Answer:
- area: 32/3
- see attached for a graph
Step-by-step explanation:
You want the area between the curves x = y² and x +2y = 3.
Integral
The area is found by integrating a differential of area between the two curves. In this case, that can be written ...
dA = ((3 -2y) -(y²))dy
The graph shows us this is non-negative between y = -3 and y = 1, so our area is ...
[tex]\displaystyle\int_{-3}^1{(3-2y-y^2)}\,dy=\left[3y-y^2-\dfrac{y^3}{3}\right]_{-3}^1\\\\\\=3(1)-(1)^2-\dfrac{(1)^3}{3}-\left(3(-3)-(-3)^2-\dfrac{(-3)^3}{3}\right)\\\\\\=3-1-\dfrac{1}{3}+9+9-9=10\dfrac{2}{3}=\boxed{\dfrac{32}{3}}[/tex]
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Additional comment
The attached graph shows the two curves and the area of interest in blue. The difference between the two curves is shown in red. The shaded areas are the same.
The red graph clearly shows the area of interest is the area under a parabola bounded by a 4×4 square. It is fairly easy to show that the area under a parabola is 2/3 of the area of the bounding rectangle. Here, that means the area is (2/3)(4²) = 32/3.
