Question
Find the area enclosed between the curves
y=(x−4)2 and y=16−x2over their intersection interval.Solution
To find the area enclosed between the curvesy=(x−4)2 and y=16−x2, we first need to find the points of intersection of these two curves. We set the y values equal to each other: (x−4)2 = 16−x2 x2 +16 – 8x = 16 – x2 2x2 – 8x = 0 2x(x-4) = 0 The solutions are x=0 and x=4. These are the limits of our integration interval. Now we need to determine which function is greater in the interval [0,4]. Let's test a point within the interval, say x=2: Fory = (x−4)2, when x=2, y=(2−4)2 = 4 Fory = 16−x2, when x=2, y=16−(2)2 = 12 Since 12>4 at x=2, the curvey = 16−x2is abovey = (x−4)2 in the interval [0,4]. The area enclosed between the curves is given by the integral of the difference between the upper and lower functions over the intersection interval: 
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