question the shaded region shown below is bounded by the functions f(x)=-3x² + 10 and g(x)=-1.25x + 8 and…

Explanation:

Step1: Find intersection point

Set $-3x^{2}+10=-1.25x + 8$. Rearrange to $3x^{2}-1.25x - 2=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 3$, $b=-1.25$, $c=-2$, we get the positive - valued intersection point $x$ (since we are dealing with the region in the first - quadrant).

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 $A=\int_{0}^{x_0}((-3x^{2}+10)-(-1.25x + 8))dx=\int_{0}^{x_0}(-3x^{2}+1.25x + 2)dx$.

Step3: Integrate the function

$\int(-3x^{2}+1.25x + 2)dx=-x^{3}+\frac{1.25}{2}x^{2}+2x+C$.

Step4: Evaluate the definite integral

$A=\left[-x^{3}+\frac{1.25}{2}x^{2}+2x\right]_0^{x_0}$. Calculate $x_0$ from the quadratic formula: $x=\frac{1.25\pm\sqrt{(-1.25)^{2}-4\times3\times(-2)}}{2\times3}=\frac{1.25\pm\sqrt{1.5625 + 24}}{6}=\frac{1.25\pm\sqrt{25.5625}}{6}=\frac{1.25\pm5.056}{6}$. The positive root is $x_0=\frac{1.25 + 5.056}{6}\approx1.051$. Then $A=- (1.051)^{3}+\frac{1.25}{2}(1.051)^{2}+2\times1.051$. $A=-1.16 + 0.69 + 2.102=1.632$.

Answer:

$1.632$