What is the area of the region bound by the graphs of f(x)=sqrt(x-2),g(x)=14-x , and x=2 ? Choose 1 answer: (A) (99)/(2) (B) (19)/(6) (C) (151)/(2) (D) (45)/(2)

Set up integral limits: To find the area of the region, we need to set up an integral with the appropriate limits of integration. The region is bounded by the graphs of f(x) and g(x), and the vertical line x=2. We need to find the points of intersection between f(x) and g(x) to determine the limits.Find points of intersection: Set f(x) equal to g(x) to find the points of intersection: sqrt(x-2) = 14-x. Square both sides to eliminate the square root: (sqrt(x-2))^2 = (14-x)^2.Simplify and rearrange equation: Simplify the equation: x-2 = (14-x)^2. Expand the right side: x-2 = 196 - 28x + x^2.Factor quadratic equation: Rearrange the equation to form a quadratic equation: x^2 - 29x + 198 = 0.Calculate area using integrals: Factor the quadratic equation: (x-18)(x-11) = 0. This gives us the points of intersection x=18 and x=11.Integrate from x=2 to x=11: The area of the region can be found by integrating the difference between g(x) and f(x) from x=2 to x=11, and then from x=11 to x=18. The integral is split because f(x) is only defined for x >= 2.Evaluate integral at bounds: Set up the integral for the first part from x=2 to x=11: ∫ from 2 to 11 of (14-x - sqrt(x-2)) dx.Integrate from x=11 to x=18: Calculate the integral: ∫ from 2 to 11 of (14-x) dx - ∫ from 2 to 11 of sqrt(x-2) dx.Evaluate integral at bounds: Integrate the first part: ∫ from 2 to 11 of (14-x) dx = [14x - x^2/2] from 2 to 11.Evaluate the first part at the bounds: [14(11) - (11)^2/2] - [14(2) - (2)^2/2] = [154 - 60.5] - [28 - 2] = 93.5 - 26 = 67.5.Integrate the second part: ∫ from 2 to 11 of sqrt(x-2) dx. Let u = x-2, then du = dx and when x=2, u=0 and when x=11, u=9. The integral becomes ∫ from 0 to 9 of sqrt(u) du.Calculate the integral: ∫ from 0 to 9 of sqrt(u) du = [2/3 * u^(3/2)] from 0 to 9.Evaluate the second part at the bounds: [2/3 * 9^(3/2)] - [2/3 * 0^(3/2)] = [2/3 * 27] - 0 = 18.Subtract the second part from the first part to get the area from x=2 to x=11: 67.5 - 18 = 49.5.Now, set up the integral for the second part from x=11 to x=18: ∫ from 11 to 18 of (14-x) dx.Calculate the integral: ∫ from 11 to 18 of (14-x) dx = [14x - x^2/2] from 11 to 18.Evaluate the integral at the bounds: [14(18) - (18)^2/2] - [14(11) - (11)^2/2] = [252 - 162] - [154 - 60.5] = 90 - 93.5 = -3.5.

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What is the area of the region bound by the graphs of
f(x)=sqrt(x-2),g(x)=14-x , and
x=2 ?
Choose 1 answer:
(A)
(99)/(2)
(B)
(19)/(6)
(C)
(151)/(2)
(D)
(45)/(2)

What is the area of the region bound by the graphs of $f(x)=\sqrt{x-2}, g(x)=14-x$ , and $x=2$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{99}{2}$ $\newline$ (B) $\frac{19}{6}$ $\newline$ (C) $\frac{151}{2}$ $\newline$ (D) $\frac{45}{2}$

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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

\newline

Choose

1

answer:

\newline

(A)

\frac{99}{2}

\newline

(B)

\frac{19}{6}

\newline

(C)

\frac{151}{2}

\newline

(D)

