Find the area bounded by y² = 4x and y + 2x = 12.
Let us express both functions as x=x(y)x = x(y)x=x(y):
f1(y)=y24,f2(y)=6−y2f_1(y) = \frac{y^2}{4}, f_2(y) = 6 - \frac{y}{2}f1(y)=4y2,f2(y)=6−2y.
The two intersect at y=−6,y=4y = -6, y = 4y=−6,y=4, hence the area between them is:
S=∫−64[(6−y2)−y24]dy=(6y−y24−y312)∣−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}S=∫−64[(6−2y)−4y2]dy=(6y−4y2−12y3)∣−64=3125
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