If $f(x)=2\ln (\sqrt{e^x})$, what is the area bounded by f(x) for the interval [0, 2] on the x-axis?

The correct answer is 2.

To find the area bounded by the function $f(x)$ and the $x$-axis over the interval $[0,2]$, we first simplify the expression for the function.

The given function is:
$$f(x)=2\ln (\sqrt{e^x})$$
We can rewrite the square root as an exponent of $\frac{1}{2}$:
$$\sqrt{e^x}=(e^x{)}^{1/2}=e^{x/2}$$
Substituting this back into the function, we get:
$$f(x)=2\ln (e^{x/2})$$
Since the natural logarithm $\ln (y)$ and the exponential function $e^y$ are inverse functions, $\ln (e^y)=y$. Applying this property:
$$f(x)=2\cdot \left(\frac{x}{2}\right)=x$$

So, the function simplifies to the simple identity function $f(x)=x$.

To find the area under this curve from $x=0$ to $x=2$, we calculate the definite integral:
$$\text{Area}={\int }_0^{2}xdx$$
Integrating $x$ gives:
$$\text{Area}={\left[\frac{x^2}{2}\right]}_0^2$$
Now, we evaluate this expression at the upper and lower limits:
$$\text{Area}=\left(\frac{2^2}{2}\right)-\left(\frac{0^2}{2}\right)$$
$$\text{Area}=\frac{4}{2}-0=2$$

Alternatively, since $f(x)=x$ is a straight line, the area under the curve on the interval $[0,2]$ forms a right-angled triangle with a base of length 2 (from $x=0$ to $x=2$) and a height of 2 (since $f(2)=2$). Using the triangle area formula:
$$\text{Area}=\frac{1}{2}\cdot \text{base}\cdot \text{height}=\frac{1}{2}\cdot 2\cdot 2=2$$