\frac{45}{2}

Set up integral limits: To find the area of the region, we need to set up an integral with the appropriate limits of integration. The region is bounded by the graphs of $f(x)$ and $g(x)$ , and the vertical line $x=2$ . We need to find the points of intersection between $f(x)$ and $g(x)$ to determine the limits.
Find points of intersection: Set $f(x)$ equal to $g(x)$ to find the points of intersection: $\sqrt{x-2} = 14-x$ . Square both sides to eliminate the square root: $(\sqrt{x-2})^2 = (14-x)^2$ .
Simplify and rearrange equation: Simplify the equation: $x-2 = (14-x)^2$ . Expand the right side: $x-2 = 196 - 28x + x^2$ .
Factor quadratic equation: Rearrange the equation to form a quadratic equation: $x^2 - 29x + 198 = 0$ .
Calculate area using integrals: Factor the quadratic equation: $(x-18)(x-11) = 0$ . This gives us the points of intersection $x=18$ and $x=11$ .
Integrate from $x=2$ to $x=11$ : The area of the region can be found by integrating the difference between $g(x)$ and $f(x)$ from $x=2$ to $x=11$ , and then from $x=11$ to $x=18$ . The integral is split because $f(x)$ is only defined for $x \geq 2$ .
Evaluate integral at bounds: Set up the integral for the first part from $x=2$ to $x=11$ : $\int_{2}^{11} (14-x - \sqrt{x-2}) \, dx$ .
Integrate from $x=11$ to $x=18$ : Calculate the integral: $\int_{2}^{11} (14-x) \,dx - \int_{2}^{11} \sqrt{x-2} \,dx$ .
Evaluate integral at bounds: Integrate the first part: $\int_{2}^{11} (14-x) \, dx = [14x - \frac{x^2}{2}]$ from $2$ to $11$ .
Evaluate integral at bounds: Integrate the first part: $\int_{2}^{11} (14-x) \, dx = [14x - \frac{x^2}{2}]$ from $2$ to $11$ .Evaluate the first part at the bounds: $[14(11) - \frac{(11)^2}{2}] - [14(2) - \frac{(2)^2}{2}] = [154 - 60.5] - [28 - 2] = 93.5 - 26 = 67.5$ .
Evaluate integral at bounds: Integrate the first part: $\int_{2}^{11} (14-x) \, dx = [14x - \frac{x^2}{2}]$ from $2$ to $11$ .Evaluate the first part at the bounds: $[14(11) - \frac{(11)^2}{2}] - [14(2) - \frac{(2)^2}{2}] = [154 - 60.5] - [28 - 2] = 93.5 - 26 = 67.5$ .Integrate the second part: $\int_{2}^{11} \sqrt{x-2} \, dx$ . Let $u = x-2$ , then $du = dx$ and when $x=2$ , $u=0$ and when $x=11$ , $2$ $0$ . The integral becomes $2$ $1$ .
Evaluate integral at bounds: Integrate the first part: $\int_{2}^{11} (14-x) \, dx = [14x - \frac{x^2}{2}]$ from $2$ to $11$ .Evaluate the first part at the bounds: $[14(11) - \frac{(11)^2}{2}] - [14(2) - \frac{(2)^2}{2}] = [154 - 60.5] - [28 - 2] = 93.5 - 26 = 67.5$ .Integrate the second part: $\int_{2}^{11} \sqrt{x-2} \, dx$ . Let $u = x-2$ , then $du = dx$ and when $x=2$ , $u=0$ and when $x=11$ , $2$ $0$ . The integral becomes $2$ $1$ .Calculate the integral: $2$ $2$ from $2$ $3$ to $2$ $4$ .
Evaluate integral at bounds: Integrate the first part: $\int_{2}^{11} (14-x) \, dx = [14x - \frac{x^2}{2}]$ from $2$ to $11$ .Evaluate the first part at the bounds: $[14(11) - \frac{(11)^2}{2}] - [14(2) - \frac{(2)^2}{2}] = [154 - 60.5] - [28 - 2] = 93.5 - 26 = 67.5$ .Integrate the second part: $\int_{2}^{11} \sqrt{x-2} \, dx$ . Let $u = x-2$ , then $du = dx$ and when $x=2$ , $u=0$ and when $x=11$ , $2$ $0$ . The integral becomes $2$ $1$ .Calculate the integral: $2$ $2$ from $2$ $3$ to $2$ $4$ .Evaluate the second part at the bounds: $2$ $5$ .
