Find the horizontal line y=m that divided the region enclosed by the graphs of y=x^2 and y=16 into two regions of equal area.
To find the horizontal line that will divide the region enclosed by the graphs of y=x^2 and y=16 into two regions of equal area, we need to first find the points of intersection of the two graphs.
Setting y=x^2 and y=16 equal to each other, we get:
x^2 = 16
Taking the square root of both sides, we get:
x = ±4
So the points of intersection are (-4, 16) and (4, 16).
Now, we need to find the area of the region enclosed by the two graphs. This can be done by finding the area under the curve y=x^2 between x=-4 and x=4, and subtracting this from the area of the rectangle with height 16 and width 8 (the distance between x=-4 and x=4).
The area under the curve y=x^2 between x=-4 and x=4 is:
∫[4,4] x^2 dx = [1/3x^3] [from -4 to 4]
= 1/3 * (64 - (-64))
= 1/3 * 128
= 128/3
So the area enclosed by the two graphs is:
Area = 16*8 - 128/3
= 128 - 42.666...
= 85.333...
Since we want to divide this region into two equal parts, the area of each part should be 85.333... / 2 = 42.666...
So, with the line y = m dividing the region, the area of the region above the line would be 42.666..., and the area below the line would also be 42.666....
Therefore, the horizontal line that will divide the region enclosed by the graphs of y=x^2 and y=16 into two regions of equal area is y=42.666... (or approximately y=42.67).