Question #183871

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


1
Expert's answer
2021-04-23T08:25:46-0400


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

f1(y)=y24,f2(y)=6y2f_1(y) = \frac{y^2}{4}, f_2(y) = 6 - \frac{y}{2}.

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

S=64[(6y2)y24]dy=(6yy24y312)64=1253S = \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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