the shaded region shown below is bounded by the functions f(x)=-2x² + 3.75x + 7 and g(x)=0.75x + 4 and the x…

Explanation:

Step1: Find intersection point

Set $f(x)=g(x)$, so $-2x^{2}+3.75x + 7=0.75x + 4$. Rearrange to $-2x^{2}+3x + 3 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a=-2$, $b = 3$, $c = 3$, we get $x=\frac{-3\pm\sqrt{9+24}}{-4}=\frac{-3\pm\sqrt{33}}{-4}$. We take the positive root $x=\frac{3 + \sqrt{33}}{4}\approx2.186$. Also, find the $x$-intercept of $f(x)$ by setting $f(x)=0$, $-2x^{2}+3.75x + 7=0$, and the positive root gives the right - hand bound of the region.

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 $x = a$ is $A=\int_{0}^{a}((-2x^{2}+3.75x + 7)-(0.75x + 4))dx=\int_{0}^{a}(-2x^{2}+3x + 3)dx$.

Step3: Evaluate integral

Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int(-2x^{2}+3x + 3)dx=-\frac{2}{3}x^{3}+\frac{3}{2}x^{2}+3x+C$. Then $A=\left[-\frac{2}{3}x^{3}+\frac{3}{2}x^{2}+3x\right]_{0}^{a}$, where $a=\frac{3+\sqrt{33}}{4}$. Using a calculator to evaluate $\left(-\frac{2}{3}\left(\frac{3+\sqrt{33}}{4}\right)^{3}+\frac{3}{2}\left(\frac{3+\sqrt{33}}{4}\right)^{2}+3\left(\frac{3+\sqrt{33}}{4}\right)\right)-0\approx8.167$.

Answer:

$8.167$