g. use the substitution ( u = 5x + 4 ) to evaluate the integral ( int \frac{10x + 11}{sqrt{5x + 4}} dx…

Explanation:

Step1: Find the intersection points

Set (-(x^{2}-4)=-2x + 4). [ \begin{align*} -x^{2}+4&=-2x + 4\ -x^{2}+2x&=0\ x(-x + 2)&=0 \end{align*} ] So (x = 0) or (x=2).

Step2: Set up the integral for the area

The area (A=\int_{a}^{b}\left[\text{top - bottom}\right]dx). Here, the top function is (y=-(x^{2}-4)=-x^{2}+4) and the bottom function is (y=-2x + 4). So (A=\int_{0}^{2}\left[(-x^{2}+4)-(-2x + 4)\right]dx=\int_{0}^{2}(-x^{2}+2x)dx)

Step3: Integrate the function

[ \begin{align*} \int_{0}^{2}(-x^{2}+2x)dx&=\left[-\frac{x^{3}}{3}+x^{2}\right]_{0}^{2}\ &=\left(-\frac{2^{3}}{3}+2^{2}\right)-\left(-\frac{0^{3}}{3}+0^{2}\right)\ &=\left(-\frac{8}{3}+4\right)-0\ &=\frac{- 8 + 12}{3}\ &=\frac{4}{3} \end{align*} ]

Answer:

The area enclosed by the two graphs is (\frac{4}{3}) square units.