the shaded region shown below is bounded by the functions f(x)=-x² + 4 and g(x)=-0.75x + 3, the y - axis and…

Explanation:

Step1: Recall area - between - curves formula

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, $f(x)=-x^{2}+4$, $g(x)=- 0.75x + 3$, $a = 0$, and $b = 2$. Since $f(x)\geq g(x)$ on the interval $[0,2]$, the area formula becomes $A=\int_{0}^{2}[(-x^{2}+4)-(-0.75x + 3)]dx$.

Step2: Simplify the integrand

[ \begin{align*} (-x^{2}+4)-(-0.75x + 3)&=-x^{2}+4 + 0.75x-3\ &=-x^{2}+0.75x + 1 \end{align*} ]

Step3: Integrate the simplified function

We know that $\int(-x^{2}+0.75x + 1)dx=-\frac{1}{3}x^{3}+\frac{0.75}{2}x^{2}+x+C$. Evaluating the definite - integral $\int_{0}^{2}(-x^{2}+0.75x + 1)dx=\left[-\frac{1}{3}x^{3}+\frac{3}{8}x^{2}+x\right]_{0}^{2}$.

Step4: Evaluate the definite - integral

[ \begin{align*} \left(-\frac{1}{3}(2)^{3}+\frac{3}{8}(2)^{2}+2\right)-\left(-\frac{1}{3}(0)^{3}+\frac{3}{8}(0)^{2}+0\right)&=-\frac{8}{3}+\frac{3}{8}\times4 + 2-0\ &=-\frac{8}{3}+\frac{3}{2}+2\ &=-\frac{16}{6}+\frac{9}{6}+\frac{12}{6}\ &=\frac{-16 + 9+12}{6}\ &=\frac{5}{6}\approx0.833 \end{align*} ]

Answer:

$0.833$