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

Explanation:

Step1: Find the difference of functions

We need to find $g(x)-f(x)$. So, $(-5x + 7)-(x^{2}-5x + 3)=-x^{2}+4$.

Step2: Use definite - integral for area

The area $A$ between two functions over the interval $[a,b]$ is given by $A=\int_{a}^{b}[g(x)-f(x)]dx$. Here, $a = - 2$, $b = 2$, and $g(x)-f(x)=-x^{2}+4$. So, $A=\int_{-2}^{2}(-x^{2}+4)dx$.

Step3: Integrate term - by - term

We know that $\int(-x^{2}+4)dx=-\frac{1}{3}x^{3}+4x + C$.

Step4: Evaluate the definite integral

$A=\left[-\frac{1}{3}x^{3}+4x\right]_{-2}^{2}=\left(-\frac{1}{3}(2)^{3}+4(2)\right)-\left(-\frac{1}{3}(-2)^{3}+4(-2)\right)$. First, calculate $-\frac{1}{3}(2)^{3}+4(2)=-\frac{8}{3}+8=\frac{-8 + 24}{3}=\frac{16}{3}$. Second, calculate $-\frac{1}{3}(-2)^{3}+4(-2)=\frac{8}{3}-8=\frac{8 - 24}{3}=-\frac{16}{3}$. Then $A=\frac{16}{3}-\left(-\frac{16}{3}\right)=\frac{16}{3}+\frac{16}{3}=\frac{32}{3}$.

Answer:

$\frac{32}{3}$