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 ? Choose 1 answer: (A) (320)/(3) (B) 128 (C) 160 (D) (64)/(3)

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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 ? Choose 1 answer: (A)  (320)/(3) (B) 128 (C)  160 (D)  (64)/(3)

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Understand the problem: Understand the problem. We need to find the area between two curves, f(x) = sqrt(x+10) and g(x) = x-2, from x = -10 to x = 6. The area between two curves is found by integrating the absolute difference of the functions over the given interval.Set up the integral: Set up the integral to find the area. The area A between the two curves from x = a to x = b is given by the integral A = ∫ from a to b of |f(x) - g(x)| dx. In this case, we need to find ∫ from -10 to 6 of |sqrt(x+10) - (x-2)| dx.Determine the points of intersection: Determine the points of intersection. To properly set up the integral, we need to know where the graphs of f(x) and g(x) intersect, as this will affect the limits of integration if they intersect within the interval [-10, 6]. We set f(x) = g(x) and solve for x: sqrt(x+10) = x-2 Squaring both sides, we get: x + 10 = (x - 2)^2 x + 10 = x^2 - 4x + 4 0 = x^2 - 5x - 6 Factoring, we get: 0 = (x - 6)(x + 1) So, the points of intersection are x = 6 and x = -1.Split the integral if necessary: Split the integral if necessary. Since the intersection points are within the interval [-10, 6], we need to split the integral at x = -1. We will integrate from -10 to -1 where g(x) is greater than f(x), and from -1 to 6 where f(x) is greater than g(x). This gives us two integrals: A = ∫ from -10 to -1 of (x-2 - sqrt(x+10)) dx + ∫ from -1 to 6 of (sqrt(x+10) - (x-2)) dx.Calculate the first integral: Calculate the first integral from -10 to -1. A1 = ∫ from -10 to -1 of (x-2 - sqrt(x+10)) dx = ∫ from -10 to -1 of (x - 2) dx - ∫ from -10 to -1 of sqrt(x+10) dx = [1/2 * x^2 - 2x] from -10 to -1 - [2/3 * (x+10)^(3/2)] from -10 to -1 = [(1/2 * (-1)^2 - 2*(-1)) - (1/2 * (-10)^2 - 2*(-10))] - [(2/3 * (-1+10)^(3/2)) - (2/3 * (0)^(3/2))] = [(1/2 - 2) - (50 + 20)] - [(2/3 * 9^(3/2)) - 0] = [(-3/2) - 70] - [(2/3 * 27) - 0] = [-71.5 - 18] = -89.5Calculate the second integral: Calculate the second integral from -1 to 6. A2 = ∫ from -1 to 6 of (sqrt(x+10) - (x-2)) dx = ∫ from -1 to 6 of sqrt(x+10) dx - ∫ from -1 to 6 of (x - 2) dx = [2/3 * (x+10)^(3/2)] from -1 to 6 - [1/2 * x^2 - 2x] from -1 to 6 = [(2/3 * (6+10)^(3/2)) - (2/3 * (-1+10)^(3/2))] - [(1/2 * 6^2 - 2*6) - (1/2 * (-1)^2 - 2*(-1))] = [(2/3 * 16^(3/2)) - (2/3 * 9^(3/2))] - [(18 - 12) - (1/2 + 2)] = [(2/3 * 64) - (2/3 * 27)] - [6 - 2.5] = [128/3 - 54/3] - 3.5 = [74/3] - 3.5 = [74/3 - 10.5/3] = 63.5/3 = 21.167Add the absolute values: Add the absolute values of the two areas. Since we are looking for the area between the curves, we take the absolute value of each integral result (if negative) and sum them up to get the total area. Total Area = |A1| + |A2| = |-89.5| + |21.167| = 89.5 + 21.167 = 110.667Choose the correct answer: Choose the correct answer. The total area calculated does not match any of the provided answer choices, which suggests there might be a mistake in the calculations. We need to re-evaluate the integrals and ensure that the calculations are correct.

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) $\frac{320}{3}$ $\newline$ (B) $128$ $\newline$ (C) $\mathbf{1 6 0}$ $\newline$ (D) $\frac{64}{3}$

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