Evaluate integral at bounds: Integrate the first part: $\int_{2}^{11} (14-x) \, dx = [14x - \frac{x^2}{2}]$ from $2$ to $11$ .Evaluate the first part at the bounds: $[14(11) - \frac{(11)^2}{2}] - [14(2) - \frac{(2)^2}{2}] = [154 - 60.5] - [28 - 2] = 93.5 - 26 = 67.5$ .Integrate the second part: $\int_{2}^{11} \sqrt{x-2} \, dx$ . Let $u = x-2$ , then $du = dx$ and when $x=2$ , $u=0$ and when $x=11$ , $2$ $0$ . The integral becomes $2$ $1$ .Calculate the integral: $2$ $2$ from $2$ $3$ to $2$ $4$ .Evaluate the second part at the bounds: $2$ $5$ .Subtract the second part from the first part to get the area from $x=2$ to $x=11$ : $2$ $8$ .
Evaluate integral at bounds: Integrate the first part: $\int_{2}^{11} (14-x) \, dx = [14x - \frac{x^2}{2}]$ from $2$ to $11$ .Evaluate the first part at the bounds: $[14(11) - \frac{(11)^2}{2}] - [14(2) - \frac{(2)^2}{2}] = [154 - 60.5] - [28 - 2] = 93.5 - 26 = 67.5$ .Integrate the second part: $\int_{2}^{11} \sqrt{x-2} \, dx$ . Let $u = x-2$ , then $du = dx$ and when $x=2$ , $u=0$ and when $x=11$ , $2$ $0$ . The integral becomes $2$ $1$ .Calculate the integral: $2$ $2$ from $2$ $3$ to $2$ $4$ .Evaluate the second part at the bounds: $2$ $5$ .Subtract the second part from the first part to get the area from $x=2$ to $x=11$ : $2$ $8$ .Now, set up the integral for the second part from $x=11$ to $11$ $0$ : $11$ $1$ .
Evaluate integral at bounds: Integrate the first part: $\int_{2}^{11} (14-x) \, dx = [14x - \frac{x^2}{2}]$ from $2$ to $11$ . Evaluate the first part at the bounds: $[14(11) - \frac{(11)^2}{2}] - [14(2) - \frac{(2)^2}{2}] = [154 - 60.5] - [28 - 2] = 93.5 - 26 = 67.5$ . Integrate the second part: $\int_{2}^{11} \sqrt{x-2} \, dx$ . Let $u = x-2$ , then $du = dx$ and when $x=2$ , $u=0$ and when $x=11$ , $2$ $0$ . The integral becomes $2$ $1$ . Calculate the integral: $2$ $2$ from $2$ $3$ to $2$ $4$ . Evaluate the second part at the bounds: $2$ $5$ . Subtract the second part from the first part to get the area from $x=2$ to $x=11$ : $2$ $8$ . Now, set up the integral for the second part from $x=11$ to $11$ $0$ : $11$ $1$ . Calculate the integral: $11$ $2$ from $11$ to $11$ $4$ .
Evaluate integral at bounds: Integrate the first part: $\int_{2}^{11} (14-x) \, dx = [14x - \frac{x^2}{2}]$ from $2$ to $11$ .Evaluate the first part at the bounds: $[14(11) - \frac{(11)^2}{2}] - [14(2) - \frac{(2)^2}{2}] = [154 - 60.5] - [28 - 2] = 93.5 - 26 = 67.5$ .Integrate the second part: $\int_{2}^{11} \sqrt{x-2} \, dx$ . Let $u = x-2$ , then $du = dx$ and when $x=2$ , $u=0$ and when $x=11$ , $2$ $0$ . The integral becomes $2$ $1$ .Calculate the integral: $2$ $2$ from $2$ $3$ to $2$ $4$ .Evaluate the second part at the bounds: $2$ $5$ .Subtract the second part from the first part to get the area from $x=2$ to $x=11$ : $2$ $8$ .Now, set up the integral for the second part from $x=11$ to $11$ $0$ : $11$ $1$ .Calculate the integral: $11$ $2$ from $11$ to $11$ $4$ .Evaluate the integral at the bounds: $11$ $5$ .