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) (64)/(3) (B) 160 (C) (320)/(3) (D) 128

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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)  (64)/(3) (B) 160 (C)  (320)/(3) (D) 128

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Set up integral bounds: To find the area between the two curves, we need to set up an integral from the lower bound of x = -10 to the upper bound of x = 6. The area A can be found by integrating the difference between the two functions over this interval.Determine function order: First, we need to determine which function is above the other in the interval from x = -10 to x = 6. Since f(x) = sqrt(x+10) is always positive or zero and g(x) = x-2 can be negative or positive, we can conclude that f(x) is above g(x) for the entire interval.Calculate integral: The area A between the two curves is given by the integral from -10 to 6 of (f(x) - g(x)) dx, which is the integral from -10 to 6 of (sqrt(x+10) - (x-2)) dx.Integrate sqrt(x+10): Now we calculate the integral: A = ∫ from -10 to 6 of (sqrt(x+10) - (x-2)) dx This requires us to integrate term by term.Integrate (x-2): First, we integrate sqrt(x+10) with respect to x from -10 to 6. The antiderivative of sqrt(x+10) is (2/3)*(x+10)^(3/2).Evaluate antiderivatives: Next, we integrate (x-2) with respect to x from -10 to 6. The antiderivative of (x-2) is (1/2)*x^2 - 2x.Evaluate sqrt(x+10): Now we evaluate the antiderivatives at the bounds x = 6 and x = -10 and subtract the lower bound from the upper bound for each term. For sqrt(x+10), we have [(2/3)*(6+10)^(3/2)] - [(2/3)*(-10+10)^(3/2)]. For (x-2), we have [(1/2)*6^2 - 2*6] - [(1/2)*(-10)^2 - 2*(-10)].Evaluate (x-2): Evaluating the antiderivatives, we get: For sqrt(x+10), [(2/3)*(16)^(3/2)] - [(2/3)*(0)^(3/2)] = (2/3)*64 - 0 = 128/3. For (x-2), [(1/2)*36 - 12] - [(1/2)*100 + 20] = 18 - 12 - 50 - 20 = -64.Add antiderivatives: Now we add the results of the two antiderivatives to find the total area: A = (128/3) - (-64) = (128/3) + (192/3) = (320/3).Final answer: The final answer is (320/3), which corresponds to answer choice (C).

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) $\frac{64}{3}$ $\newline$ (B) $160$ $\newline$ (C) $\frac{320}{3}$ $\newline$ (D) $128$

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