What is the area of the region between the graphs of $f(x)=\sqrt{x+1}$ and $g(x)=2 x-4$ from $x=0$ to $x=3$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{23}{3}$ $\newline$ (B) $\frac{5}{3}$ $\newline$ (C) $\frac{14}{3}$ $\newline$ (D) $-3$

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Q. What is the area of the region between the graphs of

f(x)=\sqrt{x+1}

and

g(x)=2 x-4

from

x=0

x=3

\newline

Choose

1

answer:

\newline

(A)

\frac{23}{3}

\newline

(B)

\frac{5}{3}

\newline

(C)

\frac{14}{3}

\newline

(D)

-3

Understand the problem: Understand the problem. $\newline$ We need to find the area between two curves, $f(x)$ and $g(x)$ , from $x=0$ to $x=3$ . The area between two curves is found by integrating the difference between the functions over the given interval.
Set up the integral: Set up the integral to find the area between the curves. $\newline$ The area $A$ is given by the integral from $x=0$ to $x=3$ of the top function minus the bottom function. We need to determine which function is on top (greater $y$ -value) for the interval $[0,3]$ .
Compare the functions: Compare the functions at a point in the interval to determine which is on top. Let's evaluate both functions at $x=1$ (a point within the interval $[0,3]$ ): $f(1) = \sqrt{1+1} = \sqrt{2}$ $g(1) = 2(1)-4 = -2$ Since \sqrt{2} > -2, $f(x)$ is on top of $g(x)$ in the interval $[0,3]$ .
Write the integral: Write the integral for the area. $\newline$ $A = \int_{0}^{3} (f(x) - g(x)) \, dx$ $\newline$ $A = \int_{0}^{3} (\sqrt{x+1} - (2x-4)) \, dx$
Calculate the integral: Calculate the integral. $\newline$ $A = \int_{0}^{3} (\sqrt{x+1} - 2x + 4) \, dx$ $\newline$ This requires us to integrate term by term.
Integrate each term separately: Integrate each term separately. $\newline$ The integral of $\sqrt{x+1}$ with respect to $x$ is $\frac{2}{3}\cdot(x+1)^{\frac{3}{2}}$ . $\newline$ The integral of $-2x$ with respect to $x$ is $-x^2$ . $\newline$ The integral of $4$ with respect to $x$ is $4x$ . $\newline$ So, $A = \left[\frac{2}{3}\cdot(x+1)^{\frac{3}{2}} - x^2 + 4x\right]$ from $x$ $0$ to $x$ $1$ .
Evaluate the antiderivative: Evaluate the antiderivative at the upper and lower bounds and subtract. $\newline$ $A = \left[\left(\frac{2}{3}\right)\cdot\left(3+1\right)^{\frac{3}{2}} - 3^2 + 4\cdot3\right] - \left[\left(\frac{2}{3}\right)\cdot\left(0+1\right)^{\frac{3}{2}} - 0^2 + 4\cdot0\right]$ $\newline$ $A = \left[\left(\frac{2}{3}\right)\cdot4^{\frac{3}{2}} - 9 + 12\right] - \left[\left(\frac{2}{3}\right)\cdot1^{\frac{3}{2}} - 0 + 0\right]$ $\newline$ $A = \left[\left(\frac{2}{3}\right)\cdot8 - 9 + 12\right] - \left[\left(\frac{2}{3}\right)\cdot1 - 0 + 0\right]$ $\newline$ $A = \left[\frac{16}{3} - 9 + 12\right] - \left[\frac{2}{3}\right]$ $\newline$ $A = \left(\frac{16}{3} + \frac{36}{3} - \frac{27}{3}\right) - \left(\frac{2}{3}\right)$ $\newline$ $A = \left(\frac{25}{3}\right) - \left(\frac{2}{3}\right)$ $\newline$ $A = \frac{23}{3}$