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

Explanation:

Step1: Set up the integral

The area $A$ between two functions $g(x)$ and $f(x)$ over the interval $[a,b]$ is given by $A=\int_{a}^{b}[g(x)-f(x)]dx$. Here, $a = - 10$, $b=-7$, $g(x)=-5x + 9$ and $f(x)=x^{2}+12x + 79$. So, $A=\int_{-10}^{-7}[(-5x + 9)-(x^{2}+12x + 79)]dx=\int_{-10}^{-7}(-x^{2}-17x - 70)dx$.

Step2: Integrate term - by - term

Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int(-x^{2}-17x - 70)dx=-\frac{x^{3}}{3}-\frac{17x^{2}}{2}-70x+C$.

Step3: Evaluate the definite integral

[ \begin{align*} &\left(-\frac{(-7)^{3}}{3}-\frac{17(-7)^{2}}{2}-70(-7)\right)-\left(-\frac{(-10)^{3}}{3}-\frac{17(-10)^{2}}{2}-70(-10)\right)\ =&\left(\frac{343}{3}-\frac{833}{2}+490\right)-\left(\frac{1000}{3}-850 + 700\right)\ =&\frac{343}{3}-\frac{833}{2}+490-\frac{1000}{3}+850 - 700\ =&\left(\frac{343}{3}-\frac{1000}{3}\right)-\frac{833}{2}+(490 + 850-700)\ =&-\frac{657}{3}-\frac{833}{2}+640\ =&- 219-\frac{833}{2}+640\ =&\frac{-438 - 833+1280}{2}\ =&\frac{-1271 + 1280}{2}\ =&\frac{9}{2} \end{align*} ]

Answer:

$\frac{9}{2}$