What is the area of the region between the graphs of $f(x)=\sqrt{x+10}$ and $g(x)=x-2$ from $x=-10$ to $x=6$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\mathbf{1 2 8}$ $\newline$ (B) $\frac{64}{3}$ $\newline$ (C) $\frac{320}{3}$ $\newline$ (D) $160$

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

f(x)=\sqrt{x+10}

and

g(x)=x-2

from

x=-10

x=6

\newline

Choose

1

answer:

\newline

(A)

\mathbf{1 2 8}

\newline

(B)

\frac{64}{3}

\newline

(C)

\frac{320}{3}

\newline

(D)

160

Understand the problem: Understand the problem. $\newline$ We need to find the area between two curves, $f(x)$ and $g(x)$ , over the interval from $x=-10$ to $x=6$ . The area between two curves is given by the integral of the upper function minus the lower function over the interval of interest.
Determine upper and lower functions: Determine which function is the upper function and which is the lower function on the interval from $x=-10$ to $x=6$ . Since $f(x) = \sqrt{x+10}$ is always positive or zero for $x$ in the interval $[-10, 6]$ , and $g(x) = x-2$ can be negative, zero, or positive in this interval, we need to compare the values of $f(x)$ and $g(x)$ to determine which one is the upper function. For $x=-10$ , $f(x)=0$ and $x=6$ $0$ , so $f(x)$ is above $g(x)$ . For $x=6$ , $x=6$ $4$ and $x=6$ $5$ , so $f(x)$ is still above $g(x)$ . Therefore, $f(x)$ is the upper function and $g(x)$ is the lower function on the entire interval.
Set up integral: Set up the integral to find the area between the curves. $\newline$ The area $A$ between the curves is given by the integral from $x=-10$ to $x=6$ of $(f(x) - g(x)) \, dx$ , which is: $\newline$ $A = \int_{-10}^{6} (\sqrt{x+10} - (x-2)) \, dx$
Calculate integral: Calculate the integral. $\newline$ We need to calculate the integral $A = \int_{-10}^{6} (\sqrt{x+10} - (x-2)) \, dx$ . This requires us to integrate each term separately and then combine the results.
Integrate $\sqrt{x+10}$ : Integrate the first term, $\sqrt{x+10}$ . The integral of $\sqrt{x+10}$ with respect to $x$ from $-10$ to $6$ is: $\int_{-10}^{6} \sqrt{x+10} \, dx = \left[\frac{2}{3} \cdot (x+10)^{\frac{3}{2}}\right]_{-10}^{6}$ Evaluating this from $-10$ to $6$ gives: $\left[\frac{2}{3} \cdot (6+10)^{\frac{3}{2}}\right] - \left[\frac{2}{3} \cdot (0)\right] = \left[\frac{2}{3} \cdot 16^{\frac{3}{2}}\right] = \left[\frac{2}{3} \cdot 64\right] = \frac{128}{3}$
Integrate $(x-2)$ : Integrate the second term, $(x-2)$ . The integral of $(x-2)$ with respect to $x$ from $-10$ to $6$ is: $\int_{-10}^{6} (x-2) \, dx = \left[\frac{1}{2} \cdot x^2 - 2x\right]_{-10}^{6}$ Evaluating this from $-10$ to $6$ gives: $\left[\frac{1}{2} \cdot 6^2 - 2\cdot6\right] - \left[\frac{1}{2} \cdot (-10)^2 - 2\cdot(-10)\right] = [18 - 12] - [50 + 20] = 6 - 70 = -64$
Combine integral results: Combine the results of the integrals to find the total area. $\newline$ The total area $A$ is the difference between the two integrals we just calculated: $\newline$ $A = (\frac{128}{3}) - (-64) = \frac{128}{3} + 64 = \frac{128}{3} + \frac{192}{3} = \frac{320}{3}$
Choose correct answer: Choose the correct answer from the given options. $\newline$ The calculated area is $\frac{320}{3}$ , which corresponds to option (C).