The problem asks us to find the value of the function $f(1)$ given the area under the curve $y=f(x)$ from $x=0$ to $x=b$. We are given that the area, let's call it $A(b)$, is defined as:

$A(b) = \int_0^b f(x) dx = \sqrt{3+b^2} - \sqrt{3}$

Here, $f(x)$ is a continuous, non-negative function on $[0, 2]$, and $b \in (0, 2]$.

Relating Area and the Function using Calculus

According to the Fundamental Theorem of Calculus Part 1, the derivative of the definite integral function $A(b) = \int_0^b f(x) dx$ with respect to $b$ gives us the function $f(b)$.

So, we need to find the derivative of $A(b)$ with respect to $b$. $f(b) = \frac{d}{db} A(b) = \frac{d}{db} (\sqrt{3+b^2} - \sqrt{3})$

Calculating the Derivative

We differentiate the expression term by term:

• The derivative of the constant term $\sqrt{3}$ is 0.
• For the term $\sqrt{3+b^2}$, we use the chain rule. Let $u = 3+b^2$, then $\frac{du}{db} = 2b$. The derivative is $\frac{d}{db}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{db}$.
• Substituting back: $\frac{d}{db}(\sqrt{3+b^2}) = \frac{1}{2\sqrt{3+b^2}} \cdot (2b) = \frac{b}{\sqrt{3+b^2}}$.

Combining these, the derivative $f(b)$ is:

$f(b) = \frac{b}{\sqrt{3+b^2}} - 0 = \frac{b}{\sqrt{3+b^2}}$

Finding the Value of f(1)

Now we substitute $b=1$ into the expression for $f(b)$ to find $f(1)$.

$f(1) = \frac{1}{\sqrt{3+1^2}} = \frac{1}{\sqrt{3+1}} = \frac{1}{\sqrt{4}} = \frac{1}{2}$

The value of $f(1)$ is $\frac{1}{2}$, which is equal to 0.5.

Rounding to 1 decimal place, $f(1) = 0.5$.