The given problem asks for the area of the shaded region bounded by the functions $f(x) = -x^2 + 4$, $g(x) = -0.75x + 3$, the $y$-axis, and the vertical line $x = 2$. I'll solve this step by step by calculating the definite integral of the difference between the two functions over the given interval.

Step 1: Set up the limits of integration

The area is between $x = 0$ (the $y$-axis) and $x = 2$ (the given vertical line). The area between two curves is found by integrating the difference of the top function and the bottom function.

• The top function is $f(x) = -x^2 + 4$.
• The bottom function is $g(x) = -0.75x + 3$.

Step 2: Write the area formula

The formula for the area is:

$\text{Area} = \int_0^2 \left[(-x^2 + 4) - (-0.75x + 3)\right] \, dx$

Simplifying the expression inside the integral:

$\int_0^2 \left(-x^2 + 4 + 0.75x - 3\right) \, dx$

$= \int_0^2 \left(-x^2 + 0.75x + 1\right) \, dx$

Step 3: Calculate the integral

We'll integrate each term separately:

$\int_0^2 -x^2 \, dx + \int_0^2 0.75x \, dx + \int_0^2 1 \, dx$

1. Integral of $-x^2$: $\int_0^2 -x^2 \, dx = \left[-\frac{x^3}{3}\right]_0^2 = -\frac{8}{3}$

2. Integral of $0.75x$: $\int_0^2 0.75x \, dx = \left[0.75 \times \frac{x^2}{2}\right]_0^2 = 0.75 \times 2 = 1.5$

3. Integral of $1$: $\int_0^2 1 \, dx = \left[x\right]_0^2 = 2$

Step 4: Add the results

$-\frac{8}{3} + 1.5 + 2 = \frac{-8}{3} + \frac{3.5 \times 3}{3} = \frac{-8 + 10.5}{3} = \frac{2.5}{3}$

Let me double-check calculations.