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Okay, I figured it out, but I need someone to help explain to me how I got to where I did.
The answer I got for the approximation of the area of the vertical slice with respect to x is [ ( 12 - 6x ) - ( x^2 - 4 ) ] delta(x). Originally I kept putting a positive sign between the functions but then it occurred to me that Integral ( x^2 -4 ) dx is negative over the range of integration. My line of reasoning is that what the question was asking was for the area, which must be positive, thus I needed the absolute value of the integral, which means I needed to subtract the negative to make it positive. Is that right, or am I just a blind squirrel that managed to find a nut?
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When integrating, you already know that you are finding the area under the curve. Area ABOVE the x-axis is positive, and BELOW the x-axis is negative.
What you end up finding is the NET area, with an integral. You're still adding the two areas together, except one is negative, which of course leads to subtraction.
Area above x + (- Area below x)
Area under the curve is a good visual representation of integration, but the term, "area," comes with a limitation that integration doesn't have. By common sense, "area," can't be negative, right? An integral can.
Try integrating x^2 - 6 from 0 to 1, just to convince yourself that integrals can be negative, as well
You're on the right path. Area above, positive, area below, negative.