Answer
$\dfrac{16}{3}$
Work Step by Step
We are given that $y=\sqrt x$
Here, the area $(A)$ can be expressed as: $A=\int_0^{4} \sqrt x \ dx$
In order to solve the above integral, we will use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Now, we have $A=\int_0^{4} \sqrt x \ dx\\ =\int_0^{4} x^{1/2} \ dx \\=[\dfrac{x^{3/2}}{3/2}]_0^4$
or, $=[\dfrac{(4)^{3/2}}{3/2}]-[\dfrac{(0)^{3/2}}{3/2}]$
or, $=\dfrac{16}{3}$
Therefore, the required area is: $Area=\dfrac{16}{3}$
