The area of the region is -2/3 square units
The area bounded by the region
Since the region is bounded on the left by x = y²/2 and on the right by x = 3y - 4 for 2 ≤ y ≤ 4, the area of the region is given by
[tex]A = \int\limits^4_2 {\frac{y^{2}}{2} } \, dy - \int\limits^4_2 {(3y - 4)} \, dy \\[/tex]
Integrating, we have
A = [y³/6]₂⁴ - [3y²/2 - 4y]₂⁴
A = [4³/6 - 2³/6] - [3(4)²/2 - 4(4) - {3(2)²/2 - 4(2)]}]
A = [64/6 - 8/6] - [3(16)/2 - 16 - {3(4)/2 - 8]}]
A = [64/6 - 8/6] - [3(16)/2 - 16 - {3(4)/2 - 8]}]
A = [64/6 - 8/6] - [3(8) - 16 - {3(2) - 8]}]
A = [(64 - 8)/6] - [3(8) - 16 - {3(2) - 8]}]
A = [56/6] - [24 - 16 - {6 - 8}]
A = [56/6] - [8 - {-2}]
A = [56/6] - [8 + 2]
A = [56/6] - [10]
A = 56/6 - 10
Taking the L.C.M, we have
A = (56 - 10 × 6)/6
A = (56 - 60)/6
A = -4/6
A = -2/3 square units
So, the area of the region is -2/3 square units
Find the graph in the attachment
Learn more about area bounded by region here:
https://brainly.com/question/14510627
