anonymous
  • anonymous
Find the area under the curve y = (x – 2)^2 from x = 0 to x = 2
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Before finding 'area' under a curve, you have to check to see if, over the interval of integration, the function cuts through the x-axis and enters into negative y territory. If it does, you have to break the integration up by integrating over x where y(x) is positive, and integrating over x where y(x) is negative and then take the absolute value. Here, luckily, your function does not cut the x-axis, since the minimum value occurs at x=2 and y(2) = (2-2)^2 = 0. So we may 'blindly' integrate as\[\int\limits_{0}^{2}(x-2)^2 dx = \frac{1}{3}(x-2)^3|_0^2=\frac{1}{3}(0-(-8))=\frac{8}{3}\]
anonymous
  • anonymous
I keep getting strange answers. What's the correct answer?
anonymous
  • anonymous
sorry, didn't see the other half of the screen.

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