what is the area of the region between the graphs of f(x) = √x + 1 and g(x) = 2x - 4 from x = 0 to x = 3…

Explanation:

Step1: Recall area - between - curves formula

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. First, we need to determine which function is greater on the interval $[0,3]$. Let's find the difference $h(x)=f(x)-g(x)=\sqrt{x + 1}-(2x - 4)=\sqrt{x + 1}-2x + 4$. Evaluate $h(x)$ at some points in the interval $[0,3]$. When $x = 0$, $h(0)=\sqrt{0 + 1}-2\times0 + 4=1 + 4=5>0$. When $x = 3$, $h(3)=\sqrt{3+1}-2\times3 + 4=2-6 + 4=0$. So, $f(x)\geq g(x)$ on the interval $[0,3]$, and the area formula becomes $A=\int_{0}^{3}(\sqrt{x + 1}-(2x - 4))dx=\int_{0}^{3}(\sqrt{x + 1}-2x + 4)dx$.

Step2: Integrate term - by - term

We know that $\int\sqrt{x + 1}dx=\int(x + 1)^{\frac{1}{2}}dx$. Using the power - rule for integration $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$), with $u=x + 1$ and $n=\frac{1}{2}$, we get $\int(x + 1)^{\frac{1}{2}}dx=\frac{2}{3}(x + 1)^{\frac{3}{2}}+C$. $\int(-2x)dx=-x^{2}+C$ and $\int4dx=4x+C$. So, $\int_{0}^{3}(\sqrt{x + 1}-2x + 4)dx=\left[\frac{2}{3}(x + 1)^{\frac{3}{2}}-x^{2}+4x\right]_{0}^{3}$.

Step3: Evaluate the definite integral

First, substitute $x = 3$: $\frac{2}{3}(3 + 1)^{\frac{3}{2}}-3^{2}+4\times3=\frac{2}{3}\times8-9 + 12=\frac{16}{3}+3=\frac{16 + 9}{3}=\frac{25}{3}$. Then substitute $x = 0$: $\frac{2}{3}(0 + 1)^{\frac{3}{2}}-0^{2}+4\times0=\frac{2}{3}$. Now, subtract: $\frac{25}{3}-\frac{2}{3}=\frac{23}{3}$.

Answer:

B. $\frac{23}{3}$