Set up the definite integral required to find the area of the region between the graph of y = 16

Question

Ma. Theresa Alin · Accepted Answer

Step 1: Find the points of intersection between the two functions y=16-x^(2) and y=2x-47 by sett

Set up the definite integral required to find the area of the region between the graph of $y = 16 - x^2$ and $y = 2x - 47$ over the interval $-3 \le x \le 6$. \begin{equation*} \int_{-3}^{6} \boxed{16 - x^2 - (2x - 47)} dx \end{equation*}