2y - y^2 - (y^2 - 4y) = 6y - 2y^2
6y - 2y^2 = 0
2y(3 - y) = 0
y = 0, 3
Using I[a,b] for the integral from 0 to 3 and E[a,b] for the evaluation from a to b
I[0, 3] 6y - 2y^2 dy = 3y^2 - 2/3y^3 E[0, 3] = 3*3^2 - 2/3 * 3^3 = 27 - 2/3*27 = 27 - 18 = 9
Kyle P.
asked 05/08/20Find the area bounded by the curves x = y2 - 4y and x = 2y - y2. Your work must include an integral in one variable
2y - y^2 - (y^2 - 4y) = 6y - 2y^2
6y - 2y^2 = 0
2y(3 - y) = 0
y = 0, 3
Using I[a,b] for the integral from 0 to 3 and E[a,b] for the evaluation from a to b
I[0, 3] 6y - 2y^2 dy = 3y^2 - 2/3y^3 E[0, 3] = 3*3^2 - 2/3 * 3^3 = 27 - 2/3*27 = 27 - 18 = 9
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