the shaded region shown below is bounded by the functions f(x)=-4x² + 9 and g(x)=3x + 4, the y - axis and…

Explanation:

Step1: Set up the integral for area

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, $f(x)=- 4x^{2}+9$, $g(x)=3x + 4$, $a = 0$, and $b=1.5$. Since $-4x^{2}+9\geq3x + 4$ on the interval $[0,1.5]$, the area formula becomes $A=\int_{0}^{1.5}((-4x^{2}+9)-(3x + 4))dx=\int_{0}^{1.5}(-4x^{2}-3x + 5)dx$.

Step2: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have: $\int(-4x^{2}-3x + 5)dx=-4\times\frac{x^{3}}{3}-3\times\frac{x^{2}}{2}+5x+C=-\frac{4x^{3}}{3}-\frac{3x^{2}}{2}+5x+C$.

Step3: Evaluate the definite integral

$A=\left[-\frac{4x^{3}}{3}-\frac{3x^{2}}{2}+5x\right]{0}^{1.5}$. First, substitute $x = 1.5$: $-\frac{4\times(1.5)^{3}}{3}-\frac{3\times(1.5)^{2}}{2}+5\times1.5=-\frac{4\times3.375}{3}-\frac{3\times2.25}{2}+7.5$. $=-\frac{13.5}{3}-\frac{6.75}{2}+7.5=-4.5 - 3.375+7.5$. Then substitute $x = 0$ (which gives $0$). $A=-4.5-3.375 + 7.5= - 7.875+7.5=- 0.375$. But since area is non - negative, we made a mistake above. We should have $A=\int{0}^{1.5}((-4x^{2}+9)-(3x + 4))dx=\int_{0}^{1.5}(-4x^{2}-3x + 5)dx$. $A=\left[-\frac{4x^{3}}{3}-\frac{3x^{2}}{2}+5x\right]_{0}^{1.5}=-\frac{4\times(1.5)^{3}}{3}-\frac{3\times(1.5)^{2}}{2}+5\times1.5$ $=-\frac{4\times3.375}{3}-\frac{3\times2.25}{2}+7.5=-4.5-3.375 + 7.5 = 0.375$.

Answer:

$0.375$