Posted by Anonymous on Thursday, October 28, 2010 at 11:05am.
Find the area under the curve below from x = 0 to x = 2. Give your answer correct to 3 decimal places.
y = 2x  x2

calc  jai, Thursday, October 28, 2010 at 11:15am
is 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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