question the shaded region shown below is bounded by the functions f(x)=-4x² - 6x + 10 and g(x)=-1.5x + 7…

Explanation:

Step1: Find intersection point

Set $f(x)=g(x)$, so $-4x^{2}-6x + 10=-1.5x + 7$. Rearrange to $4x^{2}+4.5x - 3 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 4$, $b=4.5$, $c=-3$, we get the positive - valued intersection point $x$ (since we are in the first - quadrant considering the $x$ and $y$ axes bounds).

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x = 0$ to the intersection point $x_0$ is given by $A=\int_{0}^{x_0}((-4x^{2}-6x + 10)-(-1.5x + 7))dx=\int_{0}^{x_0}(-4x^{2}-4.5x + 3)dx$.

Step3: Evaluate integral

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(-4x^{2}-4.5x + 3)dx=-4\times\frac{x^{3}}{3}-4.5\times\frac{x^{2}}{2}+3x+C$. Evaluate the definite integral $[- \frac{4}{3}x^{3}-\frac{9}{4}x^{2}+3x]_0^{x_0}$. Using a calculator to solve the quadratic equation $4x^{2}+4.5x - 3 = 0$ for the positive root $x_0=\frac{-4.5+\sqrt{4.5^{2}-4\times4\times(-3)}}{2\times4}\approx0.549$. Then evaluate the definite integral: [ \begin{align*} &\left(-\frac{4}{3}(0.549)^{3}-\frac{9}{4}(0.549)^{2}+3(0.549)\right)-0\ &\approx-\frac{4}{3}(0.164)-\frac{9}{4}(0.301)+1.647\ &\approx - 0.219-0.677 + 1.647\ &\approx0.751 \end{align*} ]

Answer:

$0.751$