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 $h(x)=f(x)-g(x)=(-x^{2}+21x - 44)-(5x + 19)=-x^{2}+16x - 63$.

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, on the interval $[7,9]$, $f(x)-g(x)\geq0$. So $A=\int_{7}^{9}(-x^{2}+16x - 63)dx$.

Step3: Integrate term - by - term

We know that $\int(-x^{2}+16x - 63)dx=-\frac{1}{3}x^{3}+8x^{2}-63x + C$.

Step4: Evaluate the definite integral

$A=\left(-\frac{1}{3}x^{3}+8x^{2}-63x\right)\big|_{7}^{9}$ $=\left(-\frac{1}{3}(9)^{3}+8(9)^{2}-63(9)\right)-\left(-\frac{1}{3}(7)^{3}+8(7)^{2}-63(7)\right)$ $=\left(-243 + 648-567\right)-\left(-\frac{343}{3}+392 - 441\right)$ $=(-162)-\left(-\frac{343}{3}-49\right)$ $=-162+\frac{343}{3}+49$ $=-113+\frac{343}{3}$ $=\frac{-339 + 343}{3}=\frac{4}{3}$.

Answer:

$\frac{4}{3}$