If you want the yellow area, why not just use the integral calculus? Or do you want to get the yellow from first principles using a Riemann sum?I'm not sure if this is below uni level, but I've been having a bit of trouble with it. What I'm trying to do is calculate the area surrounded by two curves, p = ½x² - 1½x - 1, and s = ½x + 1. I calculate the intercept points of the two graphs to be (2 - √8, 2 - √2) and (2 + √8, 2 + √2), distance between them being 5.6568, although I'm open to any corrections. The y difference between curves must surely be ½x² -2x -2, and for each value of X (being infinitely small), the area is (½x² -2x -2)(5.6568/n), where n is the number of rectangles, as described by this site:
The area under a curve
I don't really understand the line "In this case, the height of the rectangles are given by xi² and so the area of a rectangle is xi/n = i²/n³ We then find that the sum of the areas of all the rectangles is..."
I'm not sure how to incorporate my expression into this method, since the area calculation will still be left with x's in the expression even though terms where 1/n ≈ 0 are removed. I don't see that changing the length of the area from 1 to 5.6568 should affect the method, but I could be wrong. Describing myself as a novice in these areas would be an overstatement.
Can someone please show me the method?