then evaluate x^2 - (x^3)/3 from 0 to 2.

hope this helps. :)

- calc -
**jai**, Thursday, October 28, 2010 at 11:18am**i'll just retype it to make the integration part clearer:

integral 2x = [2x^(1+1)]/(1+1) = (2x^2)/2 = x^2

integral x^2 = [x^(2+1)]/(2+1) = (x^3)/3

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