Question

Find the area under the curve y = (x – 2)^2 from x = 0 to x = 2

Answer 1

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\[\int\limits_{0}^{2}(x-2)^2 dx = \frac{1}{3}(x-2)^3|_0^2=\frac{1}{3}(0-(-8))=\frac{8}{3}\]

Answer 2

I keep getting strange answers. What's the correct answer?

Answer 3

sorry, didn't see the other half of the screen.