What is the area of the region bound by the graphs of $f(x)=\sqrt{x-2}$ , $g(x)=14-x$ , and $x=2$ ?

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Write a linear equation from two points

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Q. What is the area of the region bound by the graphs of

f(x)=\sqrt{x-2}

g(x)=14-x

, and

x=2

Find Intersection Points: First, we need to find the points of intersection between the two curves $f(x)$ and $g(x)$ . To do this, we set $f(x)$ equal to $g(x)$ and solve for $x$ .
$f(x) = \sqrt{x-2}$
$g(x) = 14-x$
$\sqrt{x-2} = 14-x$
Squaring both sides to eliminate the square root gives us:
$x - 2 = (14 - x)^2$
$x - 2 = 196 - 28x + x^2$
Rearrange the equation to standard quadratic form:
$g(x)$ $0$
Find Y-Coordinates: Next, we factor the quadratic equation to find the values of $x$ . $(x - 11)(x - 18) = 0$ This gives us two solutions for $x$ : $x = 11$ and $x = 18$ These are the $x$ -coordinates of the points where the two curves intersect.
Calculate Area Integral: Now, we need to find the corresponding $y$ -coordinates for these $x$ -values by substituting them back into either of the original equations. We'll use $f(x) = \sqrt{x-2}$ . $\newline$ For $x = 11$ : $\newline$ $f(11) = \sqrt{11 - 2} = \sqrt{9} = 3$ $\newline$ For $x = 18$ : $\newline$ $f(18) = \sqrt{18 - 2} = \sqrt{16} = 4$ $\newline$ So, the points of intersection are $(11, 3)$ and $(18, 4)$ .
Integrate $g(x)$ : To find the area of the region, we need to integrate the difference between the two functions from the leftmost boundary, $x=2$ , to the rightmost intersection point, $x=18$ . The area $A$ is given by the integral from $2$ to $18$ of $(g(x) - f(x)) \, dx$ . $A = \int_{2}^{18} (14 - x - \sqrt{x - 2}) \, dx$
Integrate $f(x)$ : We calculate the integral separately for each function and then subtract the results. $\newline$ First, for $g(x) = 14 - x$ , the integral is: $\newline$ $\int(14 - x) \, dx$ from $2$ to $18$ = $[14x - (x^2)/2]$ from $2$ to $18$ $\newline$ = $(14\cdot18 - (18^2)/2) - (14\cdot2 - (2^2)/2)$ $\newline$ = $(252 - 162) - (28 - 2)$ $\newline$ = $g(x) = 14 - x$ $0$ $\newline$ = $g(x) = 14 - x$ $1$
Subtract Integrals: Next, for $f(x) = \sqrt{x - 2}$ , the integral is: $\newline$ $\int_{2}^{18}\sqrt{x - 2} \, dx$ $\newline$ This is an integral of a square root function, which can be solved by a substitution method or by looking up the integral in a table. The antiderivative of $\sqrt{x - 2}$ is: $\newline$ $(2/3)(x - 2)^{(3/2)}$ $\newline$ Evaluating this from $2$ to $18$ gives us: $\newline$ $(2/3)(18 - 2)^{(3/2)} - (2/3)(2 - 2)^{(3/2)}$ $\newline$ $= (2/3)(16)^{(3/2)} - 0$ $\newline$ $= (2/3)(64)$ $\newline$ $= 128/3$
Subtract Integrals: Next, for $f(x) = \sqrt{x - 2}$ , the integral is: $\newline$ $\int_{2}^{18}\sqrt{x - 2} \, dx$ $\newline$ This is an integral of a square root function, which can be solved by a substitution method or by looking up the integral in a table. The antiderivative of $\sqrt{x - 2}$ is: $\newline$ $(2/3)(x - 2)^{(3/2)}$ $\newline$ Evaluating this from $2$ to $18$ gives us: $\newline$ $(2/3)(18 - 2)^{(3/2)} - (2/3)(2 - 2)^{(3/2)}$ $\newline$ $= (2/3)(16)^{(3/2)} - 0$ $\newline$ $= (2/3)(64)$ $\newline$ $= 128/3$ Now, we subtract the integral of $\int_{2}^{18}\sqrt{x - 2} \, dx$ $0$ from the integral of $\int_{2}^{18}\sqrt{x - 2} \, dx$ $1$ to find the area of the region. $\newline$ Area $\int_{2}^{18}\sqrt{x - 2} \, dx$ $2$ $\newline$ To combine these, we need a common denominator, which is $\int_{2}^{18}\sqrt{x - 2} \, dx$ $3$ . $\newline$ Area $\int_{2}^{18}\sqrt{x - 2} \, dx$ $4$ $\newline$ $\int_{2}^{18}\sqrt{x - 2} \, dx$ $5$ $\newline$ $\int_{2}^{18}\sqrt{x - 2} \, dx$ $6$