question the shaded region shown below is bounded by the functions f(x)=-x² - 1.25x + 7 and g(x)=x + 4 and…

Explanation:

Step1: Find intersection point

Set $f(x)=g(x)$, so $-x^{2}-1.25x + 7=x + 4$. Rearrange to $x^{2}+2.25x - 3=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=2.25$, $c=-3$, we get the positive - root (since we are in the first - quadrant considering the $x$ and $y$ axes) $x=\frac{-2.25+\sqrt{(2.25)^{2}-4\times1\times(-3)}}{2\times1}\approx1.12$.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x = 0$ to $x = 1.12$ is given by $A=\int_{0}^{1.12}((-x^{2}-1.25x + 7)-(x + 4))dx=\int_{0}^{1.12}(-x^{2}-2.25x + 3)dx$.

Step3: Integrate

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(-x^{2}-2.25x + 3)dx=-\frac{x^{3}}{3}-2.25\times\frac{x^{2}}{2}+3x+C$.

Step4: Evaluate definite integral

$A=\left[-\frac{x^{3}}{3}-1.125x^{2}+3x\right]_{0}^{1.12}=-\frac{(1.12)^{3}}{3}-1.125\times(1.12)^{2}+3\times1.12-0$. Using a calculator: $-\frac{(1.12)^{3}}{3}-1.125\times(1.12)^{2}+3\times1.12\approx-\frac{1.404928}{3}-1.125\times1.2544 + 3.36$. $= - 0.468309-1.4101 + 3.36\approx1.482$.

Answer:

$1.482$