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Answer:
- A = (0, 1)
- B = (3, -2)
- area = 4.5 square units
Step-by-step explanation:
Rewriting the equations to make x the subject, we have ...
x = y² -1 . . . . . [eq1]
x = 1 - y . . . . . .[eq2]
At the points of intersection, the difference will be zero.
y² -1 -(1 -y) = 0
y² +y -2 = 0
(y -1)(y +2) = 0
The y-coordinates of points A and B are 1 and -2.
The corresponding x-coordinates are ...
x = 1 -{1, -2} = {1 -1, 1+2} = {0, 3}
Then A = (0, 1) and B = (3, -2).
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A differential of area can be written ...
(x2 -x1)dy = ((1 -y) -(y² -1))dy = (2 -y -y²)dy
Integrating this over the interval y = [-2, 1] gives the area.
[tex]\displaystyle A=\int_{-2}^1(2-y-y^2)\,dy=\left.(2y-\dfrac{1}{2}y^2-\dfrac{1}{3}y^3)\right|_{-2}^1\\\\=\left(2-\dfrac{1}{2}-\dfrac{1}{3}\right)-\left(2(-2)-\dfrac{(-2)^2}{2}-\dfrac{(-2)^3}{3}\right)=\dfrac{7}{6}+4+2-\dfrac{8}{3}\\\\=\boxed{4.5}[/tex]
The area of the shaded region is 4.5 square units.
