find the area of the regions enclosed by the line y = 2 and the curve y = x² - 7. \n\na. ∫( )dy from 0 to…

Answer:

First, find the intersection - points of the line $y = 2$ and the curve $y=x^{2}-7$. Set $x^{2}-7 = 2$, then $x^{2}=9$, so $x=-3$ and $x = 3$. The area $A$ between two curves $y = f(x)$ and $y = g(x)$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, $f(x)=2$ and $g(x)=x^{2}-7$, and $a=-3$, $b = 3$. We can split the integral based on the symmetry of the parabola $y=x^{2}-7$ about the $y -$axis. The area $A = 2\int_{0}^{3}[(2-(x^{2}-7))]dx=2\int_{0}^{3}(9 - x^{2})dx$. We can also write the area as $\int_{-3}^{3}(2-(x^{2}-7))dx=\int_{-3}^{3}(9 - x^{2})dx$. The correct form of the integral for the area in terms of $x$ is $\int_{-3}^{3}(9 - x^{2})dx$, which can be split into $\int_{-3}^{0}(9 - x^{2})dx+\int_{0}^{3}(9 - x^{2})dx$. So the answer is C. $\int_{0}^{3}(9 - x^{2})dx+\int_{- 3}^{0}(9 - x^{2})dx$

Explanation:

Step1: Find intersection points

Set $x^{2}-7 = 2$, solve for $x$. $x^{2}=9$, so $x=\pm3$.

Step2: Determine the integrand

The upper - curve is $y = 2$ and the lower - curve is $y=x^{2}-7$. The integrand is $2-(x^{2}-7)=9 - x^{2}$.

Step3: Set up the integral

Due to symmetry about the $y -$axis, the area $A=\int_{-3}^{3}(9 - x^{2})dx=\int_{-3}^{0}(9 - x^{2})dx+\int_{0}^{3}(9 - x^{2})dx$.