In this challenging area integral problem, we are given the region bounded by y=1-x^2 and the x axis. The goal of the problem is to calculate c so the line y=c cuts this area in half, and we cut the parabola in half using calculus.
We begin by computing the total area under the parabola y=1-x^2, and this is just a standard definite integral on [-1,1] of 1-x^2. We take advantage of symmetry and integrate from 0 to 1, then double the result to get the area under the parabola as 4/3.
Next, we want to compute the area between y=1-x^2 and y=c. We start by computing the intersections of y=1-x^2 and y=c, because these will be our limits of integration for the area integral. Now we set up the area integral as the definite integral of 1-x^2-c on 0 to sqrt(1-c), just using the same trick of cutting the integration interval in half and doubling the result. This gives us a value of the area of the upper half in terms of c.
Finally, we set the area of the upper half equal to half the total area of the parabola, or 2/3, then solve for c.
After simplifying, we find c so the area under y=1-x^2 is cut in half by a horizontal line y=c, and it turns out to be 1-2^(1/3)/2 or about 0.37. This makes sense, because we know the horizontal line cuts the parabola below y=0.5.
Негізгі бет Cut the parabola in half using calculus: y=1-x^2 cut in half by a horizontal line y=c.
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