Answer
$\dfrac{1}{2}$
Work Step by Step
The area $(A)$ can be calculated as:
$A=\int_{0}^{1} [4-(4-x^2)] \ dx +\int_1^{4/3} [4-3x] \ dx \\= \int_{0}^{1} (-x^2) \ dx +\int_1^{4/3} [4-3x] \ dx \\=\dfrac{1}{3} (x^3)_0^1 +[4x-\dfrac{3x^2}{2}]_1^{4/3} \\=\dfrac{1}{3}+\dfrac{8}{3}-\dfrac{5}{2} \\=\dfrac{1}{2}$
