the shaded region shown below is bounded by the functions f(x)=-4x² + 9 and g(x)=2.75x + 2, the y - axis and…

Explanation:

Step1: Determine the upper - lower functions

On the interval $[0,1.5]$, we need to find which function is above the other. Let's find the difference $h(x)=f(x)-g(x)=-4x^{2}-2.75x + 7$. We can also test a value in the interval, say $x = 0$. $f(0)=9$ and $g(0)=2$, so $f(x)\geq g(x)$ on $[0,1.5]$. 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, $a = 0$, $b = 1.5$, $f(x)=-4x^{2}+9$ and $g(x)=2.75x + 2$.

Step2: Set up the integral

The integral for the area is $A=\int_{0}^{1.5}((-4x^{2}+9)-(2.75x + 2))dx=\int_{0}^{1.5}(-4x^{2}-2.75x + 7)dx$.

Step3: Integrate term - by - term

Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have: $\int(-4x^{2}-2.75x + 7)dx=-4\times\frac{x^{3}}{3}-2.75\times\frac{x^{2}}{2}+7x+C=-\frac{4}{3}x^{3}-\frac{2.75}{2}x^{2}+7x+C$.

Step4: Evaluate the definite integral

$A=\left[-\frac{4}{3}x^{3}-\frac{2.75}{2}x^{2}+7x\right]_{0}^{1.5}$. First, substitute $x = 1.5$: $-\frac{4}{3}(1.5)^{3}-\frac{2.75}{2}(1.5)^{2}+7(1.5)=-\frac{4}{3}\times3.375-\frac{2.75}{2}\times2.25 + 10.5$. $=-4.5-3.09375 + 10.5$. Then substitute $x = 0$ (which gives $0$). $A=-4.5-3.09375 + 10.5=2.90625\approx2.906$.

Answer:

$2.906$