current objective find the area of a region bounded between a linear function and another function switch…

Explanation:

Step1: Determine the difference function

We find $h(x)=f(x)-g(x)=(-x^{2}+11x - 3)-(5x + 5)=-x^{2}+6x - 8$.

Step2: Use the definite - integral formula for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ over the interval $[a,b]$ is $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, $a = 2$, $b = 4$, and we integrate $h(x)=-x^{2}+6x - 8$. So $A=\int_{2}^{4}(-x^{2}+6x - 8)dx$.

Step3: Integrate term - by - term

$\int(-x^{2}+6x - 8)dx=-\frac{1}{3}x^{3}+3x^{2}-8x+C$.

Step4: Evaluate the definite integral

$A=\left[-\frac{1}{3}x^{3}+3x^{2}-8x\right]_{2}^{4}=\left(-\frac{1}{3}(4)^{3}+3(4)^{2}-8(4)\right)-\left(-\frac{1}{3}(2)^{3}+3(2)^{2}-8(2)\right)$. $= \left(-\frac{64}{3}+48 - 32\right)-\left(-\frac{8}{3}+12 - 16\right)$. $=\left(-\frac{64}{3}+16\right)-\left(-\frac{8}{3}-4\right)$. $=\left(-\frac{64}{3}+\frac{48}{3}\right)-\left(-\frac{8}{3}-\frac{12}{3}\right)$. $=-\frac{16}{3}-\left(-\frac{20}{3}\right)$. $=-\frac{16}{3}+\frac{20}{3}=\frac{4}{3}$.

Answer:

$\frac{4}{3}$