What is the area of the region between the graphs of f(x)=-x^(2)+2x+12 and g(x)=x^(2)-12 from x=-3 to x=4 ? Choose 1 answer: (A) (52)/(3)sqrt13-1-32sqrt3 (B) (83)/(3) (C) 7 (D) (343)/(3)

Question

What is the area of the region between the graphs of  f(x)=-x^(2)+2x+12 and  g(x)=x^(2)-12 from  x=-3 to  x=4 ? Choose 1 answer: (A)  (52)/(3)sqrt13-1-32sqrt3 (B)  (83)/(3) (C) 7 (D)  (343)/(3)

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Write Functions and Interval: Write down the given functions and the interval for which we need to find the area between the graphs. f(x) = -x^2 + 2x + 12 g(x) = x^2 - 12 Interval: x = -3 to x = 4Set Up Integral: Set up the integral to find the area between the two curves. The area A is given by the integral of the top function minus the bottom function from the leftmost point of the interval to the rightmost point. A = ∫ from x=-3 to x=4 (f(x) - g(x)) dxSubtract Functions: Subtract the function g(x) from f(x) to find the integrand. f(x) - g(x) = (-x^2 + 2x + 12) - (x^2 - 12) f(x) - g(x) = -x^2 + 2x + 12 - x^2 + 12 f(x) - g(x) = -2x^2 + 2x + 24Integrate Function: Integrate the function -2x^2 + 2x + 24 with respect to x from -3 to 4. A = ∫ from x=-3 to x=4 (-2x^2 + 2x + 24) dx A = [-2/3 x^3 + x^2 + 24x] from x=-3 to x=4Evaluate Antiderivative: Evaluate the antiderivative at the upper limit of integration (x=4) and then at the lower limit of integration (x=-3), and subtract the latter from the former. A = [-2/3 * 4^3 + 4^2 + 24*4] - [-2/3 * (-3)^3 + (-3)^2 + 24*(-3)] A = [-2/3 * 64 + 16 + 96] - [-2/3 * (-27) + 9 - 72] A = [-128/3 + 16 + 96] - [54/3 + 9 - 72] A = [-128/3 + 48/3 + 288/3] - [54/3 + 27/3 - 216/3] A = [208/3] - [-135/3] A = 208/3 + 135/3 A = 343/3

What is the area of the region between the graphs of $f(x)=-x^{2}+2 x+12$ and $g(x)=x^{2}-12$ from $x=-3$ to $x=4$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{52}{3} \sqrt{13}-1-32 \sqrt{3}$ $\newline$ (B) $\frac{83}{3}$ $\newline$ (C) $7$ $\newline$ (D) $\frac{343}{3}$

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