Question #183871

Find the area bounded by y² = 4x and y + 2x = 12.

Expert's answer

Let us express both functions as "x = x(y)":

"f_1(y) = \\frac{y^2}{4}, f_2(y) = 6 - \\frac{y}{2}".

The two intersect at "y = -6, y = 4", hence the area between them is:

"S = \\int_{-6}^4 [(6-\\frac{y}{2}) - \\frac{y^2}{4}] dy = (6 y - \\frac{y^2}{4} - \\frac{y^3}{12})|_{-6}^4 = \\frac{125}{3}"

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