2. find the area bounded by the graphs of the indicated equations over the given interval. compute answers…

Explanation:

Step1: Determine the sign of the function

The function $y = x^{2}-4=(x - 2)(x + 2)$. In the interval $[0,3]$, when $0\leq x<2$, $y=x^{2}-4<0$; when $2\leq x\leq3$, $y=x^{2}-4\geq0$.

Step2: Split the integral for area calculation

The area $A=\int_{0}^{2}-(x^{2}-4)dx+\int_{2}^{3}(x^{2}-4)dx$.

Step3: Integrate term - by - term

For $\int-(x^{2}-4)dx=-\left(\frac{x^{3}}{3}-4x\right)+C=-\frac{x^{3}}{3}+4x + C$. Evaluating from $0$ to $2$: $\left(-\frac{2^{3}}{3}+4\times2\right)-\left(-\frac{0^{3}}{3}+4\times0\right)=-\frac{8}{3}+8=\frac{- 8 + 24}{3}=\frac{16}{3}$. For $\int(x^{2}-4)dx=\frac{x^{3}}{3}-4x + C$. Evaluating from $2$ to $3$: $\left(\frac{3^{3}}{3}-4\times3\right)-\left(\frac{2^{3}}{3}-4\times2\right)=(9 - 12)-\left(\frac{8}{3}-8\right)=-3-\frac{8}{3}+8=\frac{-9 - 8+24}{3}=\frac{7}{3}$.

Step4: Calculate the total area

$A=\frac{16}{3}+\frac{7}{3}=\frac{16 + 7}{3}=\frac{23}{3}\approx7.667$.

Answer:

$7.667$