What is the area of the region between the graphs of {:[f(x)=2x^(2)+5x" and "],[g(x)=-x^(2)-6x+4" from "],[x=-4" to "x=0" ? "]:} Choose 1 answer: (A) 40 (B) (355)/(12) (C) 8 (D) (128)/(3)

Question

What is the area of the region between the graphs of  {:[f(x)=2x^(2)+5x" and "],[g(x)=-x^(2)-6x+4" from "],[x=-4" to "x=0" ? "]:} Choose 1 answer: (A) 40 (B)  (355)/(12) (C) 8 (D)  (128)/(3)

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Find Difference of Functions: To find the area between the two curves, we need to integrate the difference of the functions over the given interval from x = -4 to x = 0. First, we find the difference between the functions f(x) and g(x): Difference (h(x)) = f(x) - g(x) = (2x^2 + 5x) - (-x^2 - 6x + 4)Simplify the Difference: Now, we simplify the difference h(x): h(x) = 2x^2 + 5x + x^2 + 6x - 4 h(x) = 3x^2 + 11x - 4Integrate Difference Over Interval: Next, we integrate h(x) from x = -4 to x = 0 to find the area between the curves: Area = ∫ from -4 to 0 of (3x^2 + 11x - 4) dxFind Antiderivative of Difference: We find the antiderivative of h(x): Antiderivative of h(x) = (3/3)x^3 + (11/2)x^2 - 4x Antiderivative of h(x) = x^3 + (11/2)x^2 - 4xEvaluate Antiderivative at Bounds: We evaluate the antiderivative at the bounds x = 0 and x = -4: At x = 0: F(0) = (0)^3 + (11/2)(0)^2 - 4(0) = 0 At x = -4: F(-4) = (-4)^3 + (11/2)(-4)^2 - 4(-4) F(-4) = -64 + 11(8) + 16 F(-4) = -64 + 88 + 16 F(-4) = 40Calculate Area Between Curves: Finally, we subtract the lower bound value from the upper bound value to find the area: Area = F(0) - F(-4) Area = 0 - 40 Area = -40 Since area cannot be negative, we take the absolute value: Area = 40

What is the area of the region between the graphs of $f(x)=2 x^{2}+5 x$ and $g(x)=-x^{2}-6 x+4$ from $x=-4$ to $x=0$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $40$ $\newline$ (B) $\frac{355}{12}$ $\newline$ (C) $8$ $\newline$ (D) $\frac{128}{3}$

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