Find the finite area enclosed between the curves y=x^2-5x+6 and y=4-x^2

Starting by factorising the curve equations: the first one factorises to y = (x-3)(x-2) and the second one becomes y=(2-x)(2+x). From this, a rough sketch of the curves can be drawn and it can be seen that for the area in question, y=4-x² is always above the other curve. This will become important for the integration step. The next step is to find the points where the two curves intersect (we only care about the x-coordinate here). Equating the two curves gives x²-5x+6=4-x² which can be rearranged and factorised to get (2x-1)(x-2)=0, so the required coordinates are x = 1/2 and x=2.By looking at the rough sketch, we can see that we want to subtract the area below y=x²-5x+6 from the area below y=4-x² between x = 0.5 and x=2. To do this, we compute integral from 0.5 to 2 of 4-x²-(x²-5x+6) to get the integral from 0.5 to 2 of -2x²+5x-2, which is [-2x³/3 + 5x²/2 - 2x] from 0.5 to 2. Substituting in 0.5 and 2 gives 9/8 (which is 1.125).