find the area of the region bounded by the graphs of the given equations. y = x^2 - 2, y = -x - 4, x = -1, x…

Explanation:

Step1: Determine the upper - lower functions

On the interval $[-1,1]$, we need to find which function is on top. Let's find the difference $f(x)=(x^{2}-2)-(-x - 4)=x^{2}+x + 2$. The discriminant of $y = x^{2}+x + 2$ is $\Delta=1^{2}-4\times2=-7<0$ and the coefficient of $x^{2}$ is positive, so $x^{2}-2\geq -x - 4$ on $[-1,1]$.

Step2: Use the area formula

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}[f(x)-g(x)]dx$. Here, $f(x)=x^{2}-2$, $g(x)=-x - 4$, $a=-1$, $b = 1$. So $A=\int_{-1}^{1}[(x^{2}-2)-(-x - 4)]dx=\int_{-1}^{1}(x^{2}+x + 2)dx$.

Step3: Integrate term - by - term

We know that $\int(x^{2}+x + 2)dx=\frac{1}{3}x^{3}+\frac{1}{2}x^{2}+2x+C$. Then $\int_{-1}^{1}(x^{2}+x + 2)dx=\left[\frac{1}{3}x^{3}+\frac{1}{2}x^{2}+2x\right]_{-1}^{1}$.

Step4: Evaluate the definite integral

$\left(\frac{1}{3}(1)^{3}+\frac{1}{2}(1)^{2}+2(1)\right)-\left(\frac{1}{3}(-1)^{3}+\frac{1}{2}(-1)^{2}+2(-1)\right)=\left(\frac{1}{3}+\frac{1}{2}+2\right)-\left(-\frac{1}{3}+\frac{1}{2}-2\right)=\frac{2 + 3+12}{6}-\left(-\frac{2 + 3 - 12}{6}\right)=\frac{17}{6}-\left(-\frac{7}{6}\right)=\frac{17 + 7}{6}=\frac{24}{6}=4$.

Answer:

4