find the area of the region bounded by the graphs of the given equations. y = x^2 - 6x - 55, y = 1 - x^2

Explanation:

Step1: Find intersection points

Set $x^{2}-6x - 55=1 - x^{2}$. $2x^{2}-6x-56 = 0$. $x^{2}-3x - 28=0$. Factor: $(x - 7)(x + 4)=0$. So $x=-4$ and $x = 7$.

Step2: Determine upper - lower functions

Let $f(x)=1 - x^{2}$ and $g(x)=x^{2}-6x - 55$. $f(x)-g(x)=(1 - x^{2})-(x^{2}-6x - 55)=-2x^{2}+6x + 56$.

Step3: Calculate the area using integral

The area $A=\int_{-4}^{7}[(1 - x^{2})-(x^{2}-6x - 55)]dx=\int_{-4}^{7}(-2x^{2}+6x + 56)dx$. Using the integral rules $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$: $\int(-2x^{2}+6x + 56)dx=-2\times\frac{x^{3}}{3}+6\times\frac{x^{2}}{2}+56x+C=-\frac{2}{3}x^{3}+3x^{2}+56x+C$. Evaluate the definite integral: $\left(-\frac{2}{3}(7)^{3}+3(7)^{2}+56(7)\right)-\left(-\frac{2}{3}(-4)^{3}+3(-4)^{2}+56(-4)\right)$. $=\left(-\frac{2}{3}\times343 + 3\times49+392\right)-\left(\frac{2}{3}\times64 + 3\times16-224\right)$. $=\left(-\frac{686}{3}+147 + 392\right)-\left(\frac{128}{3}+48-224\right)$. $=\left(-\frac{686}{3}+539\right)-\left(\frac{128}{3}-176\right)$. $=\left(\frac{-686 + 1617}{3}\right)-\left(\frac{128-528}{3}\right)$. $=\frac{931}{3}-\left(-\frac{400}{3}\right)=\frac{931 + 400}{3}=\frac{1331}{3}$.

Answer:

$\frac{1331}{3}$