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) 7 (B) (343)/(3) (C) (83)/(3) (D) (52)/(3)sqrt13-1-32sqrt3

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) 7 (B)  (343)/(3) (C)  (83)/(3) (D)  (52)/(3)sqrt13-1-32sqrt3

Bytelearn · Accepted Answer

Set up integral: To find the area between the two curves, we need to integrate the difference of the functions over the given interval from x = -3 to x = 4.Substitute functions: First, we need to set up the integral. The area A is given by the integral from -3 to 4 of the difference between the functions f(x) and g(x). A = ∫ from -3 to 4 of (f(x) - g(x)) dxSimplify integrand: Substitute the given functions into the integral. A = ∫ from -3 to 4 of ((-x^2 + 2x + 12) - (x^2 - 12)) dxIntegrate with respect: Simplify the integrand by combining like terms. A = ∫ from -3 to 4 of (-x^2 + 2x + 12 - x^2 + 12) dx A = ∫ from -3 to 4 of (-2x^2 + 2x + 24) dxEvaluate upper limit: Integrate the function with respect to x. A = [(-2/3)x^3 + x^2 + 24x] from -3 to 4Evaluate lower limit: Evaluate the antiderivative at the upper limit of integration (x = 4). A(4) = (-2/3)(4)^3 + (4)^2 + 24(4) A(4) = (-2/3)(64) + 16 + 96 A(4) = (-128/3) + 16 + 96 A(4) = (-128/3) + 16(3/3) + 96(3/3) A(4) = (-128 + 48 + 288) / 3 A(4) = 208 / 3Find difference: Evaluate the antiderivative at the lower limit of integration (x = -3). A(-3) = (-2/3)(-3)^3 + (-3)^2 + 24(-3) A(-3) = (-2/3)(-27) + 9 - 72 A(-3) = (54/3) + 9 - 72 A(-3) = (54 + 27 - 216) / 3 A(-3) = (-135) / 3 A(-3) = -45Find the difference between the antiderivative evaluated at the upper and lower limits. Area = A(4) - A(-3) Area = (208/3) - (-45) Area = (208/3) + (45)(3/3) Area = (208 + 135) / 3 Area = 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) $7$ $\newline$ (B) $\frac{343}{3}$ $\newline$ (C) $\frac{83}{3}$ $\newline$ (D) $\frac{52}{3} \sqrt{13}-1-32 \sqrt{3}$

Full solution

More problems from Roots of rational numbers

Related topics