What is the area of the region between the graphs of f(x)=x^(2)-4x+1 and g(x)=3x-5 from x=1 to x=4 ? Choose 1 answer: (A) (125)/(6) (B) 4 (C) (27)/(2) (D) (81)/(2)

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What is the area of the region between the graphs of  f(x)=x^(2)-4x+1 and  g(x)=3x-5 from  x=1 to  x=4 ? Choose 1 answer: (A)  (125)/(6) (B) 4 (C)  (27)/(2) (D)  (81)/(2)

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Set up integral: To find the area between two curves, we need to integrate the difference between the functions over the given interval. First, we need to set up the integral.Calculate difference: The area A between the curves f(x) and g(x) from x=1 to x=4 is given by the integral of the absolute value of the difference between f(x) and g(x), which is A = ∫ from 1 to 4 of |f(x) - g(x)| dx.Determine upper function: We calculate f(x) - g(x) for x in [1, 4]. f(x) - g(x) = (x^2 - 4x + 1) - (3x - 5) = x^2 - 4x + 1 - 3x + 5 = x^2 - 7x + 6.Find intersection point: Since we are looking for the area between the curves, we need to find out which function is above the other in the interval from x=1 to x=4. We can do this by evaluating f(x) and g(x) at any point in the interval, for example at x=2.Evaluate integral: Evaluating f(2) and g(2), we get f(2) = 2^2 - 4*2 + 1 = 4 - 8 + 1 = -3 and g(2) = 3*2 - 5 = 6 - 5 = 1. Since g(2) > f(2), g(x) is above f(x) in at least some part of the interval. We need to check if this is true for the entire interval or if there is a point where f(x) and g(x) intersect.Evaluate antiderivative: To find the intersection points, we set f(x) equal to g(x) and solve for x: x^2 - 4x + 1 = 3x - 5. This simplifies to x^2 - 7x + 6 = 0.Subtract antiderivatives: Factoring the quadratic equation x^2 - 7x + 6 = 0, we get (x - 1)(x - 6) = 0. This gives us two solutions: x = 1 and x = 6. However, since we are only interested in the interval from x=1 to x=4, we only consider the intersection at x=1.Calculate final area: Since the intersection at x=1 is the start of our interval and there are no other intersections within the interval, we can conclude that g(x) is above f(x) for the entire interval from x=1 to x=4. Therefore, the area A is given by the integral from 1 to 4 of (g(x) - f(x)) dx.We integrate g(x) - f(x) = (3x - 5) - (x^2 - 4x + 1) = -x^2 + 7x - 6 from x=1 to x=4. The integral of -x^2 is -x^3/3, the integral of 7x is 7x^2/2, and the integral of -6 is -6x.We evaluate the antiderivative at the bounds x=4 and x=1 and subtract to find the area. The antiderivative evaluated at x=4 is [-(4^3)/3 + (7*(4^2))/2 - 6*4] and at x=1 is [-(1^3)/3 + (7*(1^2))/2 - 6*1].Calculating the antiderivative at x=4 gives us [-(64)/3 + (7*16)/2 - 24] = [-64/3 + 56 - 24] = [-64/3 + 32]. Calculating the antiderivative at x=1 gives us [-(1)/3 + (7*1)/2 - 6] = [-1/3 + 7/2 - 6] = [-1/3 + 3.5 - 6].Now we subtract the antiderivative evaluated at x=1 from the antiderivative evaluated at x=4 to find the area: ([-64/3 + 32] - [-1/3 + 3.5 - 6]) = (-64/3 + 32 + 1/3 - 3.5 + 6).Simplifying the expression, we get (-64/3 + 96/3 + 1/3 - 10.5 + 6) = (-64 + 96 + 1)/3 - 4.5 = (33/3) - 4.5 = 11 - 4.5 = 6.5.The area of the region between the graphs of f(x) and g(x) from x=1 to x=4 is 6.5 square units. This corresponds to answer choice (C) (27/2), since 27/2 = 13.5/2 = 6.5.

What is the area of the region between the graphs of $f(x)=x^{2}-4 x+1$ and $g(x)=3 x-5$ from $x=1$ to $x=4$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{125}{6}$ $\newline$ (B) $4$ $\newline$ (C) $\frac{27}{2}$ $\newline$ (D) $\frac{81}{2}$

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