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) (27)/(2) (C) 4 (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)  (27)/(2) (C) 4 (D)  (81)/(2)

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Define Area Integral: To find the area between two curves, we need to integrate the difference of the functions over the given interval. The area A is given by the integral from x=1 to x=4 of (g(x) - f(x)) dx.Find Difference of Functions: First, we need to find the expression for (g(x) - f(x)). We have g(x) = 3x - 5 and f(x) = x^2 - 4x + 1. So, (g(x) - f(x)) = (3x - 5) - (x^2 - 4x + 1).Simplify Expression: Simplify the expression for (g(x) - f(x)). This gives us (3x - 5) - (x^2 - 4x + 1) = -x^2 + 7x - 6.Set Up Integral: Now we can set up the integral to find the area A: A = ∫ from x=1 to x=4 of (-x^2 + 7x - 6) dx.Calculate Antiderivative: Calculate the integral of -x^2 + 7x - 6 with respect to x from 1 to 4. The antiderivative of -x^2 is -x^3/3, the antiderivative of 7x is 7x^2/2, and the antiderivative of -6 is -6x. So the integral becomes: A = [(-x^3/3) + (7x^2/2) - 6x] from x=1 to x=4.Evaluate Limits: Evaluate the antiderivative at the upper limit of integration (x=4) and then at the lower limit of integration (x=1), and subtract the latter from the former. This gives us A = [(-4^3/3) + (7*4^2/2) - 6*4] - [(-1^3/3) + (7*1^2/2) - 6*1].Perform Calculations: Perform the calculations for each term. For x=4, we have (-64/3) + (7*16/2) - 24 = (-64/3) + (56) - 24. For x=1, we have (-1/3) + (7/2) - 6 = (-1/3) + (3.5) - 6.Simplify Results: Simplify the calculations. For x=4, we have (-64/3) + 56 - 24 = (-64/3) + (168/3) - (72/3) = (168 - 64 - 72)/3 = 32/3. For x=1, we have (-1/3) + (3.5) - 6 = (-1/3) + (10.5/3) - (18/3) = (10.5 - 1 - 18)/3 = -8.5/3.Subtract Values: Subtract the value at x=1 from the value at x=4 to find the area A. This gives us A = (32/3) - (-8.5/3) = (32 + 8.5)/3 = 40.5/3.Convert to Improper Fraction: Convert the mixed number to an improper fraction to get the final answer. The area A = 40.5/3 = (121.5)/3 = 40.5/3 = 13.5.

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) $\frac{27}{2}$ $\newline$ (C) $4$ $\newline$ (D) $\frac{81}{2}$

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